Bash: How to reference a variable that contains another variable in its name?

Question

I have a running variable, let's say it value is INDEX = 1122 for the current iteration. I have exported:

export PRO"$INDEX"=/some/fancy/dirry

Now I want to reference to this variable, but if I do:

echo $"PRO$INDEX"

I get:

PRO1122

Instead I would like to see:

/some/fancy/dirry

Much like the result of echo $PRO1122.

However, I cannot of course hard code the value 1122 in the loop.

anubhava · Accepted Answer · 2014-02-21 15:25:27Z

6

BASH already has a variable reference mechanism.

You can do:

v="PRO$INDEX"
echo "${!v}"
/some/fancy/dirry

answered Feb 21, 2014 at 15:25

anubhava

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Comments

Barmar · Accepted Answer · 2014-02-21 15:20:05Z

3

Use eval:

eval echo '$PRO'$INDEX

But it seems like you should be using an array rather than embedding the index into the variable name.

answered Feb 21, 2014 at 15:20

Barmar

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eval should only be used for shells (like POSIX sh, which also lacks array support) that don't support some form of indirect parameter expansion.