Checking for the existence of multiple directories
I want to check for the existence of multiple directories, say, dir1
, dir2
and dir3
, in the working directory.
I have the following
if [ -d "$PWD/dir1" ] && [ -d "$PWD/dir2" ] && [ -d "$PWD/dir3" ]; then
echo True
else
echo False
fi
But I suspect there is a more elegant way of doing this. Do not assume that there is a rule in the names of the directories.
The goal is to check for the existence of a few directories and for the nonexistence of others.
I'm using Bash, but portable code is preferred.
shell-script shell files directory control-flow
New contributor
add a comment |
I want to check for the existence of multiple directories, say, dir1
, dir2
and dir3
, in the working directory.
I have the following
if [ -d "$PWD/dir1" ] && [ -d "$PWD/dir2" ] && [ -d "$PWD/dir3" ]; then
echo True
else
echo False
fi
But I suspect there is a more elegant way of doing this. Do not assume that there is a rule in the names of the directories.
The goal is to check for the existence of a few directories and for the nonexistence of others.
I'm using Bash, but portable code is preferred.
shell-script shell files directory control-flow
New contributor
“The goal is to check for the existence of a few directories and for the nonexistence of others.” Well, your example checks for existence of all listed directories. Can you elaborate on this part ? Is it all listed or any listed ? Do you need a check that passed values are in fact directories and not other types of files ?
– Sergiy Kolodyazhnyy
14 mins ago
add a comment |
I want to check for the existence of multiple directories, say, dir1
, dir2
and dir3
, in the working directory.
I have the following
if [ -d "$PWD/dir1" ] && [ -d "$PWD/dir2" ] && [ -d "$PWD/dir3" ]; then
echo True
else
echo False
fi
But I suspect there is a more elegant way of doing this. Do not assume that there is a rule in the names of the directories.
The goal is to check for the existence of a few directories and for the nonexistence of others.
I'm using Bash, but portable code is preferred.
shell-script shell files directory control-flow
New contributor
I want to check for the existence of multiple directories, say, dir1
, dir2
and dir3
, in the working directory.
I have the following
if [ -d "$PWD/dir1" ] && [ -d "$PWD/dir2" ] && [ -d "$PWD/dir3" ]; then
echo True
else
echo False
fi
But I suspect there is a more elegant way of doing this. Do not assume that there is a rule in the names of the directories.
The goal is to check for the existence of a few directories and for the nonexistence of others.
I'm using Bash, but portable code is preferred.
shell-script shell files directory control-flow
shell-script shell files directory control-flow
New contributor
New contributor
edited 10 hours ago
Elegance
New contributor
asked 12 hours ago
EleganceElegance
283
283
New contributor
New contributor
“The goal is to check for the existence of a few directories and for the nonexistence of others.” Well, your example checks for existence of all listed directories. Can you elaborate on this part ? Is it all listed or any listed ? Do you need a check that passed values are in fact directories and not other types of files ?
– Sergiy Kolodyazhnyy
14 mins ago
add a comment |
“The goal is to check for the existence of a few directories and for the nonexistence of others.” Well, your example checks for existence of all listed directories. Can you elaborate on this part ? Is it all listed or any listed ? Do you need a check that passed values are in fact directories and not other types of files ?
– Sergiy Kolodyazhnyy
14 mins ago
“The goal is to check for the existence of a few directories and for the nonexistence of others.” Well, your example checks for existence of all listed directories. Can you elaborate on this part ? Is it all listed or any listed ? Do you need a check that passed values are in fact directories and not other types of files ?
– Sergiy Kolodyazhnyy
14 mins ago
“The goal is to check for the existence of a few directories and for the nonexistence of others.” Well, your example checks for existence of all listed directories. Can you elaborate on this part ? Is it all listed or any listed ? Do you need a check that passed values are in fact directories and not other types of files ?
