Remove everything between specified characters including multi lines
I have a file with contents like so:
## this must go ##
## also
this
must go
##
hello world
##and this one
too##
I want to remove all between ##, including multi lines, so I am left just with hello world
This removes only part that is on one line:
sed -i.bak 's/##.*##//g' myfile
How to remove multi line stuff too?
P.S Im on MAC
regex perl awk sed
edited Nov 22 '18 at 15:47
RavinderSingh13
28.9k41638
asked Nov 22 '18 at 14:16
bukowskibukowski
91862146
add a comment |
I have a file with contents like so:
## this must go ##
## also
this
must go
##
hello world
##and this one
too##
I want to remove all between ##, including multi lines, so I am left just with hello world
This removes only part that is on one line:
sed -i.bak 's/##.*##//g' myfile
How to remove multi line stuff too?
P.S Im on MAC
regex perl awk sed
edited Nov 22 '18 at 15:47
RavinderSingh13
28.9k41638
asked Nov 22 '18 at 14:16
bukowskibukowski
91862146
Probably perl solution will be simpler:perl -0pe 's/##.*?##R*//gs' file > newfile(demo).
– Wiktor Stribiżew
Nov 22 '18 at 14:26
@WiktorStribiżew hey your solution works the best - post your answer & I'll accept it. Thanks!
– bukowski
Nov 22 '18 at 14:51
Posted with explanation.
– Wiktor Stribiżew
Nov 22 '18 at 14:55
add a comment |
I have a file with contents like so:
## this must go ##
## also
this
must go
##
hello world
##and this one
too##
I want to remove all between ##, including multi lines, so I am left just with hello world
This removes only part that is on one line:
sed -i.bak 's/##.*##//g' myfile
How to remove multi line stuff too?
P.S Im on MAC
regex perl awk sed
edited Nov 22 '18 at 15:47
RavinderSingh13
28.9k41638
asked Nov 22 '18 at 14:16
bukowskibukowski
91862146
I have a file with contents like so:
## this must go ##
## also
this
must go
##
hello world
##and this one
too##
I want to remove all between ##, including multi lines, so I am left just with hello world
This removes only part that is on one line:
sed -i.bak 's/##.*##//g' myfile
How to remove multi line stuff too?
P.S Im on MAC
regex perl awk sed
regex perl awk sed
edited Nov 22 '18 at 15:47
RavinderSingh13
28.9k41638
asked Nov 22 '18 at 14:16
bukowskibukowski
91862146
edited Nov 22 '18 at 15:47
RavinderSingh13
28.9k41638
asked Nov 22 '18 at 14:16
bukowskibukowski
91862146
edited Nov 22 '18 at 15:47
RavinderSingh13
28.9k41638
edited Nov 22 '18 at 15:47
RavinderSingh13
28.9k41638
edited Nov 22 '18 at 15:47
RavinderSingh13
28.9k41638
28.9k41638
asked Nov 22 '18 at 14:16
bukowskibukowski
91862146
asked Nov 22 '18 at 14:16
bukowskibukowski
91862146
asked Nov 22 '18 at 14:16
bukowskibukowski
91862146
91862146
Probably perl solution will be simpler:perl -0pe 's/##.*?##R*//gs' file > newfile(demo).
– Wiktor Stribiżew
Nov 22 '18 at 14:26
@WiktorStribiżew hey your solution works the best - post your answer & I'll accept it. Thanks!
– bukowski
Nov 22 '18 at 14:51
Posted with explanation.
– Wiktor Stribiżew
Nov 22 '18 at 14:55
add a comment |
Probably perl solution will be simpler:perl -0pe 's/##.*?##R*//gs' file > newfile(demo).
– Wiktor Stribiżew
Nov 22 '18 at 14:26
@WiktorStribiżew hey your solution works the best - post your answer & I'll accept it. Thanks!
– bukowski
Nov 22 '18 at 14:51
Posted with explanation.
– Wiktor Stribiżew
Nov 22 '18 at 14:55
Probably perl solution will be simpler:
perl -0pe 's/##.*?##R*//gs' file > newfile (demo).– Wiktor Stribiżew
Nov 22 '18 at 14:26
Probably perl solution will be simpler:
perl -0pe 's/##.*?##R*//gs' file > newfile (demo).– Wiktor Stribiżew
Nov 22 '18 at 14:26
@WiktorStribiżew hey your solution works the best - post your answer & I'll accept it. Thanks!
– bukowski
Nov 22 '18 at 14:51
@WiktorStribiżew hey your solution works the best - post your answer & I'll accept it. Thanks!
