Reference bash variable from a variable [duplicate]
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How to do indirect variable evaluation
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How can I reference a variable in bash based on another variable? Let me setup the example:
package="foobar"
# the variable I wish to reference is $foobar_darwin_amd64
# thus trying:
echo "$package_darwin_amd64"
But this does not work.
bash variable-substitution
asked Jan 31 at 23:27
Justin
1112
marked as duplicate by Barmar, Jeff Schaller, G-Man, Kusalananda bash
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Feb 1 at 6:44
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up vote
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This question already has an answer here:
How to do indirect variable evaluation
5 answers
How can I reference a variable in bash based on another variable? Let me setup the example:
package="foobar"
# the variable I wish to reference is $foobar_darwin_amd64
# thus trying:
echo "$package_darwin_amd64"
But this does not work.
bash variable-substitution
asked Jan 31 at 23:27
Justin
1112
marked as duplicate by Barmar, Jeff Schaller, G-Man, Kusalananda bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.
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Feb 1 at 6:44
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 |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
How to do indirect variable evaluation
5 answers
How can I reference a variable in bash based on another variable? Let me setup the example:
package="foobar"
# the variable I wish to reference is $foobar_darwin_amd64
# thus trying:
echo "$package_darwin_amd64"
But this does not work.
bash variable-substitution
asked Jan 31 at 23:27
Justin
1112
This question already has an answer here:
How to do indirect variable evaluation
5 answers
How can I reference a variable in bash based on another variable? Let me setup the example:
package="foobar"
# the variable I wish to reference is $foobar_darwin_amd64
# thus trying:
echo "$package_darwin_amd64"
But this does not work.
This question already has an answer here:
How to do indirect variable evaluation
5 answers
bash variable-substitution
asked Jan 31 at 23:27
Justin
1112
asked Jan 31 at 23:27
Justin
1112
asked Jan 31 at 23:27
Justin
1112
asked Jan 31 at 23:27
Justin
1112
1112
marked as duplicate by Barmar, Jeff Schaller, G-Man, Kusalananda bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.
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3 Answers
3
active
oldest
votes
up vote
2
down vote
If you shell supports the $!varname form of indirect references, you can do (as suggested by @Barmar):
$ foobar_darwin_amd64=pinto
$ package=foobar
$ varname="$package_darwin_amd64"
$ echo $!varname
pinto
Otherwise, you can use eval:
$ foobar_darwin_amd64=pinto
$ package=foobar
$ eval echo $$package_darwin_amd64
pinto
That said, using eval has some risks associated with it, see this link for more discussion.
answered Jan 31 at 23:31
Andy Dalton
4,7561520
Why useevalwhen you can use an indirect reference?
– Barmar
Jan 31 at 23:33
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up vote
1
down vote
Here is the solution that worked for me:
VARNAME="$package_darwin_amd64"
echo "$!VARNAME"
answered Jan 31 at 23:49
Justin
1112
add a comment |Â
up vote
1
down vote
In bash v4 you can use a "nameref"
$ foobar_darwin_amd64=pinto
$ package=foobar
$ declare -n var=$package_darwin_amd64
$ echo $var
pinto
answered Feb 1 at 0:32
glenn jackman
46.5k265103
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If you shell supports the $!varname form of indirect references, you can do (as suggested by @Barmar):
$ foobar_darwin_amd64=pinto
$ package=foobar
$ varname="$package_darwin_amd64"
$ echo $!varname
pinto
Otherwise, you can use eval:
$ foobar_darwin_amd64=pinto
$ package=foobar
$ eval echo $$package_darwin_amd64
pinto
That said, using eval has some risks associated with it, see this link for more discussion.
answered Jan 31 at 23:31
Andy Dalton
4,7561520
Why useevalwhen you can use an indirect reference?
– Barmar
Jan 31 at 23:33
add a comment |Â
up vote
2
down vote
If you shell supports the $!varname form of indirect references, you can do (as suggested by @Barmar):
$ foobar_darwin_amd64=pinto
$ package=foobar
$ varname="$package_darwin_amd64"
$ echo $!varname
pinto
Otherwise, you can use eval:
$ foobar_darwin_amd64=pinto
$ package=foobar
$ eval echo $$package_darwin_amd64
pinto
That said, using eval has some risks associated with it, see this link for more discussion.
answered Jan 31 at 23:31
Andy Dalton
4,7561520
Why useevalwhen you can use an indirect reference?
