Parse all strings of specific length?
I've exported my email archive of 10 years which is very large.
I want to parse all the text for any string that is 64 characters long in search of a bitcoin private key.
How can I parse strings of a certain length in characters?
text-processing files wildcards pattern-matching
|
show 1 more comment
I've exported my email archive of 10 years which is very large.
I want to parse all the text for any string that is 64 characters long in search of a bitcoin private key.
How can I parse strings of a certain length in characters?
text-processing files wildcards pattern-matching
What format are the emails in? Plain text? Maildir?
– Sparhawk
Jan 6 at 0:13
@Sparhawk I'm still downloading the files, I'm hoping they are intxt
or something I cancat
and parse with a pipe.
– Philip Kirkbride
Jan 6 at 0:14
@Sparhawk file type is mbox, hoping to convert it intotxt
for easy parsing
– Philip Kirkbride
Jan 6 at 0:34
1
What do you mean by "parse" here? Do you just want to find all strings of exactly 64 characters and then parse them? And how are strings defined? Can we assume you mean things that are delineated with whitespace? So anything with a space, a tab, a newline etc on either side of it? And what operating system are you using? Is it Linux? Can we assume you have access to GNU tools?
– terdon♦
Jan 6 at 0:43
1
The entire email file is a "string of certain length"; IMHO, the string you're looking for consists of certain characters and is delimited in some way. What can you say about the string besides it being 64 of some character?
– Jeff Schaller
Jan 6 at 3:31
|
show 1 more comment
I've exported my email archive of 10 years which is very large.
I want to parse all the text for any string that is 64 characters long in search of a bitcoin private key.
How can I parse strings of a certain length in characters?
text-processing files wildcards pattern-matching
I've exported my email archive of 10 years which is very large.
I want to parse all the text for any string that is 64 characters long in search of a bitcoin private key.
How can I parse strings of a certain length in characters?
text-processing files wildcards pattern-matching
text-processing files wildcards pattern-matching
edited Jan 6 at 1:09
terdon♦
129k31253428
129k31253428
asked Jan 6 at 0:06
Philip KirkbridePhilip Kirkbride
2,4212984
2,4212984
What format are the emails in? Plain text? Maildir?
– Sparhawk
Jan 6 at 0:13
@Sparhawk I'm still downloading the files, I'm hoping they are intxt
or something I cancat
and parse with a pipe.
– Philip Kirkbride
Jan 6 at 0:14
@Sparhawk file type is mbox, hoping to convert it intotxt
for easy parsing
– Philip Kirkbride
Jan 6 at 0:34
1
What do you mean by "parse" here? Do you just want to find all strings of exactly 64 characters and then parse them? And how are strings defined? Can we assume you mean things that are delineated with whitespace? So anything with a space, a tab, a newline etc on either side of it? And what operating system are you using? Is it Linux? Can we assume you have access to GNU tools?
– terdon♦
Jan 6 at 0:43
1
The entire email file is a "string of certain length"; IMHO, the string you're looking for consists of certain characters and is delimited in some way. What can you say about the string besides it being 64 of some character?
– Jeff Schaller
Jan 6 at 3:31
|
show 1 more comment
What format are the emails in? Plain text? Maildir?
– Sparhawk
Jan 6 at 0:13
@Sparhawk I'm still downloading the files, I'm hoping they are intxt
or something I cancat
and parse with a pipe.
– Philip Kirkbride
Jan 6 at 0:14
@Sparhawk file type is mbox, hoping to convert it intotxt
for easy parsing
– Philip Kirkbride
Jan 6 at 0:34
1
What do you mean by "parse" here? Do you just want to find all strings of exactly 64 characters and then parse them? And how are strings defined? Can we assume you mean things that are delineated with whitespace? So anything with a space, a tab, a newline etc on either side of it? And what operating system are you using? Is it Linux? Can we assume you have access to GNU tools?
– terdon♦
Jan 6 at 0:43
1
The entire email file is a "string of certain length"; IMHO, the string you're looking for consists of certain characters and is delimited in some way. What can you say about the string besides it being 64 of some character?
– Jeff Schaller
Jan 6 at 3:31
What format are the emails in? Plain text? Maildir?
– Sparhawk
Jan 6 at 0:13
What format are the emails in? Plain text? Maildir?