– bukowski
Nov 22 '18 at 14:51
Posted with explanation.
– Wiktor Stribiżew
Nov 22 '18 at 14:55
Posted with explanation.
– Wiktor Stribiżew
Nov 22 '18 at 14:55
add a comment |
4 Answers
4
active
oldest
votes
You may use perl to achieve what you want:
perl -0pe 's/##.*?##R*//gs' file > newfile
See the online demo
The 0 argument makes it possible to find matches across lines.
The pattern matches
##- two#symbols
.*?- any 0+ chars (even line break chars due to thesmodifier) as few as possible
##- two#symbols
R*- any 0+ line break sequences.
answered Nov 22 '18 at 14:53
Wiktor StribiżewWiktor Stribiżew
320k16140222
add a comment |
It should be very easy task for awk(in case you are ok with it). Could you please try following, will add explanation shortly too.
awk '/^##.*##$/{next} /^##$/{flag="";next} /^##/ && !/##$/{flag=1} flag{next} 1' Input_file
Adding a non-one liner form of solution too now.
awk '
/^##.*##$/{
next
}
/^##$/{
flag=""
next
}
/^##/ && !/##$/{
flag=1
}
flag{
next
}
1
' Input_file
answered Nov 22 '18 at 14:27
RavinderSingh13RavinderSingh13
28.9k41638
add a comment |
Give a try to this:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
Wise men read this excellent tutorial: Sed - An Introduction and Tutorial by Bruce Barnett
The test:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
hello world
answered Nov 22 '18 at 14:31
Jay jargotJay jargot
1,9491510
add a comment |
This might work for you (GNU sed):
sed -z 's/##[^#]*(#[^#][^#]*)*##n?//g' file
The -z option allows the whole of the file to be slurped into sed's pattern space. The regexp match is in three parts. The first part matches ## followed by zero or more non-#'s. The second part matches a zero or more group of characters consisting of a single # followed by a non-# followed by zero or more non-#'s. The third part matches ## and a possible newline. This regexp removes such matches globally throughout the file.
This can by shortened slightly by using the -r option to sweeten the final offering to:
sed -rz 's/##[^#]*(#[^#]+)*##n?//g' file
If the version of sed does not offer either options, then another solution is:
sed 'H;$!d;x;s/.//;s/##[^#]*(#[^#][^#]*)*##n?//g' file
It should by noted, that in the example above, all ##'s either start or end at the beginning or end of a line and so the solution below might also fit the bill:
sed 's/^##/,/##$/d' file
answered Nov 23 '18 at 8:42
potongpotong
35.9k43061
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
You may use perl to achieve what you want:
perl -0pe 's/##.*?##R*//gs' file > newfile
See the online demo
The 0 argument makes it possible to find matches across lines.
The pattern matches
##- two#symbols
.*?- any 0+ chars (even line break chars due to thesmodifier) as few as possible
##- two#symbols
R*- any 0+ line break sequences.
answered Nov 22 '18 at 14:53
Wiktor StribiżewWiktor Stribiżew
320k16140222
add a comment |
You may use perl to achieve what you want:
perl -0pe 's/##.*?##R*//gs' file > newfile
See the online demo
The 0 argument makes it possible to find matches across lines.
The pattern matches
##- two#symbols
.*?- any 0+ chars (even line break chars due to thesmodifier) as few as possible
##- two#symbols
R*- any 0+ line break sequences.
answered Nov 22 '18 at 14:53
Wiktor StribiżewWiktor Stribiżew
320k16140222
add a comment |
You may use perl to achieve what you want:
perl -0pe 's/##.*?##R*//gs' file > newfile
See the online demo
The 0 argument makes it possible to find matches across lines.
The pattern matches
##- two#symbols
.*?- any 0+ chars (even line break chars due to thesmodifier) as few as possible
##- two#symbols
R*- any 0+ line break sequences.
answered Nov 22 '18 at 14:53
Wiktor StribiżewWiktor Stribiżew
320k16140222
You may use perl to achieve what you want:
perl -0pe 's/##.*?##R*//gs' file > newfile
See the online demo
The 0 argument makes it possible to find matches across lines.
