Bash - variable variables [duplicate]
32
This question already has an answer here:
Dynamic variable names in Bash
10 answers
I have the variable $foo="something" and would like to use:
bar="foo"; echo $($bar)
to get "something" echoed.
bash variables
asked May 25 '12 at 15:40
user1165454user1165454
181138
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Feb 4 at 22:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
32
This question already has an answer here:
Dynamic variable names in Bash
10 answers
I have the variable $foo="something" and would like to use:
bar="foo"; echo $($bar)
to get "something" echoed.
bash variables
asked May 25 '12 at 15:40
user1165454user1165454
181138
marked as duplicate by codeforester bash
Users with the bash badge can single-handedly close bash questions as duplicates and reopen them as needed.
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Feb 4 at 22:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
Please see BashFAQ/006. Also, you shouldn't try to use a dollar sign on the left side of an assignment.
– Dennis Williamson
May 25 '12 at 15:56
add a comment |
32
32
32
12
This question already has an answer here:
Dynamic variable names in Bash
10 answers
I have the variable $foo="something" and would like to use:
bar="foo"; echo $($bar)
to get "something" echoed.
bash variables
asked May 25 '12 at 15:40
user1165454user1165454
181138
This question already has an answer here:
Dynamic variable names in Bash
10 answers
I have the variable $foo="something" and would like to use:
bar="foo"; echo $($bar)
to get "something" echoed.
This question already has an answer here:
Dynamic variable names in Bash
10 answers
bash variables
bash variables
asked May 25 '12 at 15:40
user1165454user1165454
181138
asked May 25 '12 at 15:40
user1165454user1165454
181138
edited May 25 '12 at 21:17
user1165454
asked May 25 '12 at 15:40
user1165454user1165454
181138
asked May 25 '12 at 15:40
user1165454user1165454
181138
asked May 25 '12 at 15:40
user1165454user1165454
181138
181138
marked as duplicate by codeforester bash
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Feb 4 at 22:37
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Feb 4 at 22:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
Please see BashFAQ/006. Also, you shouldn't try to use a dollar sign on the left side of an assignment.
– Dennis Williamson
May 25 '12 at 15:56
add a comment |
3
Please see BashFAQ/006. Also, you shouldn't try to use a dollar sign on the left side of an assignment.
– Dennis Williamson
May 25 '12 at 15:56
3
3
Please see BashFAQ/006. Also, you shouldn't try to use a dollar sign on the left side of an assignment.
– Dennis Williamson
May 25 '12 at 15:56
Please see BashFAQ/006. Also, you shouldn't try to use a dollar sign on the left side of an assignment.
– Dennis Williamson
May 25 '12 at 15:56
add a comment |
4 Answers
4
active
oldest
votes
63
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
edited Mar 26 '16 at 13:15
Gilles
75.5k19161205
answered May 25 '12 at 15:49
dAm2KdAm2K
7,95843039
Much better than my answer.
– mkb
May 25 '12 at 16:15
@mkb I didn't know eval :-)
– dAm2K
May 25 '12 at 22:28
1
inshit says bad substitution. Any idea how to do it insh?
– Shiplu Mokaddim
Sep 6 '13 at 6:00
how does this work with arrays?
– Edison
Apr 14 '14 at 21:14
@Edisonfoo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]}
– dAm2K
Apr 14 '14 at 23:23
|
show 3 more comments
7
The accepted answer is great. However, @Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[@]", so that the array is expanded with the "!". Check out this function to dump variables:
$ function dump_variables() {
for var in "$@"; do
echo "$var=${!var}"
done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[@]
This outputs:
STRING=Hello World
ARRAY=ab
ARRAY[@]=ab cd
When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[@] format, you get the array and all its values expanded.
answered Aug 6 '14 at 18:10
bishopbishop
24.5k56790
Good point about handling arrays. Any idea how to get the indices of an array? The manual indicates this is normally done with${!ARRAY[@]}, which seems to conflict with the variable indirection syntax.
– dimo414
Aug 28 '14 at 5:33
@dimo414 Yeah, getting the keys through indirection is trickier. You'd have to pass just the name, then do the expansion in the method:local -a 'keys=("${!'"$var"'[@]}")'. The indirection article on Bash Hackers goes into more depth.
– bishop
Aug 28 '14 at 13:20
add a comment |
5
eval echo "$$bar" would do it.
edited Mar 26 '16 at 13:16
Gilles
75.5k19161205
answered May 25 '12 at 15:44
mkbmkb
11k12246
6
Be aware of the security implications ofeval.