– Sparhawk
Jan 6 at 0:13
@Sparhawk I'm still downloading the files, I'm hoping they are in
txt
or something I can cat
and parse with a pipe.– Philip Kirkbride
Jan 6 at 0:14
@Sparhawk I'm still downloading the files, I'm hoping they are in
txt
or something I can cat
and parse with a pipe.– Philip Kirkbride
Jan 6 at 0:14
@Sparhawk file type is mbox, hoping to convert it into
txt
for easy parsing– Philip Kirkbride
Jan 6 at 0:34
@Sparhawk file type is mbox, hoping to convert it into
txt
for easy parsing– Philip Kirkbride
Jan 6 at 0:34
1
1
What do you mean by "parse" here? Do you just want to find all strings of exactly 64 characters and then parse them? And how are strings defined? Can we assume you mean things that are delineated with whitespace? So anything with a space, a tab, a newline etc on either side of it? And what operating system are you using? Is it Linux? Can we assume you have access to GNU tools?
– terdon♦
Jan 6 at 0:43
What do you mean by "parse" here? Do you just want to find all strings of exactly 64 characters and then parse them? And how are strings defined? Can we assume you mean things that are delineated with whitespace? So anything with a space, a tab, a newline etc on either side of it? And what operating system are you using? Is it Linux? Can we assume you have access to GNU tools?
– terdon♦
Jan 6 at 0:43
1
1
The entire email file is a "string of certain length"; IMHO, the string you're looking for consists of certain characters and is delimited in some way. What can you say about the string besides it being 64 of some character?
– Jeff Schaller
Jan 6 at 3:31
The entire email file is a "string of certain length"; IMHO, the string you're looking for consists of certain characters and is delimited in some way. What can you say about the string besides it being 64 of some character?
– Jeff Schaller
Jan 6 at 3:31
|
show 1 more comment
4 Answers
4
active
oldest
votes
If you have GNU grep
(default on Linux), you can do:
grep -Po '(^|s)S{64}(s|$)' file
The -P
enables Perl Compatible Regular Expressions, which give us b
(word-boundaries) S
(non-whitespace) and {N}
(find exactly N characters), and the -o
means "print only the matching part of the line. Then, we look for stretches of non-whitespace that are exactly 64 characters long that are either at the beginning of the line (^
) or after whitespace ('s
) and which end either at the end of the line ($
) or with another whitespace character.
Note that the result will include any whitespace characters at the beginning and end of the string, so if you want to parse this further, you might want to use this instead:
grep -Po '(^|s)KS{64}(?=s|$)'
That will look for a whitespace character or the beginning of the string (s|^)
, then discard it K
and then look for 64 non-whitespace characters followed by (the (?=foo)
is called a "lookahead" and will not be included in the match) either a whitespace character, or the end of the line.
pcre also gives us negative lookahead and lookbehind assertions:grep -Po '(?<!S)S{64}(?!S)'
is enough to find runs of 64 non-spaces; but please read my answer for why that's probably not what's intended.
– pizdelect
Jan 6 at 1:23
@pizdelect yes, I know negative lookbehinds, but I find theK
approach much more readable and elegant.
– terdon♦
Jan 6 at 3:09
elegance is not my strong point, but your 1st example will also include the delimiting spaces in the output, and the 2nd example will not match words which start at the beginning of the line.
– pizdelect
Jan 6 at 3:31
@pizdelect well yes, that's why I included the second example with the lookarounds. But you're right about the second, that's a typo (see the description which I ncludes the start), thanks for pointing it out. Fixed now.
– terdon♦
Jan 6 at 3:37
add a comment |
If you mean to search for a 256-bit number in hexadecimal form (64 chars from the range 0-9
and A-F
-- one of the formats in which a bitcoin private key could appear), this should do:
egrep -aro '<[A-F0-9]{64}>' files and dirs ...
Add the -i
option or also include the a-f
range if some of the keys are in lowercase.
For the general problem of finding runs of characters from the same class having a specified length, you would better use pcre regexps, which could be used with GNU grep with the -P
option. For instance, to find runs of uppercase letters from any charset, of min length of 2 and max length of 4, and which are delimited by chars which are not uppercase letters:
echo ÁRVÍZtűrő tükörFÚRÓgép |
LC_CTYPE=en_US.UTF-8 grep -Po '(?<!p{Lu})p{Lu}{2,4}(?!p{Lu})'
FÚRÓ
Replace p{Lu}
with p{Ll}
for lowercase letters, S
for non-spaces, etc. See here and here for the full list.
(?<!...)
and (?!...)
are negative lookbehind and lookahead zero-width assertions; e.g. (?<!<)w(?!>)
will match a "word" character when not bracketed by <
and >
. The <
zero-width assertion from vi
could be implemented by (?<!w)(?=w)
.
add a comment |
If you want to find all words of length 64 from /path/to/file
, you can use
tr -c '[[:alnum:]]' 'n' < /path/to/file | grep '^.{64}$'
This replaces all non-alphanumeric characters by newlines, so each word is on its own line. Then it filters this result to include only the words of length 64.
