I am trying to do compare a string in bash to a regex pattern and have found something odd. For starters I am using GNU bash, version 5.0.17(1)-release (x86_64-pc-linux-gnu). This is within WSL.
For example here is sample program demonstrating the problem:
#!/bin/env bash
name="John"
if [[ "${name}" =~ "John"* ]]; then
echo "found"
else
echo "not found"
fi
exit
As expected this will echo found since the name "John" matches the regex pattern described. Now what I find odd is if I drop the n in John, it still echos found. Imo "Joh" does match the pattern of "John"*.
If you drop the "hn" and just set $name to "Jo" then it echos not found. It seems to only affect the last character in the Regex pattern (aside from the wildcard).
I am converting an old csh script to bash and this behavior is not happening in csh. What is causing bash to do this?
>Solution :
You’re mixing up syntax for shell patterns and regular expressions. Your regular expression, after stripping the quoting, is John*: Joh followed by any number of n, including 0. Matches Joh, John, Johnn, Johnnn, …
It’s not anchored, so it also matches any string containing one of the matches above.
Since it’s not anchored, depending on what you want, you could do any of these:
- Any string containing
Johnshould match:- Regex:
[[ $name =~ John ]] - Shell pattern:
[[ $name == *John* ]]
- Regex:
- Any string that begins with
Johnshould match:- Regex:
[[ $name =~ ^John ]] - Shell pattern:
[[ $name == John* ]]
- Regex:
Notice that shell patterns, unlike the regular expressions, must match the entire string.
A note on quoting: within [[ ... ]], the left-hand side doesn’t have to be quoted; on the right-hand side, quoted parts are interpreted literally. For regular expressions, it’s a good practice to define it in a separate variable:
re='^John'
if [[ $name =~ $re ]]; then
This avoids a few edge cases with special characters in the regex.