– Dennis Williamson
May 25 '12 at 15:58
1
This solution has the benefit of being POSIX-compatible for non-Bash shells (for example, lightweight environments like embedded systems or Docker containers). And you can assign the value to another variable like so:sh var=$(eval echo "$$bar")
– Jason Suárez
Feb 3 '17 at 4:29
add a comment |
1
To make it more clear how to do it with arrays:
arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation
# of the variable (array) as its value:
var=arr[@]
echo "${!var}"
answered May 16 '16 at 5:56
JahidJahid
13.7k46084
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
63
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
edited Mar 26 '16 at 13:15
Gilles
75.5k19161205
answered May 25 '12 at 15:49
dAm2KdAm2K
7,95843039
Much better than my answer.
– mkb
May 25 '12 at 16:15
@mkb I didn't know eval :-)
– dAm2K
May 25 '12 at 22:28
1
inshit says bad substitution. Any idea how to do it insh?
– Shiplu Mokaddim
Sep 6 '13 at 6:00
how does this work with arrays?
– Edison
Apr 14 '14 at 21:14
@Edisonfoo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]}
– dAm2K
Apr 14 '14 at 23:23
|
show 3 more comments
63
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
edited Mar 26 '16 at 13:15
Gilles
75.5k19161205
answered May 25 '12 at 15:49
dAm2KdAm2K
7,95843039
Much better than my answer.
– mkb
May 25 '12 at 16:15
@mkb I didn't know eval :-)
– dAm2K
May 25 '12 at 22:28
1
inshit says bad substitution. Any idea how to do it insh?
– Shiplu Mokaddim
Sep 6 '13 at 6:00
how does this work with arrays?
– Edison
Apr 14 '14 at 21:14
@Edisonfoo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]}
– dAm2K
Apr 14 '14 at 23:23
|
show 3 more comments
63
63
63
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
edited Mar 26 '16 at 13:15
Gilles
75.5k19161205
answered May 25 '12 at 15:49
dAm2KdAm2K
7,95843039
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
edited Mar 26 '16 at 13:15
Gilles
75.5k19161205
answered May 25 '12 at 15:49
dAm2KdAm2K
7,95843039
edited Mar 26 '16 at 13:15
Gilles
75.5k19161205
edited Mar 26 '16 at 13:15
Gilles
75.5k19161205
edited Mar 26 '16 at 13:15
Gilles
75.5k19161205
75.5k19161205
answered May 25 '12 at 15:49
dAm2KdAm2K
7,95843039
answered May 25 '12 at 15:49
dAm2KdAm2K
7,95843039
answered May 25 '12 at 15:49
dAm2KdAm2K
7,95843039
7,95843039
Much better than my answer.
– mkb
May 25 '12 at 16:15
@mkb I didn't know eval :-)
– dAm2K
May 25 '12 at 22:28
1
inshit says bad substitution. Any idea how to do it insh?
– Shiplu Mokaddim
Sep 6 '13 at 6:00
how does this work with arrays?
– Edison
Apr 14 '14 at 21:14
@Edisonfoo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]}
– dAm2K
Apr 14 '14 at 23:23
|
show 3 more comments
Much better than my answer.
– mkb
May 25 '12 at 16:15
@mkb I didn't know eval :-)
– dAm2K
May 25 '12 at 22:28
1
inshit says bad substitution. Any idea how to do it insh?
– Shiplu Mokaddim
Sep 6 '13 at 6:00
how does this work with arrays?
– Edison
Apr 14 '14 at 21:14
@Edisonfoo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]}
– dAm2K
Apr 14 '14 at 23:23
Much better than my answer.
– mkb
May 25 '12 at 16:15
Much better than my answer.
– mkb
May 25 '12 at 16:15
@mkb I didn't know eval :-)
– dAm2K
May 25 '12 at 22:28
@mkb I didn't know eval :-)
– dAm2K
May 25 '12 at 22:28
1
1
in
sh it says bad substitution. Any idea how to do it in sh?– Shiplu Mokaddim
Sep 6 '13 at 6:00
in
sh it says bad substitution. Any idea how to do it in sh?– Shiplu Mokaddim
Sep 6 '13 at 6:00
how does this work with arrays?
– Edison
Apr 14 '14 at 21:14
how does this work with arrays?
– Edison
Apr 14 '14 at 21:14
@Edison
foo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]} – dAm2K
Apr 14 '14 at 23:23
@Edison
foo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]} – dAm2K
Apr 14 '14 at 23:23
|
show 3 more comments
7
The accepted answer is great. However, @Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[@]", so that the array is expanded with the "!". Check out this function to dump variables:
$ function dump_variables() {
for var in "$@"; do
echo "$var=${!var}"
done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[@]
This outputs:
STRING=Hello World
ARRAY=ab
ARRAY[@]=ab cd
When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[@] format, you get the array and all its values expanded.
answered Aug 6 '14 at 18:10
bishopbishop
24.5k56790
Good point about handling arrays. Any idea how to get the indices of an array? The manual indicates this is normally done with${!ARRAY[@]}, which seems to conflict with the variable indirection syntax.
