Creating Variable Name with Variable

Question

I have something like this:

$CORSAIR_TYPE="good"
$CASE="CORSAIR"
...
$local temp_type=${CASE}_TYPE
echo ${!temp_type}

This gives me the value of the variable CORSAIR_TYPE, which is "good". Is there a way of doing this without creating the temp_type variable? What would I replace echo ${!temp_type} with?

Thanks

With `${!foo}`, `foo` has to be a variable. So, no, at least with the code you've given, not really. Also, there is no situation where I'd go with any `eval` solution just to avoid creating a single temporary variable. — jedwards, Mar 26 '13 at 15:09

score 0 · Answer 1 · answered Mar 26 '13 at 13:33

0

For anyone else stumbling across this post, this is one way I found of doing it:

echo $(eval echo $`echo ${CASE}_TYPE`)

answered Mar 26 '13 at 13:33

halexh

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All you need is `eval 'echo "$'${CASE}'_TYPE"'` – William Mar 26 '13 at 13:53

Creating Variable Name with Variable

1 Answers1