The pattern matches
##- two#symbols
.*?- any 0+ chars (even line break chars due to thesmodifier) as few as possible
##- two#symbols
R*- any 0+ line break sequences.
answered Nov 22 '18 at 14:53
Wiktor StribiżewWiktor Stribiżew
320k16140222
answered Nov 22 '18 at 14:53
Wiktor StribiżewWiktor Stribiżew
320k16140222
answered Nov 22 '18 at 14:53
Wiktor StribiżewWiktor Stribiżew
320k16140222
answered Nov 22 '18 at 14:53
Wiktor StribiżewWiktor Stribiżew
320k16140222
320k16140222
add a comment |
add a comment |
It should be very easy task for awk(in case you are ok with it). Could you please try following, will add explanation shortly too.
awk '/^##.*##$/{next} /^##$/{flag="";next} /^##/ && !/##$/{flag=1} flag{next} 1' Input_file
Adding a non-one liner form of solution too now.
awk '
/^##.*##$/{
next
}
/^##$/{
flag=""
next
}
/^##/ && !/##$/{
flag=1
}
flag{
next
}
1
' Input_file
answered Nov 22 '18 at 14:27
RavinderSingh13RavinderSingh13
28.9k41638
add a comment |
It should be very easy task for awk(in case you are ok with it). Could you please try following, will add explanation shortly too.
awk '/^##.*##$/{next} /^##$/{flag="";next} /^##/ && !/##$/{flag=1} flag{next} 1' Input_file
Adding a non-one liner form of solution too now.
awk '
/^##.*##$/{
next
}
/^##$/{
flag=""
next
}
/^##/ && !/##$/{
flag=1
}
flag{
next
}
1
' Input_file
answered Nov 22 '18 at 14:27
RavinderSingh13RavinderSingh13
28.9k41638
add a comment |
It should be very easy task for awk(in case you are ok with it). Could you please try following, will add explanation shortly too.
awk '/^##.*##$/{next} /^##$/{flag="";next} /^##/ && !/##$/{flag=1} flag{next} 1' Input_file
Adding a non-one liner form of solution too now.
awk '
/^##.*##$/{
next
}
/^##$/{
flag=""
next
}
/^##/ && !/##$/{
flag=1
}
flag{
next
}
1
' Input_file
answered Nov 22 '18 at 14:27
RavinderSingh13RavinderSingh13
28.9k41638
It should be very easy task for awk(in case you are ok with it). Could you please try following, will add explanation shortly too.
awk '/^##.*##$/{next} /^##$/{flag="";next} /^##/ && !/##$/{flag=1} flag{next} 1' Input_file
Adding a non-one liner form of solution too now.
awk '
/^##.*##$/{
next
}
/^##$/{
flag=""
next
}
/^##/ && !/##$/{
flag=1
}
flag{
next
}
1
' Input_file
answered Nov 22 '18 at 14:27
RavinderSingh13RavinderSingh13
28.9k41638
answered Nov 22 '18 at 14:27
RavinderSingh13RavinderSingh13
28.9k41638
answered Nov 22 '18 at 14:27
RavinderSingh13RavinderSingh13
28.9k41638
answered Nov 22 '18 at 14:27
RavinderSingh13RavinderSingh13
28.9k41638
28.9k41638
add a comment |
add a comment |
Give a try to this:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
Wise men read this excellent tutorial: Sed - An Introduction and Tutorial by Bruce Barnett
The test:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
hello world
answered Nov 22 '18 at 14:31
Jay jargotJay jargot
1,9491510
add a comment |
Give a try to this:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
Wise men read this excellent tutorial: Sed - An Introduction and Tutorial by Bruce Barnett
The test:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
hello world
answered Nov 22 '18 at 14:31
Jay jargotJay jargot
1,9491510
add a comment |
Give a try to this:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
Wise men read this excellent tutorial: Sed - An Introduction and Tutorial by Bruce Barnett
The test:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
hello world
answered Nov 22 '18 at 14:31
Jay jargotJay jargot
1,9491510
Give a try to this:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
Wise men read this excellent tutorial: Sed - An Introduction and Tutorial by Bruce Barnett
The test:
sed -n '/^##/ { :1 ; /##$/ { d } ; n ; b 1 } ; p' myfile
hello world
answered Nov 22 '18 at 14:31
Jay jargotJay jargot
1,9491510
edited Nov 22 '18 at 16:26
answered Nov 22 '18 at 14:31
Jay jargotJay jargot
1,9491510
answered Nov 22 '18 at 14:31
Jay jargotJay jargot
1,9491510
answered Nov 22 '18 at 14:31
Jay jargotJay jargot
1,9491510
1,9491510
add a comment |
add a comment |
This might work for you (GNU sed):
sed -z 's/##[^#]*(#[^#][^#]*)*##n?//g' file
The -z option allows the whole of the file to be slurped into sed's pattern space. The regexp match is in three parts. The first part matches ## followed by zero or more non-#'s. The second part matches a zero or more group of characters consisting of a single # followed by a non-# followed by zero or more non-#'s. The third part matches ## and a possible newline. This regexp removes such matches globally throughout the file.