– dimo414
Aug 28 '14 at 5:33
@dimo414 Yeah, getting the keys through indirection is trickier. You'd have to pass just the name, then do the expansion in the method:local -a 'keys=("${!'"$var"'[@]}")'. The indirection article on Bash Hackers goes into more depth.
– bishop
Aug 28 '14 at 13:20
add a comment |
7
The accepted answer is great. However, @Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[@]", so that the array is expanded with the "!". Check out this function to dump variables:
$ function dump_variables() {
for var in "$@"; do
echo "$var=${!var}"
done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[@]
This outputs:
STRING=Hello World
ARRAY=ab
ARRAY[@]=ab cd
When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[@] format, you get the array and all its values expanded.
answered Aug 6 '14 at 18:10
bishopbishop
24.5k56790
Good point about handling arrays. Any idea how to get the indices of an array? The manual indicates this is normally done with${!ARRAY[@]}, which seems to conflict with the variable indirection syntax.
– dimo414
Aug 28 '14 at 5:33
@dimo414 Yeah, getting the keys through indirection is trickier. You'd have to pass just the name, then do the expansion in the method:local -a 'keys=("${!'"$var"'[@]}")'. The indirection article on Bash Hackers goes into more depth.
– bishop
Aug 28 '14 at 13:20
add a comment |
7
7
7
The accepted answer is great. However, @Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[@]", so that the array is expanded with the "!". Check out this function to dump variables:
$ function dump_variables() {
for var in "$@"; do
echo "$var=${!var}"
done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[@]
This outputs:
STRING=Hello World
ARRAY=ab
ARRAY[@]=ab cd
When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[@] format, you get the array and all its values expanded.
answered Aug 6 '14 at 18:10
bishopbishop
24.5k56790
The accepted answer is great. However, @Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[@]", so that the array is expanded with the "!". Check out this function to dump variables:
$ function dump_variables() {
for var in "$@"; do
echo "$var=${!var}"
done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[@]
This outputs:
STRING=Hello World
ARRAY=ab
ARRAY[@]=ab cd
When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[@] format, you get the array and all its values expanded.
answered Aug 6 '14 at 18:10
bishopbishop
24.5k56790
answered Aug 6 '14 at 18:10
bishopbishop
24.5k56790
answered Aug 6 '14 at 18:10
bishopbishop
24.5k56790
answered Aug 6 '14 at 18:10
bishopbishop
24.5k56790
24.5k56790
Good point about handling arrays. Any idea how to get the indices of an array? The manual indicates this is normally done with${!ARRAY[@]}, which seems to conflict with the variable indirection syntax.
– dimo414
Aug 28 '14 at 5:33
@dimo414 Yeah, getting the keys through indirection is trickier. You'd have to pass just the name, then do the expansion in the method:local -a 'keys=("${!'"$var"'[@]}")'. The indirection article on Bash Hackers goes into more depth.
– bishop
Aug 28 '14 at 13:20
add a comment |
Good point about handling arrays. Any idea how to get the indices of an array? The manual indicates this is normally done with${!ARRAY[@]}, which seems to conflict with the variable indirection syntax.
– dimo414
Aug 28 '14 at 5:33
@dimo414 Yeah, getting the keys through indirection is trickier. You'd have to pass just the name, then do the expansion in the method:local -a 'keys=("${!'"$var"'[@]}")'. The indirection article on Bash Hackers goes into more depth.
– bishop
Aug 28 '14 at 13:20
Good point about handling arrays. Any idea how to get the indices of an array? The manual indicates this is normally done with
${!ARRAY[@]}, which seems to conflict with the variable indirection syntax.– dimo414
Aug 28 '14 at 5:33
Good point about handling arrays. Any idea how to get the indices of an array? The manual indicates this is normally done with
${!ARRAY[@]}, which seems to conflict with the variable indirection syntax.– dimo414
Aug 28 '14 at 5:33
@dimo414 Yeah, getting the keys through indirection is trickier. You'd have to pass just the name, then do the expansion in the method:
local -a 'keys=("${!'"$var"'[@]}")'. The indirection article on Bash Hackers goes into more depth.– bishop
Aug 28 '14 at 13:20
@dimo414 Yeah, getting the keys through indirection is trickier. You'd have to pass just the name, then do the expansion in the method:
local -a 'keys=("${!'"$var"'[@]}")'. The indirection article on Bash Hackers goes into more depth.– bishop
Aug 28 '14 at 13:20
add a comment |
5
eval echo "$$bar" would do it.
edited Mar 26 '16 at 13:16
Gilles
75.5k19161205
answered May 25 '12 at 15:44
mkbmkb
11k12246
6
Be aware of the security implications ofeval.