What about dot (.
), comma (,
), colon (:
), semicolon (;
) and many other usual punctuation characters, shoudln't those also be converted to a newline ?
– Isaac
Jan 6 at 1:52
@Isaac why? Why are you assuming they can't appear inside the target string?
– terdon♦
Jan 6 at 3:08
1
@terdon Because I assume that the target string is a bitcoin private key which is usually a 64 hex character string. Hmmm, ooops, sorry, the OP stated that: I am not assuming then.
– Isaac
Jan 6 at 5:05
1
@Isaac You're not wrong
– Fox
2 days ago
add a comment |
It seems that grep is the correct tool to "search" for an string. What is left to do is to define such string with a regex. The first issue is to define the limits of a word. It is not as simple as "an space", as a book, a lamp
use ,
as word delimiter, in the same concept, many other characters, or even the start or end of a line could act as word delimiter. There are some word delimiters in GNU grep:
<
word start.
>
word end.
b
word boundary.
All of them assume that a word is a sequence of [a-zA-Z0-9_]
characters. If that is ok for you, this regex could work:
grep -o '<.{64}>' file
If you could use extended regex, the could be reduced:
grep -oE '<.{64}>' file
That selects from a "word start" (<
), 64 ({64}
) characters (.
), to a "word end" (>
) and prints only the matching (-o
) parts.
However, the dot (.
) will match any character, that may be too much.
If you want to be more strict on the selection (hex digits), use:
grep -oE '<[0-9a-fA-F]{64}>' file
Which will allow hex digits in lowercase or uppercase. But if you really want to be strict, as some non-ASCII characters might be included, use:
LC_ALL=C grep -oE '<[0-9a-fA-F]{64}>' file
Some implementations of grep (as grep -P) do not have a "start of word" or "end of word" (as <
and >
) but have "word boundary" (as b
):
grep -oP 'b[0-9a-fA-F]{64}b' file
There are some languages that accept the POSIX word boundaries [[:<:]]
and [[:>:]]
, but not perl, and only from PCRE 8.34.
And there are a lot more flavors of "word boundaries".
add a comment |
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4 Answers
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4 Answers
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votes
If you have GNU grep
(default on Linux), you can do:
grep -Po '(^|s)S{64}(s|$)' file
The -P
enables Perl Compatible Regular Expressions, which give us b
(word-boundaries) S
(non-whitespace) and {N}
(find exactly N characters), and the -o
means "print only the matching part of the line. Then, we look for stretches of non-whitespace that are exactly 64 characters long that are either at the beginning of the line (^
) or after whitespace ('s
) and which end either at the end of the line ($
) or with another whitespace character.
Note that the result will include any whitespace characters at the beginning and end of the string, so if you want to parse this further, you might want to use this instead:
grep -Po '(^|s)KS{64}(?=s|$)'
That will look for a whitespace character or the beginning of the string (s|^)
, then discard it K
and then look for 64 non-whitespace characters followed by (the (?=foo)
is called a "lookahead" and will not be included in the match) either a whitespace character, or the end of the line.
pcre also gives us negative lookahead and lookbehind assertions:grep -Po '(?<!S)S{64}(?!S)'
is enough to find runs of 64 non-spaces; but please read my answer for why that's probably not what's intended.
– pizdelect
Jan 6 at 1:23
@pizdelect yes, I know negative lookbehinds, but I find theK
approach much more readable and elegant.
– terdon♦
Jan 6 at 3:09
elegance is not my strong point, but your 1st example will also include the delimiting spaces in the output, and the 2nd example will not match words which start at the beginning of the line.
– pizdelect
Jan 6 at 3:31
@pizdelect well yes, that's why I included the second example with the lookarounds. But you're right about the second, that's a typo (see the description which I ncludes the start), thanks for pointing it out. Fixed now.
– terdon♦
Jan 6 at 3:37
add a comment |
If you have GNU grep
(default on Linux), you can do:
grep -Po '(^|s)S{64}(s|$)' file
The -P
enables Perl Compatible Regular Expressions, which give us b
(word-boundaries) S
(non-whitespace) and {N}
(find exactly N characters), and the -o
means "print only the matching part of the line. Then, we look for stretches of non-whitespace that are exactly 64 characters long that are either at the beginning of the line (^
) or after whitespace ('s
) and which end either at the end of the line ($
) or with another whitespace character.
