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I have the following string /path/to/my-jar-1.0.jar for which I am trying to write a bash regex to pull out my-jar.
Now I believe the following regex would work: ([^\/]*?)-\d but I don't know how to get bash to run it.
The following: echo '/path/to/my-jar-1.0.jar' | grep -Po '([^\/]*?)-\d' captures my-jar-1
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Regular expressions allow us to not just match text but also to extract information for further processing. This is done by defining groups of characters and capturing them using the special parentheses ( and ) metacharacters. Any subpattern inside a pair of parentheses will be captured as a group.
capturing in regexps means indicating that you're interested not only in matching (which is finding strings of characters that match your regular expression), but you're also interested in using specific parts of the matched string later on.
Groups group multiple patterns as a whole, and capturing groups provide extra submatch information when using a regular expression pattern to match against a string. Backreferences refer to a previously captured group in the same regular expression.
In BASH you can do:
s='/path/to/my-jar-1.0.jar'
[[ $s =~ .*/([^/[:digit:]]+)-[[:digit:]] ]] && echo "${BASH_REMATCH[1]}"
my-jar
Here "${BASH_REMATCH[1]}" will print captured group #1 which is expression inside first (...).
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