How can I define a variable name in bash that includes a variabe?

Question

I want to define the following:

COMMIT_ID_${ARCH}=$(git log --format="%H" -n 1)

But it fails:

-bash: COMMIT_ID_amd64=a7c9e0a53972a3d42a8035f45469f1959a0475f8: command not found

How will you reference your variable later? Because while you can use your dynamic variable with declare+indirect expansion, using associative arrays is generally simpler & more readable. — Aaron
– Aaron, Commented Oct 21, 2019 at 14:23
Unless you will have two variables like COMMIT_ID_x86 and COMMIT_ID_arm, there's nothing gained by defining COMMIT_ID_x86 instead of simply COMMIT_ID. — chepner
– chepner, Commented Oct 21, 2019 at 14:24
I'm not actually echoing it as it is in a script and is used in a sed substitution. — mickt
– mickt, Commented Oct 21, 2019 at 14:33
All your questions about indirection are answered by BashFAQ/006 — kvantour
– kvantour, Commented Oct 21, 2019 at 14:33

chepner · Accepted Answer · 2019-10-21 14:25:54Z

2

Use an associative array.

declare -A commit_id

commit_id[$ARCH]=$(git log --format="%H" -n 1)

...

echo "${commit_id[$ARCH]}"

answered Oct 21, 2019 at 14:25

chepner

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3 Comments

mickt Over a year ago

It appears that the declaration does not persist across functions. Would you know how I can resolve this?

chepner Over a year ago

In a function, add the -g flag to the declare command to make the array a global variable.

mickt Over a year ago

I moved it outside the function and it worked but I think -g and keeping it within s better. Thanks for your assistance.

anubhava · Accepted Answer · 2019-10-21 14:38:47Z

0

You can use declare directive for this:

declare "COMMIT_ID_${ARCH}"=$(git log --format="%H" -n 1)

To examine new variable use:

declare -p "COMMIT_ID_${ARCH}"

Based on comments below by OP:

# create your variable
var="COMMIT_ID_${ARCH}"

# set variable by calling git log
declare "$var"=$(git log --format="%H" -n 1)

# examine value of $var
echo "${!var}"

edited Oct 21, 2019 at 14:38

answered Oct 21, 2019 at 14:20

anubhava

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6 Comments

mickt Over a year ago

I tried this earlier but it did not return the correct value: echo $COMMIT_ID_${ARCH} returns amd64. I'd prefer not to have to run declare -p each time I want the actual value if possible. I'm not actually echoing it as it is in a script and is used in a sed substitution.

mickt Over a year ago

Clear explanation of declare usage, thank you. Unfortunately though it still leaves me with the problem of not having $ARCH ref in variable name as I will have amd64 & arm64v8 and need to differentiate between them.

anubhava Over a year ago

You will have 2 different variables created COMMIT_ID_amd64 and COMMIT_ID_amd64v8 after above run.

mickt Over a year ago

Would they both not be called var?

anubhava Over a year ago

var is just a temporary holder. Where and how are you going to use these 2 variables?

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