Note that the result will include any whitespace characters at the beginning and end of the string, so if you want to parse this further, you might want to use this instead:
grep -Po '(^|s)KS{64}(?=s|$)'
That will look for a whitespace character or the beginning of the string (s|^)
, then discard it K
and then look for 64 non-whitespace characters followed by (the (?=foo)
is called a "lookahead" and will not be included in the match) either a whitespace character, or the end of the line.
pcre also gives us negative lookahead and lookbehind assertions:grep -Po '(?<!S)S{64}(?!S)'
is enough to find runs of 64 non-spaces; but please read my answer for why that's probably not what's intended.
– pizdelect
Jan 6 at 1:23
@pizdelect yes, I know negative lookbehinds, but I find theK
approach much more readable and elegant.
– terdon♦
Jan 6 at 3:09
elegance is not my strong point, but your 1st example will also include the delimiting spaces in the output, and the 2nd example will not match words which start at the beginning of the line.
– pizdelect
Jan 6 at 3:31
@pizdelect well yes, that's why I included the second example with the lookarounds. But you're right about the second, that's a typo (see the description which I ncludes the start), thanks for pointing it out. Fixed now.
– terdon♦
Jan 6 at 3:37
add a comment |
If you have GNU grep
(default on Linux), you can do:
grep -Po '(^|s)S{64}(s|$)' file
The -P
enables Perl Compatible Regular Expressions, which give us b
(word-boundaries) S
(non-whitespace) and {N}
(find exactly N characters), and the -o
means "print only the matching part of the line. Then, we look for stretches of non-whitespace that are exactly 64 characters long that are either at the beginning of the line (^
) or after whitespace ('s
) and which end either at the end of the line ($
) or with another whitespace character.
Note that the result will include any whitespace characters at the beginning and end of the string, so if you want to parse this further, you might want to use this instead:
grep -Po '(^|s)KS{64}(?=s|$)'
That will look for a whitespace character or the beginning of the string (s|^)
, then discard it K
and then look for 64 non-whitespace characters followed by (the (?=foo)
is called a "lookahead" and will not be included in the match) either a whitespace character, or the end of the line.
If you have GNU grep
(default on Linux), you can do:
grep -Po '(^|s)S{64}(s|$)' file
The -P
enables Perl Compatible Regular Expressions, which give us b
(word-boundaries) S
(non-whitespace) and {N}
(find exactly N characters), and the -o
means "print only the matching part of the line. Then, we look for stretches of non-whitespace that are exactly 64 characters long that are either at the beginning of the line (^
) or after whitespace ('s
) and which end either at the end of the line ($
) or with another whitespace character.
Note that the result will include any whitespace characters at the beginning and end of the string, so if you want to parse this further, you might want to use this instead:
grep -Po '(^|s)KS{64}(?=s|$)'
That will look for a whitespace character or the beginning of the string (s|^)
, then discard it K
and then look for 64 non-whitespace characters followed by (the (?=foo)
is called a "lookahead" and will not be included in the match) either a whitespace character, or the end of the line.
edited Jan 6 at 3:37
answered Jan 6 at 0:47
terdon♦terdon
129k31253428
129k31253428
pcre also gives us negative lookahead and lookbehind assertions:grep -Po '(?<!S)S{64}(?!S)'
is enough to find runs of 64 non-spaces; but please read my answer for why that's probably not what's intended.
– pizdelect
Jan 6 at 1:23
@pizdelect yes, I know negative lookbehinds, but I find theK
approach much more readable and elegant.
– terdon♦
Jan 6 at 3:09
elegance is not my strong point, but your 1st example will also include the delimiting spaces in the output, and the 2nd example will not match words which start at the beginning of the line.
– pizdelect
Jan 6 at 3:31
@pizdelect well yes, that's why I included the second example with the lookarounds. But you're right about the second, that's a typo (see the description which I ncludes the start), thanks for pointing it out. Fixed now.
– terdon♦
Jan 6 at 3:37
add a comment |
pcre also gives us negative lookahead and lookbehind assertions:grep -Po '(?<!S)S{64}(?!S)'
is enough to find runs of 64 non-spaces; but please read my answer for why that's probably not what's intended.
– pizdelect
Jan 6 at 1:23
@pizdelect yes, I know negative lookbehinds, but I find theK
approach much more readable and elegant.
– terdon♦
Jan 6 at 3:09
elegance is not my strong point, but your 1st example will also include the delimiting spaces in the output, and the 2nd example will not match words which start at the beginning of the line.
– pizdelect
Jan 6 at 3:31
@pizdelect well yes, that's why I included the second example with the lookarounds. But you're right about the second, that's a typo (see the description which I ncludes the start), thanks for pointing it out. Fixed now.
