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!

Answer 1

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]}

Answer 2

在bash中，我认为在双引号中保留元字符状态的唯一字符是美元符号（$），反向符号（`）和反斜杠（\\）。