bash truncate string Works on commandline but not in script

Question

I am trying to cut off all occurrences of a character from the end of a string.

The test script I came up with is:

#!/bin/bash
Input="/tmp/blah/bloh/////////////"

Desired="/tmp/blah/bloh"

cut='/'

result=${Input%%+(${cut})}

echo "          Input: ${Input}"
echo "Expected result: ${Desired}"
echo "         Result: ${result}"
echo "---------------------------------------"
echo -n "        Outcome: "

[ "${Desired}" = "${result}" ] && echo "Success!" || echo "Fail!"

Running this script via bash /tmp/test.sh gives the following output:

          Input: /tmp/blah/bloh/////////////
Expected result: /tmp/blah/bloh
         Result: /tmp/blah/bloh/////////////
---------------------------------------
        Outcome: Fail!

However, if I copy and paste the entire thing in my console I get the expected result of /tmp/blah/blah

What is going on here?

Replace #!/bin/bash with #!/bin/bash -i ;)

Cyrus
– Cyrus

2018-08-15 11:54:49 +00:00
Commented Aug 15, 2018 at 11:54 — Cyrus
– Cyrus, Commented Aug 15, 2018 at 11:54

chepner · Accepted Answer · 2018-08-15 12:38:13Z

3

+(${cut}) is an extended pattern, which bash does not recognize by default. You need to enable the extglob option first.

$ shopt -s extglob
$ Input="/tmp/blah/bloh/////////////"
$ cut='/'
$ echo "${Input%%+(${cut})}"
/tmp/blah/bloh

You probably have extglob enabled in your interactive shell via your .bashrc or .bash_profile configuration file, but neither file is used for the non-interactive shell started by your script.

answered Aug 15, 2018 at 12:38

chepner

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