– terdon♦
Jan 6 at 3:37
pcre also gives us negative lookahead and lookbehind assertions:
grep -Po '(?<!S)S{64}(?!S)'
is enough to find runs of 64 non-spaces; but please read my answer for why that's probably not what's intended.– pizdelect
Jan 6 at 1:23
pcre also gives us negative lookahead and lookbehind assertions:
grep -Po '(?<!S)S{64}(?!S)'
is enough to find runs of 64 non-spaces; but please read my answer for why that's probably not what's intended.– pizdelect
Jan 6 at 1:23
@pizdelect yes, I know negative lookbehinds, but I find the
K
approach much more readable and elegant.– terdon♦
Jan 6 at 3:09
@pizdelect yes, I know negative lookbehinds, but I find the
K
approach much more readable and elegant.– terdon♦
Jan 6 at 3:09
elegance is not my strong point, but your 1st example will also include the delimiting spaces in the output, and the 2nd example will not match words which start at the beginning of the line.
– pizdelect
Jan 6 at 3:31
elegance is not my strong point, but your 1st example will also include the delimiting spaces in the output, and the 2nd example will not match words which start at the beginning of the line.
– pizdelect
Jan 6 at 3:31
@pizdelect well yes, that's why I included the second example with the lookarounds. But you're right about the second, that's a typo (see the description which I ncludes the start), thanks for pointing it out. Fixed now.
– terdon♦
Jan 6 at 3:37
@pizdelect well yes, that's why I included the second example with the lookarounds. But you're right about the second, that's a typo (see the description which I ncludes the start), thanks for pointing it out. Fixed now.
– terdon♦
Jan 6 at 3:37
add a comment |
If you mean to search for a 256-bit number in hexadecimal form (64 chars from the range 0-9
and A-F
-- one of the formats in which a bitcoin private key could appear), this should do:
egrep -aro '<[A-F0-9]{64}>' files and dirs ...
Add the -i
option or also include the a-f
range if some of the keys are in lowercase.
For the general problem of finding runs of characters from the same class having a specified length, you would better use pcre regexps, which could be used with GNU grep with the -P
option. For instance, to find runs of uppercase letters from any charset, of min length of 2 and max length of 4, and which are delimited by chars which are not uppercase letters:
echo ÁRVÍZtűrő tükörFÚRÓgép |
LC_CTYPE=en_US.UTF-8 grep -Po '(?<!p{Lu})p{Lu}{2,4}(?!p{Lu})'
FÚRÓ
Replace p{Lu}
with p{Ll}
for lowercase letters, S
for non-spaces, etc. See here and here for the full list.
(?<!...)
and (?!...)
are negative lookbehind and lookahead zero-width assertions; e.g. (?<!<)w(?!>)
will match a "word" character when not bracketed by <
and >
. The <
zero-width assertion from vi
could be implemented by (?<!w)(?=w)
.
add a comment |
If you mean to search for a 256-bit number in hexadecimal form (64 chars from the range 0-9
and A-F
-- one of the formats in which a bitcoin private key could appear), this should do:
egrep -aro '<[A-F0-9]{64}>' files and dirs ...
Add the -i
option or also include the a-f
range if some of the keys are in lowercase.
For the general problem of finding runs of characters from the same class having a specified length, you would better use pcre regexps, which could be used with GNU grep with the -P
option. For instance, to find runs of uppercase letters from any charset, of min length of 2 and max length of 4, and which are delimited by chars which are not uppercase letters:
echo ÁRVÍZtűrő tükörFÚRÓgép |
LC_CTYPE=en_US.UTF-8 grep -Po '(?<!p{Lu})p{Lu}{2,4}(?!p{Lu})'
FÚRÓ
Replace p{Lu}
with p{Ll}
for lowercase letters, S
for non-spaces, etc. See here and here for the full list.
(?<!...)
and (?!...)
are negative lookbehind and lookahead zero-width assertions; e.g. (?<!<)w(?!>)
will match a "word" character when not bracketed by <
and >
. The <
zero-width assertion from vi
could be implemented by (?<!w)(?=w)
.
add a comment |
If you mean to search for a 256-bit number in hexadecimal form (64 chars from the range 0-9
and A-F
-- one of the formats in which a bitcoin private key could appear), this should do:
egrep -aro '<[A-F0-9]{64}>' files and dirs ...
Add the -i
option or also include the a-f
range if some of the keys are in lowercase.
