Bash Script - Exclamation Point within Variable Reference

Question

I'm looking over a script (which has been used successfully in the past) which contains the following:

node=1
while :
do
    userKey=WEB_${node}_USER
    userVal=`echo ${!userKey}`

I have not been able to figure out why an exclamation point would be added to a variable reference like this. What purpose does "!" serve in this context?

It's rare for me to do much scripting so if I am missing any details please let me know and I will try to provide more information. I have not been able to find this answer elsewhere.

Thanks in advance!

possible duplicated [please see](http://stackoverflow.com/questions/8515411/bash-indirect-expansion-please-explain) — bachN, Aug 01 '14 at 15:41
I much more wonder what purpose does `echo` here serve for. If you are rewriting the script, consider replacing that line it with `userVal="${!userKey}"`. — Dmitry Alexandrov, Aug 01 '14 at 16:11

chepner · Accepted Answer · 2014-08-01T16:27:36.303

It's called indirect parameter expansion. Where $userKey expands to the value of the variable userKey, ${!userKey} expands to the value of the variable whose name is the value of userKey. Since usrKey has the value WEB_1_USER (given the current value of $node, ${!userKey} expands to the same result as $WEB_1_USER.

Its use is somewhat rare, since in many cases (including, it appears, here) an array WEB_USER of user names would be clearer than a set of numbered variables.

WEB_USER=(alice bob charlie)

node=1
while :
do
  userVal=${WEB_USER[node]}

score -1 · Answer 2 · edited Aug 01 '14 at 18:08

-1

In bash, I thought that the only characters that retained their meta-character status inside double quotes were the dollar sign ($), the back-tick (`) and the backslash (\).

edited Aug 01 '14 at 18:08

rici

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answered Aug 01 '14 at 16:32

Sagar.P.Waghmare

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That's true but (a) irrelevant and (b) not an answer. (Irrelevant because the ! is part of a ${...} expansion, and the $'s metacharacter status is not in doubt.) – rici Aug 01 '14 at 18:09

Bash Script - Exclamation Point within Variable Reference

2 Answers2