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 "$PWD/dir3" ]; then
echo True
else
echo False
fi
But I suspect there is a more elegant way of doing this. Do not assume that there is a rule in the names of the directories.
The goal is to check for the existence of a few directories and for the nonexistence of others.
I'm using Bash, but portable code is preferred.
shell-script shell files directory control-flow
asked 2 hours ago
EleganceElegance
83
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I want to check for the existence of multiple directories, say, dir1, dir2 and dir3, in the working directory.
I have the following
if [ -d "$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
asked 2 hours ago
EleganceElegance
83
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 "$PWD/dir3" ]; then
echo True
else
echo False
fi
But I suspect there is a more elegant way of doing this. Do not assume that there is a rule in the names of the directories.
The goal is to check for the existence of a few directories and for the nonexistence of others.
I'm using Bash, but portable code is preferred.
shell-script shell files directory control-flow
asked 2 hours ago
EleganceElegance
83
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to check for the existence of multiple directories, say, dir1, dir2 and dir3, in the working directory.
I have the following
if [ -d "$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
asked 2 hours ago
EleganceElegance
83
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
EleganceElegance
83
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 25 mins ago
Elegance
asked 2 hours ago
EleganceElegance
83
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
EleganceElegance
83
asked 2 hours ago
EleganceElegance
83
83
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Elegance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
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" "$PWD/dir2" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null in order to make it disappear, since we only care about the exit code from ls, not its output. Anything that's written to /dev/null disappears -- it is not written to your terminal.
answered 2 hours ago
Jeff SchallerJeff Schaller
42.7k1159136
Can you help me understand this command? I know what file descriptors are. I know1is stdout,2is stderr and I know what redirecting is. I don't understand the significance of/dev/null, and I do not know how to parse the command.
– Elegance
1 hour ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
57 mins ago
Still trying to figure out how the syntax works. I read that&>filenameredirects both stdout and stderr tofilename. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there?
– Elegance
47 mins ago
It could, but not portably -- plainshdoes not understand&>; it would misinterpret that as "run me in the background and send stdout to the redirection". I spelled it out from habit and kept it there because of the "portable code" preference.
– Jeff Schaller
34 mins ago
1
I got it finally. Thanks.
– Elegance
22 mins ago
|
show 2 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 "$result"
The break causes the loop to short-circuit, just like your chain of &&
answered 2 hours ago
glenn jackmanglenn jackman
52k572112
add a comment |
A loop might be more elegant:
arr=("$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" "$PWD/dir2"
Then to loop over it use "$@".
answered 2 hours ago
TomaszTomasz
9,80152965
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
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" "$PWD/dir2" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null in order to make it disappear, since we only care about the exit code from ls, not its output. Anything that's written to /dev/null disappears -- it is not written to your terminal.
answered 2 hours ago
Jeff SchallerJeff Schaller
42.7k1159136
Can you help me understand this command? I know what file descriptors are. I know1is stdout,2is stderr and I know what redirecting is. I don't understand the significance of/dev/null, and I do not know how to parse the command.
– Elegance
1 hour ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
57 mins ago
Still trying to figure out how the syntax works. I read that&>filenameredirects both stdout and stderr tofilename. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there?
– Elegance
47 mins ago
It could, but not portably -- plainshdoes not understand&>; it would misinterpret that as "run me in the background and send stdout to the redirection". I spelled it out from habit and kept it there because of the "portable code" preference.
– Jeff Schaller
34 mins ago
1
I got it finally. Thanks.
– Elegance
22 mins ago
|
show 2 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" "$PWD/dir2" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null in order to make it disappear, since we only care about the exit code from ls, not its output. Anything that's written to /dev/null disappears -- it is not written to your terminal.
answered 2 hours ago
Jeff SchallerJeff Schaller
42.7k1159136
Can you help me understand this command? I know what file descriptors are. I know1is stdout,2is stderr and I know what redirecting is. I don't understand the significance of/dev/null, and I do not know how to parse the command.
– Elegance
1 hour ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
57 mins ago
Still trying to figure out how the syntax works. I read that&>filenameredirects both stdout and stderr tofilename. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there?
– Elegance
47 mins ago
It could, but not portably -- plainshdoes not understand&>; it would misinterpret that as "run me in the background and send stdout to the redirection". I spelled it out from habit and kept it there because of the "portable code" preference.
– Jeff Schaller
34 mins ago
1
I got it finally. Thanks.