For the general problem of finding runs of characters from the same class having a specified length, you would better use pcre regexps, which could be used with GNU grep with the -P
option. For instance, to find runs of uppercase letters from any charset, of min length of 2 and max length of 4, and which are delimited by chars which are not uppercase letters:
echo ÁRVÍZtűrő tükörFÚRÓgép |
LC_CTYPE=en_US.UTF-8 grep -Po '(?<!p{Lu})p{Lu}{2,4}(?!p{Lu})'
FÚRÓ
Replace p{Lu}
with p{Ll}
for lowercase letters, S
for non-spaces, etc. See here and here for the full list.
(?<!...)
and (?!...)
are negative lookbehind and lookahead zero-width assertions; e.g. (?<!<)w(?!>)
will match a "word" character when not bracketed by <
and >
. The <
zero-width assertion from vi
could be implemented by (?<!w)(?=w)
.
If you mean to search for a 256-bit number in hexadecimal form (64 chars from the range 0-9
and A-F
-- one of the formats in which a bitcoin private key could appear), this should do:
egrep -aro '<[A-F0-9]{64}>' files and dirs ...
Add the -i
option or also include the a-f
range if some of the keys are in lowercase.
For the general problem of finding runs of characters from the same class having a specified length, you would better use pcre regexps, which could be used with GNU grep with the -P
option. For instance, to find runs of uppercase letters from any charset, of min length of 2 and max length of 4, and which are delimited by chars which are not uppercase letters:
echo ÁRVÍZtűrő tükörFÚRÓgép |
LC_CTYPE=en_US.UTF-8 grep -Po '(?<!p{Lu})p{Lu}{2,4}(?!p{Lu})'
FÚRÓ
Replace p{Lu}
with p{Ll}
for lowercase letters, S
for non-spaces, etc. See here and here for the full list.
(?<!...)
and (?!...)
are negative lookbehind and lookahead zero-width assertions; e.g. (?<!<)w(?!>)
will match a "word" character when not bracketed by <
and >
. The <
zero-width assertion from vi
could be implemented by (?<!w)(?=w)
.
edited Jan 6 at 5:36
answered Jan 6 at 1:10
pizdelectpizdelect
44016
44016
add a comment |
add a comment |
If you want to find all words of length 64 from /path/to/file
, you can use
tr -c '[[:alnum:]]' 'n' < /path/to/file | grep '^.{64}$'
This replaces all non-alphanumeric characters by newlines, so each word is on its own line. Then it filters this result to include only the words of length 64.
What about dot (.
), comma (,
), colon (:
), semicolon (;
) and many other usual punctuation characters, shoudln't those also be converted to a newline ?
– Isaac
Jan 6 at 1:52
@Isaac why? Why are you assuming they can't appear inside the target string?
– terdon♦
Jan 6 at 3:08
1
@terdon Because I assume that the target string is a bitcoin private key which is usually a 64 hex character string. Hmmm, ooops, sorry, the OP stated that: I am not assuming then.
– Isaac
Jan 6 at 5:05
1
@Isaac You're not wrong
– Fox
2 days ago
add a comment |
If you want to find all words of length 64 from /path/to/file
, you can use
tr -c '[[:alnum:]]' 'n' < /path/to/file | grep '^.{64}$'
This replaces all non-alphanumeric characters by newlines, so each word is on its own line. Then it filters this result to include only the words of length 64.
What about dot (.
), comma (,
), colon (:
), semicolon (;
) and many other usual punctuation characters, shoudln't those also be converted to a newline ?
– Isaac
Jan 6 at 1:52
@Isaac why? Why are you assuming they can't appear inside the target string?
– terdon♦
Jan 6 at 3:08
1
@terdon Because I assume that the target string is a bitcoin private key which is usually a 64 hex character string. Hmmm, ooops, sorry, the OP stated that: I am not assuming then.
– Isaac
Jan 6 at 5:05
1
@Isaac You're not wrong
– Fox
2 days ago
add a comment |
If you want to find all words of length 64 from /path/to/file
, you can use
tr -c '[[:alnum:]]' 'n' < /path/to/file | grep '^.{64}$'
This replaces all non-alphanumeric characters by newlines, so each word is on its own line. Then it filters this result to include only the words of length 64.
If you want to find all words of length 64 from /path/to/file
, you can use
tr -c '[[:alnum:]]' 'n' < /path/to/file | grep '^.{64}$'
This replaces all non-alphanumeric characters by newlines, so each word is on its own line. Then it filters this result to include only the words of length 64.
edited 2 days ago
answered Jan 6 at 0:42
FoxFox
5,24411232
5,24411232
What about dot (.
), comma (,
), colon (:
), semicolon (;
) and many other usual punctuation characters, shoudln't those also be converted to a newline ?
– Isaac
Jan 6 at 1:52
@Isaac why? Why are you assuming they can't appear inside the target string?