– Barmar
Jan 31 at 23:33
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you shell supports the $!varname form of indirect references, you can do (as suggested by @Barmar):
$ foobar_darwin_amd64=pinto
$ package=foobar
$ varname="$package_darwin_amd64"
$ echo $!varname
pinto
Otherwise, you can use eval:
$ foobar_darwin_amd64=pinto
$ package=foobar
$ eval echo $$package_darwin_amd64
pinto
That said, using eval has some risks associated with it, see this link for more discussion.
answered Jan 31 at 23:31
Andy Dalton
4,7561520
If you shell supports the $!varname form of indirect references, you can do (as suggested by @Barmar):
$ foobar_darwin_amd64=pinto
$ package=foobar
$ varname="$package_darwin_amd64"
$ echo $!varname
pinto
Otherwise, you can use eval:
$ foobar_darwin_amd64=pinto
$ package=foobar
$ eval echo $$package_darwin_amd64
pinto
That said, using eval has some risks associated with it, see this link for more discussion.
answered Jan 31 at 23:31
Andy Dalton
4,7561520
edited Jan 31 at 23:46
answered Jan 31 at 23:31
Andy Dalton
4,7561520
answered Jan 31 at 23:31
Andy Dalton
4,7561520
answered Jan 31 at 23:31
Andy Dalton
4,7561520
4,7561520
Why useevalwhen you can use an indirect reference?
– Barmar
Jan 31 at 23:33
add a comment |Â
Why useevalwhen you can use an indirect reference?
– Barmar
Jan 31 at 23:33
Why use
eval when you can use an indirect reference?– Barmar
Jan 31 at 23:33
Why use
eval when you can use an indirect reference?– Barmar
Jan 31 at 23:33
add a comment |Â
up vote
1
down vote
Here is the solution that worked for me:
VARNAME="$package_darwin_amd64"
echo "$!VARNAME"
answered Jan 31 at 23:49
Justin
1112
add a comment |Â
up vote
1
down vote
Here is the solution that worked for me:
VARNAME="$package_darwin_amd64"
echo "$!VARNAME"
answered Jan 31 at 23:49
Justin
1112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is the solution that worked for me:
VARNAME="$package_darwin_amd64"
echo "$!VARNAME"
answered Jan 31 at 23:49
Justin
1112
Here is the solution that worked for me:
VARNAME="$package_darwin_amd64"
echo "$!VARNAME"
answered Jan 31 at 23:49
Justin
1112
answered Jan 31 at 23:49
Justin
1112
answered Jan 31 at 23:49
Justin
1112
answered Jan 31 at 23:49
Justin
1112
1112
add a comment |Â
add a comment |Â
up vote
1
down vote
In bash v4 you can use a "nameref"
$ foobar_darwin_amd64=pinto
$ package=foobar
$ declare -n var=$package_darwin_amd64
$ echo $var
pinto
answered Feb 1 at 0:32
glenn jackman
46.5k265103
add a comment |Â
up vote
1
down vote
In bash v4 you can use a "nameref"
$ foobar_darwin_amd64=pinto
$ package=foobar
$ declare -n var=$package_darwin_amd64
$ echo $var
pinto
answered Feb 1 at 0:32
glenn jackman
46.5k265103
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In bash v4 you can use a "nameref"
$ foobar_darwin_amd64=pinto
$ package=foobar
$ declare -n var=$package_darwin_amd64
$ echo $var
pinto
answered Feb 1 at 0:32
glenn jackman
46.5k265103
In bash v4 you can use a "nameref"
$ foobar_darwin_amd64=pinto
$ package=foobar
$ declare -n var=$package_darwin_amd64
$ echo $var
pinto
answered Feb 1 at 0:32
glenn jackman
46.5k265103
answered Feb 1 at 0:32
glenn jackman
46.5k265103
answered Feb 1 at 0:32
glenn jackman
46.5k265103
answered Feb 1 at 0:32
glenn jackman
46.5k265103
46.5k265103
add a comment |Â
add a comment |Â