– Sergiy Kolodyazhnyy
14 mins ago
add a comment |
5 Answers
5
active
oldest
votes
If you already expect them to be directories and are just checking whether they all exist, you could use the exit code from the ls
utility to determine whether one or more "errors occurred":
ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir3" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null
in order to make it disappear, since we only care about the exit code from ls
, not its output. Anything that's written to /dev/null
disappears — it is not written to your terminal.
Can you help me understand this command? I know what file descriptors are. I know1
is stdout,2
is stderr and I know what redirecting is. I don't understand the significance of/dev/null
, and I do not know how to parse the command.
– Elegance
10 hours ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
10 hours ago
Still trying to figure out how the syntax works. I read that&>filename
redirects both stdout and stderr tofilename
. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there
?
– Elegance
10 hours ago
1
I got it finally. Thanks.
– Elegance
10 hours ago
1
(1) You should probably use the-d
option (a) sols
needs only to stat the directories, and not read them, and (b) so the command will succeed even if the user doesn’t have read access to the directories. (2) I don’t see any reason to use"$PWD/"
except to guard against directories whose names begin with-
(and, of course, there are better ways to do that).
– G-Man
2 hours ago
|
show 3 more comments
I would loop:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
echo "$result"
The break
causes the loop to short-circuit, just like your chain of &&
add a comment |
A loop might be more elegant:
arr=("$PWD/dir1" "$PWD/dir2" "$PWD/dir2")
for d in "${arr[@]}"; do
if [ -d "$d"]; then
echo True
else
echo False
fi
done
This is Bash. A more portable one is Sh. There you can use the positional array:
set -- "$PWD/dir1" "$PWD/dir2" "$PWD/dir2"
Then to loop over it use "$@"
.
add a comment |
Why not just:
if [ -d "dir1" -a -d "dir2" -a -d "dir3" ]; then
echo True
else
echo False
fi
This is essentially what the OP started with,But I suspect there is a more elegant way of doing this
– Jeff Schaller
8 hours ago
2
Also POSIX discourages the use of-a
: "-a
and-o
binary primaries (...) operators have been marked obsolescent": pubs.opengroup.org/onlinepubs/9699919799
– Elegance
8 hours ago
@JeffSchaller It's more terse since it does it all in one call to test.
– David Conrad
5 hours ago
@Elegance They're still supported on all the systems I use, and probably will be a hundred years from now.
– David Conrad
5 hours ago
add a comment |
The goal is to check for the existence of a few directories
and for the nonexistence of others.
[Emphasis added]
Building on glenn jackman’s answer,
we can test for the nonexistence of other names like this:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
for dir in
"$PWD/dir4"
"$PWD/dir5"
"$PWD/dir6"
do
if [ -e "$dir" ]; then # Note: no “!”
result=False
break
fi
done
echo "$result"
I used
[ -e "$dir" ]
to test whether "$dir"
exists;i.e., if
dir5
exists but is a file, the result is False
. If you want only to test whether the names in the second group are directories,
use
[ -d "$dir" ]
, like in the first loop.Since we’re talking about checking for the existence of things
in the current working directory,
it’s probably not necessary to specify $PWD/
on the names; just do
for dir in
"dir1"
"dir2"
"dir3"
do
︙
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "106"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f503830%2fchecking-for-the-existence-of-multiple-directories%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you already expect them to be directories and are just checking whether they all exist, you could use the exit code from the ls
utility to determine whether one or more "errors occurred":
ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir3" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null
in order to make it disappear, since we only care about the exit code from ls
, not its output. Anything that's written to /dev/null
disappears — it is not written to your terminal.
Can you help me understand this command? I know what file descriptors are. I know1
is stdout,2
is stderr and I know what redirecting is. I don't understand the significance of/dev/null
, and I do not know how to parse the command.
– Elegance
10 hours ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
10 hours ago
Still trying to figure out how the syntax works. I read that&>filename
redirects both stdout and stderr tofilename
. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there
?
– Elegance
10 hours ago
1
I got it finally. Thanks.