– Elegance
22 mins ago
|
show 2 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" "$PWD/dir2" >/dev/null 2>&1 && echo All there
I redirect the output and stderr to /dev/null in order to make it disappear, since we only care about the exit code from ls, not its output. Anything that's written to /dev/null disappears -- it is not written to your terminal.
answered 2 hours ago
Jeff SchallerJeff Schaller
42.7k1159136
If you already expect them to be directories and are just checking whether they all exist, you could use the exit code from the ls utility to determine whether one or more "errors occurred":
ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" >/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.
answered 2 hours ago
Jeff SchallerJeff Schaller
42.7k1159136
edited 58 mins ago
answered 2 hours ago
Jeff SchallerJeff Schaller
42.7k1159136
answered 2 hours ago
Jeff SchallerJeff Schaller
42.7k1159136
answered 2 hours ago
Jeff SchallerJeff Schaller
42.7k1159136
42.7k1159136
Can you help me understand this command? I know what file descriptors are. I know1is stdout,2is 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
1 hour 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
57 mins ago
Still trying to figure out how the syntax works. I read that&>filenameredirects 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" "$PWD/dir2" &>/dev/null && echo All there?
– Elegance
47 mins ago
It could, but not portably -- plainshdoes not understand&>; it would misinterpret that as "run me in the background and send stdout to the redirection". I spelled it out from habit and kept it there because of the "portable code" preference.
– Jeff Schaller
34 mins ago
1
I got it finally. Thanks.
– Elegance
22 mins ago
|
show 2 more comments
Can you help me understand this command? I know what file descriptors are. I know1is stdout,2is stderr and I know what redirecting is. I don't understand the significance of/dev/null, and I do not know how to parse the command.
– Elegance
1 hour ago
@Elegance I added a little explanation. For more in-depth answers regarding /dev/null, see unix.stackexchange.com/questions/163352/… and unix.stackexchange.com/questions/438130/…
– Jeff Schaller
57 mins ago
Still trying to figure out how the syntax works. I read that&>filenameredirects both stdout and stderr tofilename. So couldn't the command be simplified (at least to me it is more simple) asls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there?
– Elegance
47 mins ago
It could, but not portably -- plainshdoes not understand&>; it would misinterpret that as "run me in the background and send stdout to the redirection". I spelled it out from habit and kept it there because of the "portable code" preference.
– Jeff Schaller
34 mins ago
1
I got it finally. Thanks.
– Elegance
22 mins 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
1 hour 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
1 hour 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
57 mins 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
57 mins 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" "$PWD/dir2" &>/dev/null && echo All there?– Elegance
47 mins ago
Still trying to figure out how the syntax works. I read that
&>filename redirects both stdout and stderr to filename. So couldn't the command be simplified (at least to me it is more simple) as ls "$PWD/dir1" "$PWD/dir2" "$PWD/dir2" &>/dev/null && echo All there?– Elegance
47 mins ago
It could, but not portably -- plain
sh does not understand &>; it would misinterpret that as "run me in the background and send stdout to the redirection". I spelled it out from habit and kept it there because of the "portable code" preference.– Jeff Schaller
34 mins ago
It could, but not portably -- plain
sh does not understand &>; it would misinterpret that as "run me in the background and send stdout to the redirection". I spelled it out from habit and kept it there because of the "portable code" preference.– Jeff Schaller
34 mins ago
1
1
I got it finally. Thanks.
– Elegance
22 mins ago
I got it finally. Thanks.
– Elegance
22 mins ago
|
show 2 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 "$result"
The break causes the loop to short-circuit, just like your chain of &&
answered 2 hours ago
glenn jackmanglenn jackman
52k572112
add a comment |
I would loop:
result=True
for dir in
"$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 &&
answered 2 hours ago
glenn jackmanglenn jackman
52k572112
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 "$result"
The break causes the loop to short-circuit, just like your chain of &&
answered 2 hours ago
glenn jackmanglenn jackman
52k572112
I would loop:
result=True
for dir in
"$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 2 hours ago
glenn jackmanglenn jackman
52k572112
answered 2 hours ago
glenn jackmanglenn jackman
52k572112
answered 2 hours ago
glenn jackmanglenn jackman
52k572112
answered 2 hours ago
glenn jackmanglenn jackman
52k572112
52k572112
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 "$@".
answered 2 hours ago
TomaszTomasz
9,80152965
add a comment |
A loop might be more elegant:
arr=("$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" "$PWD/dir2"
Then to loop over it use "$@".
answered 2 hours ago
TomaszTomasz
9,80152965
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 "$@".
answered 2 hours ago
TomaszTomasz
9,80152965
A loop might be more elegant:
arr=("$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" "$PWD/dir2"
Then to loop over it use "$@".
answered 2 hours ago
TomaszTomasz
9,80152965
edited 2 hours ago
answered 2 hours ago
TomaszTomasz
9,80152965
answered 2 hours ago
TomaszTomasz
9,80152965
answered 2 hours ago
TomaszTomasz
9,80152965
9,80152965
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.
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