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Shell variable inside variable

debugcn Published at Dev
10
user2230605

What is wrong in this substitution.

$ m_d_ver=0.2
$ m=mod
$ d=dom
$ echo ${$m_$d_ver}
-bash: ${$m_$d_ver}: bad substitution

Thanks,

Charles Duffy

What you're trying to do is an indirect variable lookup. The syntax for that is ${!namevar}, where namevar is a variable that contains the name you actually want to evaluate. Thus:

mod_dom_ver=0.2
m=mod
d=dom
var=${m}_${d}_ver
echo "${!var}"

See BashFAQ #006.

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