– Elegance
10 hours ago
1
(1) You should probably use the-d
option (a) sols
needs only to stat the directories, and not read them, and (b) so the command will succeed even if the user doesn’t have read access to the directories. (2) I don’t see any reason to use"$PWD/"
except to guard against directories whose names begin with-
(and, of course, there are better ways to do that).
– G-Man
2 hours ago
|
show 3 more comments
If you already expect them to be directories and are just checking whether they all exist, you could use the exit code from the ls
utility to determine whether one or more "errors occurred":
ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir3" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null
in order to make it disappear, since we only care about the exit code from ls
, not its output. Anything that's written to /dev/null
disappears — it is not written to your terminal.
Can you help me understand this command? I know what file descriptors are. I know1
is stdout,2
is stderr and I know what redirecting is. I don't understand the significance of/dev/null
, and I do not know how to parse the command.
– Elegance
10 hours ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
10 hours ago
Still trying to figure out how the syntax works. I read that&>filename
redirects both stdout and stderr tofilename
. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there
?
– Elegance
10 hours ago
1
I got it finally. Thanks.
– Elegance
10 hours ago
1
(1) You should probably use the-d
option (a) sols
needs only to stat the directories, and not read them, and (b) so the command will succeed even if the user doesn’t have read access to the directories. (2) I don’t see any reason to use"$PWD/"
except to guard against directories whose names begin with-
(and, of course, there are better ways to do that).
– G-Man
2 hours ago
|
show 3 more comments
If you already expect them to be directories and are just checking whether they all exist, you could use the exit code from the ls
utility to determine whether one or more "errors occurred":
ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir3" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null
in order to make it disappear, since we only care about the exit code from ls
, not its output. Anything that's written to /dev/null
disappears — it is not written to your terminal.
If you already expect them to be directories and are just checking whether they all exist, you could use the exit code from the ls
utility to determine whether one or more "errors occurred":
ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir3" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null
in order to make it disappear, since we only care about the exit code from ls
, not its output. Anything that's written to /dev/null
disappears — it is not written to your terminal.
edited 2 hours ago
G-Man
13.2k93566
13.2k93566
answered 12 hours ago
Jeff SchallerJeff Schaller
42.8k1159136
42.8k1159136
Can you help me understand this command? I know what file descriptors are. I know1
is stdout,2
is stderr and I know what redirecting is. I don't understand the significance of/dev/null
, and I do not know how to parse the command.
– Elegance
10 hours ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
10 hours ago
Still trying to figure out how the syntax works. I read that&>filename
redirects both stdout and stderr tofilename
. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there
?
– Elegance
10 hours ago
1
I got it finally. Thanks.
– Elegance
10 hours ago
1
(1) You should probably use the-d
option (a) sols
needs only to stat the directories, and not read them, and (b) so the command will succeed even if the user doesn’t have read access to the directories. (2) I don’t see any reason to use"$PWD/"
except to guard against directories whose names begin with-
(and, of course, there are better ways to do that).
– G-Man
2 hours ago
|
show 3 more comments
Can you help me understand this command? I know what file descriptors are. I know1
is stdout,2
is stderr and I know what redirecting is. I don't understand the significance of/dev/null
, and I do not know how to parse the command.
– Elegance
10 hours ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
10 hours ago
Still trying to figure out how the syntax works. I read that&>filename
redirects both stdout and stderr tofilename
. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there
?
– Elegance
10 hours ago
1
I got it finally. Thanks.
– Elegance
10 hours ago
1
(1) You should probably use the-d
option (a) sols
needs only to stat the directories, and not read them, and (b) so the command will succeed even if the user doesn’t have read access to the directories. (2) I don’t see any reason to use"$PWD/"
except to guard against directories whose names begin with-
(and, of course, there are better ways to do that).
