I am trying to match using a glob pattern. But it is failing when using
dgt='^+()$' [[ "$1" == $dgt ]] && echo "SUCCESS" || echo "FAILURE"
Unix & Linux Stack Exchange is a question and answer site for users of Linux, FreeBSD and other Un*x-like operating systems. It only takes a minute to sign up.Sign up to join this community
^+()$, is a mix of an extended globbing pattern and a regular expression. A globbing pattern does not need to be anchored explicitly, as it is always anchored anyway. Therefore, a globbing pattern starting with
^ and ending with
$ would match those literal characters at the start and end of a string. If you want to use a globbing pattern and don't want to match
^ at the start and
$ at the end, remove these.
You will end up with the following code:
#!/bin/bash # Bash releases earlier than 4.1 needs to enable the extglob shell # option. For release 4.1+, the pattern used in [[ ]] is assumed # to be an extended globbing pattern. # # shopt -s extglob pattern='+()' if [[ $1 == $pattern ]]; then echo 'contains only digits' else echo 'contains non-digit or is empty' fi
In a shell with no extended globbing patterns, it's easier to match non-digits:
#!/bin/sh case $1 in *[!0123456789]*) echo 'contains non-digit' ;; '') echo 'is empty' ;; *) echo 'contains only digits' esac
bash shell, you can use the above code too, as it portable and would work in all
sh-compatible shells, or you could use
#!/bin/bash pattern='*[!0123456789]*' if [[ $1 == $pattern ]]; then echo 'contains non-digit' elif [ -z "$1" ]; then echo 'is empty' else echo 'contains only digits' fi