– terdon♦
Jan 6 at 3:08
1
@terdon Because I assume that the target string is a bitcoin private key which is usually a 64 hex character string. Hmmm, ooops, sorry, the OP stated that: I am not assuming then.
– Isaac
Jan 6 at 5:05
1
@Isaac You're not wrong
– Fox
2 days ago
add a comment |
What about dot (.
), comma (,
), colon (:
), semicolon (;
) and many other usual punctuation characters, shoudln't those also be converted to a newline ?
– Isaac
Jan 6 at 1:52
@Isaac why? Why are you assuming they can't appear inside the target string?
– terdon♦
Jan 6 at 3:08
1
@terdon Because I assume that the target string is a bitcoin private key which is usually a 64 hex character string. Hmmm, ooops, sorry, the OP stated that: I am not assuming then.
– Isaac
Jan 6 at 5:05
1
@Isaac You're not wrong
– Fox
2 days ago
What about dot (
.
), comma (,
), colon (:
), semicolon (;
) and many other usual punctuation characters, shoudln't those also be converted to a newline ?– Isaac
Jan 6 at 1:52
What about dot (
.
), comma (,
), colon (:
), semicolon (;
) and many other usual punctuation characters, shoudln't those also be converted to a newline ?– Isaac
Jan 6 at 1:52
@Isaac why? Why are you assuming they can't appear inside the target string?
– terdon♦
Jan 6 at 3:08
@Isaac why? Why are you assuming they can't appear inside the target string?
– terdon♦
Jan 6 at 3:08
1
1
@terdon Because I assume that the target string is a bitcoin private key which is usually a 64 hex character string. Hmmm, ooops, sorry, the OP stated that: I am not assuming then.
– Isaac
Jan 6 at 5:05
@terdon Because I assume that the target string is a bitcoin private key which is usually a 64 hex character string. Hmmm, ooops, sorry, the OP stated that: I am not assuming then.
– Isaac
Jan 6 at 5:05
1
1
@Isaac You're not wrong
– Fox
2 days ago
@Isaac You're not wrong
– Fox
2 days ago
add a comment |
It seems that grep is the correct tool to "search" for an string. What is left to do is to define such string with a regex. The first issue is to define the limits of a word. It is not as simple as "an space", as a book, a lamp
use ,
as word delimiter, in the same concept, many other characters, or even the start or end of a line could act as word delimiter. There are some word delimiters in GNU grep:
<
word start.
>
word end.
b
word boundary.
All of them assume that a word is a sequence of [a-zA-Z0-9_]
characters. If that is ok for you, this regex could work:
grep -o '<.{64}>' file
If you could use extended regex, the could be reduced:
grep -oE '<.{64}>' file
That selects from a "word start" (<
), 64 ({64}
) characters (.
), to a "word end" (>
) and prints only the matching (-o
) parts.
However, the dot (.
) will match any character, that may be too much.
If you want to be more strict on the selection (hex digits), use:
grep -oE '<[0-9a-fA-F]{64}>' file
Which will allow hex digits in lowercase or uppercase. But if you really want to be strict, as some non-ASCII characters might be included, use:
LC_ALL=C grep -oE '<[0-9a-fA-F]{64}>' file
Some implementations of grep (as grep -P) do not have a "start of word" or "end of word" (as <
and >
) but have "word boundary" (as b
):
grep -oP 'b[0-9a-fA-F]{64}b' file
There are some languages that accept the POSIX word boundaries [[:<:]]
and [[:>:]]
, but not perl, and only from PCRE 8.34.
And there are a lot more flavors of "word boundaries".
add a comment |
It seems that grep is the correct tool to "search" for an string. What is left to do is to define such string with a regex. The first issue is to define the limits of a word. It is not as simple as "an space", as a book, a lamp
use ,
as word delimiter, in the same concept, many other characters, or even the start or end of a line could act as word delimiter. There are some word delimiters in GNU grep:
<
word start.
>
word end.
b
word boundary.
All of them assume that a word is a sequence of [a-zA-Z0-9_]
characters. If that is ok for you, this regex could work:
grep -o '<.{64}>' file
If you could use extended regex, the could be reduced:
grep -oE '<.{64}>' file
That selects from a "word start" (<
), 64 ({64}
) characters (.
), to a "word end" (>
) and prints only the matching (-o
) parts.
However, the dot (.
) will match any character, that may be too much.