– G-Man
2 hours ago
Can you help me understand this command? I know what file descriptors are. I know
1
is stdout, 2
is stderr and I know what redirecting is. I don't understand the significance of /dev/null
, and I do not know how to parse the command.– Elegance
10 hours ago
Can you help me understand this command? I know what file descriptors are. I know
1
is stdout, 2
is stderr and I know what redirecting is. I don't understand the significance of /dev/null
, and I do not know how to parse the command.– Elegance
10 hours ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
10 hours ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
10 hours ago
Still trying to figure out how the syntax works. I read that
&>filename
redirects both stdout and stderr to filename
. So couldn't the command be simplified (at least to me it is more simple) as ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there
?– Elegance
10 hours ago
Still trying to figure out how the syntax works. I read that
&>filename
redirects both stdout and stderr to filename
. So couldn't the command be simplified (at least to me it is more simple) as ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there
?– Elegance
10 hours ago
1
1
I got it finally. Thanks.
– Elegance
10 hours ago
I got it finally. Thanks.
– Elegance
10 hours ago
1
1
(1) You should probably use the
-d
option (a) so ls
needs only to stat the directories, and not read them, and (b) so the command will succeed even if the user doesn’t have read access to the directories. (2) I don’t see any reason to use "$PWD/"
except to guard against directories whose names begin with -
(and, of course, there are better ways to do that).– G-Man
2 hours ago
(1) You should probably use the
-d
option (a) so ls
needs only to stat the directories, and not read them, and (b) so the command will succeed even if the user doesn’t have read access to the directories. (2) I don’t see any reason to use "$PWD/"
except to guard against directories whose names begin with -
(and, of course, there are better ways to do that).– G-Man
2 hours ago
|
show 3 more comments
I would loop:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
echo "$result"
The break
causes the loop to short-circuit, just like your chain of &&
add a comment |
I would loop:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
echo "$result"
The break
causes the loop to short-circuit, just like your chain of &&
add a comment |
I would loop:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
echo "$result"
The break
causes the loop to short-circuit, just like your chain of &&
I would loop:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
echo "$result"
The break
causes the loop to short-circuit, just like your chain of &&
answered 12 hours ago
glenn jackmanglenn jackman
52.1k572112
52.1k572112
add a comment |
add a comment |
A loop might be more elegant:
arr=("$PWD/dir1" "$PWD/dir2" "$PWD/dir2")
for d in "${arr[@]}"; do
if [ -d "$d"]; then
echo True
else
echo False
fi
done
This is Bash. A more portable one is Sh. There you can use the positional array:
set -- "$PWD/dir1" "$PWD/dir2" "$PWD/dir2"
Then to loop over it use "$@"
.
add a comment |
A loop might be more elegant:
arr=("$PWD/dir1" "$PWD/dir2" "$PWD/dir2")
for d in "${arr[@]}"; do
if [ -d "$d"]; then
echo True
else
echo False
fi
done
This is Bash. A more portable one is Sh. There you can use the positional array:
set -- "$PWD/dir1" "$PWD/dir2" "$PWD/dir2"
Then to loop over it use "$@"
.
add a comment |
A loop might be more elegant:
arr=("$PWD/dir1" "$PWD/dir2" "$PWD/dir2")
for d in "${arr[@]}"; do
if [ -d "$d"]; then
echo True
else
echo False
fi
done
This is Bash. A more portable one is Sh. There you can use the positional array:
set -- "$PWD/dir1" "$PWD/dir2" "$PWD/dir2"
Then to loop over it use "$@"
.
A loop might be more elegant:
arr=("$PWD/dir1" "$PWD/dir2" "$PWD/dir2")
for d in "${arr[@]}"; do
if [ -d "$d"]; then
echo True
else
echo False
fi
done
This is Bash. A more portable one is Sh. There you can use the positional array:
set -- "$PWD/dir1" "$PWD/dir2" "$PWD/dir2"
Then to loop over it use "$@"
.
edited 12 hours ago
answered 12 hours ago
TomaszTomasz
9,86652966
9,86652966
add a comment |
add a comment |
Why not just:
if [ -d "dir1" -a -d "dir2" -a -d "dir3" ]; then
echo True
else
echo False
fi
This is essentially what the OP started with,But I suspect there is a more elegant way of doing this
– Jeff Schaller
8 hours ago
2
Also POSIX discourages the use of-a
: "-a
and-o
binary primaries (...) operators have been marked obsolescent": pubs.opengroup.org/onlinepubs/9699919799
– Elegance
8 hours ago
@JeffSchaller It's more terse since it does it all in one call to test.
