How to make sed do non-greedy match?

regex bash sed

19,519

Solution 1

You can't do non greedy regex in sed, but you can do something like this instead:

echo "XML-Xerces-2.7.0-0.tar.gz" | sed -e 's/^\(\([^-]\|-[^0-9]\)*\).*/\1/g'

Which will capture everything up until it finds a - followed by [0-9].

You actually don't sed when you're in bash:

shopt -s extglob
V='XML-Xerces-2.7.0-0.tar.gz'
echo "${V%%-+([0-9]).+([0-9])*}"

19,519

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Updated on June 04, 2022

Red Cricket almost 2 years
I cannot seem to figure out how to come up with the correct regex for my bash command line. Here's what I am doing:
```
echo "XML-Xerces-2.7.0-0.tar.gz" | sed -e's/^$.*$-[0-9].*/\1/g'
```
This gives me the output of ...
```
XML-Xerces-2.7.0
```
... but want I need is the output to be ...
```
XML-Xerces
```
... I guess I could do this ...
```
 echo "XML-Xerces-2.7.0-0.tar.gz" | sed -e's/^$.*$-[0-9].*/\1/g' | sed -e's/^$.*$-[0-9].*/\1/g'
```
... but I would like to know how understand sed regex a little better.

Update:

I tried this ...
```
echo "XML-Xerces-2.7.0-0.tar.gz" | sed -e's/^$[^-]*$-[0-9].*/\1/g'
```
... as suggest but that outputs XML-Xerces-2.7.0-0.tar.gz
Red Cricket over 10 years

Ah! that works! Thanks
Paul over 10 years

@RedCricket You're welcome
Red Cricket over 10 years

Pretty slick! But I am not sure I understand how to read ${V%%-+([0-9]).+([0-9])*}. Could you explain that part?
konsolebox over 10 years

@RedCricket It's an extended glob. See here. The feature is not enabled by default and we enable it through shopt -s extglob. The expansion method of the variable deletes the match found in the end of the variable's value. Expansion methods are explained here. The pattern -+([0-9]).+([0-9])* matches -2.7.0-0.tar.gz of XML-Xerces-2.7.0-0.tar.gz and so that part is deleted. In regex it's actually like -[0-9]+\.[0-9]+.*$.