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.

asked Jan 31 at 23:27

Justin

1112

marked as duplicate by Barmar, Jeff Schaller, G-Man, KusalanandaÂ bash
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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.

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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.

asked Jan 31 at 23:27

Justin

1112

marked as duplicate by Barmar, Jeff Schaller, G-Man, KusalanandaÂ bash
Users with the Â 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 commentÂ |Â

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.

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

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

marked as duplicate by Barmar, Jeff Schaller, G-Man, KusalanandaÂ bash
Users with the Â 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.

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Feb 1 at 6:44

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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.

edited Jan 31 at 23:46

answered Jan 31 at 23:31

Andy Dalton

4,7561520

Why use eval when you can use an indirect reference?
â€“Â Barmar
Jan 31 at 23:33

add a commentÂ |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 commentÂ |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 commentÂ |Â

3 Answers
3

active

oldest

votes

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.

edited Jan 31 at 23:46

answered Jan 31 at 23:31

Andy Dalton

4,7561520

Why use eval when you can use an indirect reference?
â€“Â Barmar
Jan 31 at 23:33

add a commentÂ |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.

edited Jan 31 at 23:46

answered Jan 31 at 23:31

Andy Dalton

4,7561520

Why use eval when you can use an indirect reference?
â€“Â Barmar
Jan 31 at 23:33

add a commentÂ |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.

edited Jan 31 at 23:46

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.

edited Jan 31 at 23:46

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

Why use eval when you can use an indirect reference?
â€“Â Barmar
Jan 31 at 23:33

add a commentÂ |Â

Why use eval when you can use an indirect reference?
â€“Â Barmar
Jan 31 at 23:33

Why use eval when you can use an indirect reference?
â€“Â Barmar
Jan 31 at 23:33

add a commentÂ |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 commentÂ |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 commentÂ |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

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

add a commentÂ |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 commentÂ |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 commentÂ |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

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

add a commentÂ |Â

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