– David Conrad
5 hours ago
@Elegance They're still supported on all the systems I use, and probably will be a hundred years from now.
– David Conrad
5 hours ago
add a comment |
Why not just:
if [ -d "dir1" -a -d "dir2" -a -d "dir3" ]; then
echo True
else
echo False
fi
This is essentially what the OP started with,But I suspect there is a more elegant way of doing this
– Jeff Schaller
8 hours ago
2
Also POSIX discourages the use of-a
: "-a
and-o
binary primaries (...) operators have been marked obsolescent": pubs.opengroup.org/onlinepubs/9699919799
– Elegance
8 hours ago
@JeffSchaller It's more terse since it does it all in one call to test.
– David Conrad
5 hours ago
@Elegance They're still supported on all the systems I use, and probably will be a hundred years from now.
– David Conrad
5 hours ago
add a comment |
Why not just:
if [ -d "dir1" -a -d "dir2" -a -d "dir3" ]; then
echo True
else
echo False
fi
Why not just:
if [ -d "dir1" -a -d "dir2" -a -d "dir3" ]; then
echo True
else
echo False
fi
answered 9 hours ago
David ConradDavid Conrad
1414
1414
This is essentially what the OP started with,But I suspect there is a more elegant way of doing this
– Jeff Schaller
8 hours ago
2
Also POSIX discourages the use of-a
: "-a
and-o
binary primaries (...) operators have been marked obsolescent": pubs.opengroup.org/onlinepubs/9699919799
– Elegance
8 hours ago
@JeffSchaller It's more terse since it does it all in one call to test.
– David Conrad
5 hours ago
@Elegance They're still supported on all the systems I use, and probably will be a hundred years from now.
– David Conrad
5 hours ago
add a comment |
This is essentially what the OP started with,But I suspect there is a more elegant way of doing this
– Jeff Schaller
8 hours ago
2
Also POSIX discourages the use of-a
: "-a
and-o
binary primaries (...) operators have been marked obsolescent": pubs.opengroup.org/onlinepubs/9699919799
– Elegance
8 hours ago
@JeffSchaller It's more terse since it does it all in one call to test.
– David Conrad
5 hours ago
@Elegance They're still supported on all the systems I use, and probably will be a hundred years from now.
– David Conrad
5 hours ago
This is essentially what the OP started with,
But I suspect there is a more elegant way of doing this
– Jeff Schaller
8 hours ago
This is essentially what the OP started with,
But I suspect there is a more elegant way of doing this
– Jeff Schaller
8 hours ago
2
2
Also POSIX discourages the use of
-a
: "-a
and -o
binary primaries (...) operators have been marked obsolescent": pubs.opengroup.org/onlinepubs/9699919799– Elegance
8 hours ago
Also POSIX discourages the use of
-a
: "-a
and -o
binary primaries (...) operators have been marked obsolescent": pubs.opengroup.org/onlinepubs/9699919799– Elegance
8 hours ago
@JeffSchaller It's more terse since it does it all in one call to test.
– David Conrad
5 hours ago
@JeffSchaller It's more terse since it does it all in one call to test.
– David Conrad
5 hours ago
@Elegance They're still supported on all the systems I use, and probably will be a hundred years from now.
– David Conrad
5 hours ago
@Elegance They're still supported on all the systems I use, and probably will be a hundred years from now.
– David Conrad
5 hours ago
add a comment |
The goal is to check for the existence of a few directories
and for the nonexistence of others.