This can by shortened slightly by using the -r option to sweeten the final offering to:
sed -rz 's/##[^#]*(#[^#]+)*##n?//g' file
If the version of sed does not offer either options, then another solution is:
sed 'H;$!d;x;s/.//;s/##[^#]*(#[^#][^#]*)*##n?//g' file
It should by noted, that in the example above, all ##'s either start or end at the beginning or end of a line and so the solution below might also fit the bill:
sed 's/^##/,/##$/d' file
answered Nov 23 '18 at 8:42
potongpotong
35.9k43061
add a comment |
This might work for you (GNU sed):
sed -z 's/##[^#]*(#[^#][^#]*)*##n?//g' file
The -z option allows the whole of the file to be slurped into sed's pattern space. The regexp match is in three parts. The first part matches ## followed by zero or more non-#'s. The second part matches a zero or more group of characters consisting of a single # followed by a non-# followed by zero or more non-#'s. The third part matches ## and a possible newline. This regexp removes such matches globally throughout the file.
This can by shortened slightly by using the -r option to sweeten the final offering to:
sed -rz 's/##[^#]*(#[^#]+)*##n?//g' file
If the version of sed does not offer either options, then another solution is:
sed 'H;$!d;x;s/.//;s/##[^#]*(#[^#][^#]*)*##n?//g' file
It should by noted, that in the example above, all ##'s either start or end at the beginning or end of a line and so the solution below might also fit the bill:
sed 's/^##/,/##$/d' file
answered Nov 23 '18 at 8:42
potongpotong
35.9k43061
add a comment |
This might work for you (GNU sed):
sed -z 's/##[^#]*(#[^#][^#]*)*##n?//g' file
The -z option allows the whole of the file to be slurped into sed's pattern space. The regexp match is in three parts. The first part matches ## followed by zero or more non-#'s. The second part matches a zero or more group of characters consisting of a single # followed by a non-# followed by zero or more non-#'s. The third part matches ## and a possible newline. This regexp removes such matches globally throughout the file.
This can by shortened slightly by using the -r option to sweeten the final offering to:
sed -rz 's/##[^#]*(#[^#]+)*##n?//g' file
If the version of sed does not offer either options, then another solution is:
sed 'H;$!d;x;s/.//;s/##[^#]*(#[^#][^#]*)*##n?//g' file
It should by noted, that in the example above, all ##'s either start or end at the beginning or end of a line and so the solution below might also fit the bill:
sed 's/^##/,/##$/d' file
answered Nov 23 '18 at 8:42
potongpotong
35.9k43061
This might work for you (GNU sed):
sed -z 's/##[^#]*(#[^#][^#]*)*##n?//g' file
The -z option allows the whole of the file to be slurped into sed's pattern space. The regexp match is in three parts. The first part matches ## followed by zero or more non-#'s. The second part matches a zero or more group of characters consisting of a single # followed by a non-# followed by zero or more non-#'s. The third part matches ## and a possible newline. This regexp removes such matches globally throughout the file.
This can by shortened slightly by using the -r option to sweeten the final offering to:
sed -rz 's/##[^#]*(#[^#]+)*##n?//g' file
If the version of sed does not offer either options, then another solution is:
sed 'H;$!d;x;s/.//;s/##[^#]*(#[^#][^#]*)*##n?//g' file
It should by noted, that in the example above, all ##'s either start or end at the beginning or end of a line and so the solution below might also fit the bill:
sed 's/^##/,/##$/d' file
answered Nov 23 '18 at 8:42
potongpotong
35.9k43061
edited Nov 23 '18 at 8:52
answered Nov 23 '18 at 8:42
potongpotong
35.9k43061
answered Nov 23 '18 at 8:42
potongpotong
35.9k43061
answered Nov 23 '18 at 8:42
potongpotong
35.9k43061
35.9k43061
add a comment |
add a comment |
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Probably perl solution will be simpler:
perl -0pe 's/##.*?##R*//gs' file > newfile(demo).– Wiktor Stribiżew
Nov 22 '18 at 14:26
@WiktorStribiżew hey your solution works the best - post your answer & I'll accept it. Thanks!
– bukowski
Nov 22 '18 at 14:51
Posted with explanation.
– Wiktor Stribiżew
Nov 22 '18 at 14:55