shell script , how to find a combination of ! and *

I wrote this script to find a string with only * and ! However, it can not find any string开发者_运维百科 consists only * I couldn't figure out what's wrong with my regular expression, Could someone please help? Thank you!

#!/bin/bash
arr=('!!' '!' '*' '*!' '**' '**!' '!*!' '***' 'bla!' )
echo  star is "${arr[2]}"

for i in ${arr[@]}; do if [[ "$i" == [\!\*]* ]] ; then echo match "$i"; fi; done

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You just need to quote your array in the for statement:

for i in "${arr[@]}"; do if [[ "$i" == [*!]* ]] ; then echo match "$i"; fi; done

If you put the exclamation point second, you don't need to escape it (the asterisk doesn't need escaping in any case).

By the way, that's not a regular expression. It's a shell globbing expression.

---

try

for i in "${arr[@]}"; do if [[ "$i" =~ "^[\!\*]*$" ]] ; then echo match "$i"; fi; done

Edit:(for Dennis)

$ for i in "${arr[@]}"; do if [[ "$i" =~ "^[\!\*]*$" ]] ; then echo match "$i"; fi; done
match !!
match !
match *
match *!
match **
match **!
match !*!
match ***

$ for i in "${arr[@]}"; do if [[ "$i" =~ ^[\!\*]*$ ]] ; then echo match "$i"; fi; done
match !!
match !
match *
match *!
match **
match **!
match !*!
match ***