– Dennis Williamson
May 25 '12 at 15:58
1
This solution has the benefit of being POSIX-compatible for non-Bash shells (for example, lightweight environments like embedded systems or Docker containers). And you can assign the value to another variable like so:sh var=$(eval echo "$$bar")
– Jason Suárez
Feb 3 '17 at 4:29
add a comment |
5
eval echo "$$bar" would do it.
edited Mar 26 '16 at 13:16
Gilles
75.5k19161205
answered May 25 '12 at 15:44
mkbmkb
11k12246
6
Be aware of the security implications ofeval.
– Dennis Williamson
May 25 '12 at 15:58
1
This solution has the benefit of being POSIX-compatible for non-Bash shells (for example, lightweight environments like embedded systems or Docker containers). And you can assign the value to another variable like so:sh var=$(eval echo "$$bar")
– Jason Suárez
Feb 3 '17 at 4:29
add a comment |
5
5
5
eval echo "$$bar" would do it.
edited Mar 26 '16 at 13:16
Gilles
75.5k19161205
answered May 25 '12 at 15:44
mkbmkb
11k12246
eval echo "$$bar" would do it.
edited Mar 26 '16 at 13:16
Gilles
75.5k19161205
answered May 25 '12 at 15:44
mkbmkb
11k12246
edited Mar 26 '16 at 13:16
Gilles
75.5k19161205
edited Mar 26 '16 at 13:16
Gilles
75.5k19161205
edited Mar 26 '16 at 13:16
Gilles
75.5k19161205
75.5k19161205
answered May 25 '12 at 15:44
mkbmkb
11k12246
answered May 25 '12 at 15:44
mkbmkb
11k12246
answered May 25 '12 at 15:44
mkbmkb
11k12246
11k12246
6
Be aware of the security implications ofeval.
– Dennis Williamson
May 25 '12 at 15:58
1
This solution has the benefit of being POSIX-compatible for non-Bash shells (for example, lightweight environments like embedded systems or Docker containers). And you can assign the value to another variable like so:sh var=$(eval echo "$$bar")
– Jason Suárez
Feb 3 '17 at 4:29
add a comment |
6
Be aware of the security implications ofeval.
– Dennis Williamson
May 25 '12 at 15:58
1
This solution has the benefit of being POSIX-compatible for non-Bash shells (for example, lightweight environments like embedded systems or Docker containers). And you can assign the value to another variable like so:sh var=$(eval echo "$$bar")
– Jason Suárez
Feb 3 '17 at 4:29
6
6
Be aware of the security implications of
eval.– Dennis Williamson
May 25 '12 at 15:58
Be aware of the security implications of
eval.– Dennis Williamson
May 25 '12 at 15:58
1
1
This solution has the benefit of being POSIX-compatible for non-Bash shells (for example, lightweight environments like embedded systems or Docker containers). And you can assign the value to another variable like so:
sh var=$(eval echo "$$bar") – Jason Suárez
Feb 3 '17 at 4:29
This solution has the benefit of being POSIX-compatible for non-Bash shells (for example, lightweight environments like embedded systems or Docker containers). And you can assign the value to another variable like so:
sh var=$(eval echo "$$bar") – Jason Suárez
Feb 3 '17 at 4:29
add a comment |
1
To make it more clear how to do it with arrays:
arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation
# of the variable (array) as its value:
var=arr[@]
echo "${!var}"
answered May 16 '16 at 5:56
JahidJahid
13.7k46084
add a comment |
1
To make it more clear how to do it with arrays:
arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation
# of the variable (array) as its value:
var=arr[@]
echo "${!var}"
answered May 16 '16 at 5:56
JahidJahid
13.7k46084
add a comment |
1
1
1
To make it more clear how to do it with arrays:
arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation
# of the variable (array) as its value:
var=arr[@]
echo "${!var}"
answered May 16 '16 at 5:56
JahidJahid
13.7k46084
To make it more clear how to do it with arrays:
arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation
# of the variable (array) as its value:
var=arr[@]
echo "${!var}"
answered May 16 '16 at 5:56
JahidJahid
13.7k46084
answered May 16 '16 at 5:56
JahidJahid
13.7k46084
answered May 16 '16 at 5:56
JahidJahid
13.7k46084
answered May 16 '16 at 5:56
JahidJahid
13.7k46084
13.7k46084
add a comment |
add a comment |
3
Please see BashFAQ/006. Also, you shouldn't try to use a dollar sign on the left side of an assignment.
– Dennis Williamson
May 25 '12 at 15:56