If you want to be more strict on the selection (hex digits), use:
grep -oE '<[0-9a-fA-F]{64}>' file
Which will allow hex digits in lowercase or uppercase. But if you really want to be strict, as some non-ASCII characters might be included, use:
LC_ALL=C grep -oE '<[0-9a-fA-F]{64}>' file
Some implementations of grep (as grep -P) do not have a "start of word" or "end of word" (as <
and >
) but have "word boundary" (as b
):
grep -oP 'b[0-9a-fA-F]{64}b' file
There are some languages that accept the POSIX word boundaries [[:<:]]
and [[:>:]]
, but not perl, and only from PCRE 8.34.
And there are a lot more flavors of "word boundaries".
add a comment |
It seems that grep is the correct tool to "search" for an string. What is left to do is to define such string with a regex. The first issue is to define the limits of a word. It is not as simple as "an space", as a book, a lamp
use ,
as word delimiter, in the same concept, many other characters, or even the start or end of a line could act as word delimiter. There are some word delimiters in GNU grep:
<
word start.
>
word end.
b
word boundary.
All of them assume that a word is a sequence of [a-zA-Z0-9_]
characters. If that is ok for you, this regex could work:
grep -o '<.{64}>' file
If you could use extended regex, the could be reduced:
grep -oE '<.{64}>' file
That selects from a "word start" (<
), 64 ({64}
) characters (.
), to a "word end" (>
) and prints only the matching (-o
) parts.
However, the dot (.
) will match any character, that may be too much.
If you want to be more strict on the selection (hex digits), use:
grep -oE '<[0-9a-fA-F]{64}>' file
Which will allow hex digits in lowercase or uppercase. But if you really want to be strict, as some non-ASCII characters might be included, use:
LC_ALL=C grep -oE '<[0-9a-fA-F]{64}>' file
Some implementations of grep (as grep -P) do not have a "start of word" or "end of word" (as <
and >
) but have "word boundary" (as b
):
grep -oP 'b[0-9a-fA-F]{64}b' file
There are some languages that accept the POSIX word boundaries [[:<:]]
and [[:>:]]
, but not perl, and only from PCRE 8.34.
And there are a lot more flavors of "word boundaries".
It seems that grep is the correct tool to "search" for an string. What is left to do is to define such string with a regex. The first issue is to define the limits of a word. It is not as simple as "an space", as a book, a lamp
use ,
as word delimiter, in the same concept, many other characters, or even the start or end of a line could act as word delimiter. There are some word delimiters in GNU grep:
<
word start.
>
word end.
b
word boundary.
All of them assume that a word is a sequence of [a-zA-Z0-9_]
characters. If that is ok for you, this regex could work:
grep -o '<.{64}>' file
If you could use extended regex, the could be reduced:
grep -oE '<.{64}>' file
That selects from a "word start" (<
), 64 ({64}
) characters (.
), to a "word end" (>
) and prints only the matching (-o
) parts.
However, the dot (.
) will match any character, that may be too much.
If you want to be more strict on the selection (hex digits), use:
grep -oE '<[0-9a-fA-F]{64}>' file
Which will allow hex digits in lowercase or uppercase. But if you really want to be strict, as some non-ASCII characters might be included, use:
LC_ALL=C grep -oE '<[0-9a-fA-F]{64}>' file
Some implementations of grep (as grep -P) do not have a "start of word" or "end of word" (as <
and >
) but have "word boundary" (as b
):
grep -oP 'b[0-9a-fA-F]{64}b' file
There are some languages that accept the POSIX word boundaries [[:<:]]
and [[:>:]]
, but not perl, and only from PCRE 8.34.
And there are a lot more flavors of "word boundaries".
edited 2 days ago
answered Jan 6 at 1:50
IsaacIsaac
11.3k11651
11.3k11651
add a comment |
add a comment |
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What format are the emails in? Plain text? Maildir?
– Sparhawk
Jan 6 at 0:13
@Sparhawk I'm still downloading the files, I'm hoping they are in
txt
or something I cancat
and parse with a pipe.– Philip Kirkbride
Jan 6 at 0:14
@Sparhawk file type is mbox, hoping to convert it into
txt
for easy parsing– Philip Kirkbride
Jan 6 at 0:34
1
What do you mean by "parse" here? Do you just want to find all strings of exactly 64 characters and then parse them? And how are strings defined? Can we assume you mean things that are delineated with whitespace? So anything with a space, a tab, a newline etc on either side of it? And what operating system are you using? Is it Linux? Can we assume you have access to GNU tools?
– terdon♦
Jan 6 at 0:43
1
The entire email file is a "string of certain length"; IMHO, the string you're looking for consists of certain characters and is delimited in some way. What can you say about the string besides it being 64 of some character?
– Jeff Schaller
Jan 6 at 3:31