[Emphasis added]
Building on glenn jackman’s answer,
we can test for the nonexistence of other names like this:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
for dir in
"$PWD/dir4"
"$PWD/dir5"
"$PWD/dir6"
do
if [ -e "$dir" ]; then # Note: no “!”
result=False
break
fi
done
echo "$result"
I used
[ -e "$dir" ]
to test whether "$dir"
exists;i.e., if
dir5
exists but is a file, the result is False
. If you want only to test whether the names in the second group are directories,
use
[ -d "$dir" ]
, like in the first loop.Since we’re talking about checking for the existence of things
in the current working directory,
it’s probably not necessary to specify $PWD/
on the names; just do
for dir in
"dir1"
"dir2"
"dir3"
do
︙
add a comment |
The goal is to check for the existence of a few directories
and for the nonexistence of others.
[Emphasis added]
Building on glenn jackman’s answer,
we can test for the nonexistence of other names like this:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
for dir in
"$PWD/dir4"
"$PWD/dir5"
"$PWD/dir6"
do
if [ -e "$dir" ]; then # Note: no “!”
result=False
break
fi
done
echo "$result"
I used
[ -e "$dir" ]
to test whether "$dir"
exists;i.e., if
dir5
exists but is a file, the result is False
. If you want only to test whether the names in the second group are directories,
use
[ -d "$dir" ]
, like in the first loop.Since we’re talking about checking for the existence of things
in the current working directory,
it’s probably not necessary to specify $PWD/
on the names; just do
for dir in
"dir1"
"dir2"
"dir3"
do
︙
add a comment |
The goal is to check for the existence of a few directories
and for the nonexistence of others.
[Emphasis added]
Building on glenn jackman’s answer,
we can test for the nonexistence of other names like this:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
for dir in
"$PWD/dir4"
"$PWD/dir5"
"$PWD/dir6"
do
if [ -e "$dir" ]; then # Note: no “!”
result=False
break
fi
done
echo "$result"
I used
[ -e "$dir" ]
to test whether "$dir"
exists;i.e., if
dir5
exists but is a file, the result is False
. If you want only to test whether the names in the second group are directories,
use
[ -d "$dir" ]
, like in the first loop.Since we’re talking about checking for the existence of things
in the current working directory,
it’s probably not necessary to specify $PWD/
on the names; just do
for dir in
"dir1"
"dir2"
"dir3"
do
︙
The goal is to check for the existence of a few directories
and for the nonexistence of others.
[Emphasis added]
Building on glenn jackman’s answer,
we can test for the nonexistence of other names like this:
result=True
for dir in
"$PWD/dir1"
"$PWD/dir2"
"$PWD/dir3"
do
if ! [ -d "$dir" ]; then
result=False
break
fi
done
for dir in
"$PWD/dir4"
"$PWD/dir5"
"$PWD/dir6"
do
if [ -e "$dir" ]; then # Note: no “!”
result=False
break
fi
done
echo "$result"
I used
[ -e "$dir" ]
to test whether "$dir"
exists;i.e., if
dir5
exists but is a file, the result is False
. If you want only to test whether the names in the second group are directories,
use
[ -d "$dir" ]
, like in the first loop.Since we’re talking about checking for the existence of things
in the current working directory,
it’s probably not necessary to specify $PWD/
on the names; just do
for dir in
"dir1"
"dir2"
"dir3"
do
︙
answered 1 hour ago
G-ManG-Man
13.2k93566
13.2k93566
add a comment |
add a comment |
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
Elegance is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Unix & Linux Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f503830%2fchecking-for-the-existence-of-multiple-directories%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
“The goal is to check for the existence of a few directories and for the nonexistence of others.” Well, your example checks for existence of all listed directories. Can you elaborate on this part ? Is it all listed or any listed ? Do you need a check that passed values are in fact directories and not other types of files ?
– Sergiy Kolodyazhnyy
14 mins ago