sed can't match 0 or 1 time

Question

I'm writing a bash script on Cent OS7. Now I need to use sed to remove all lines which don't contain .jpg or .jpeg.

Here is my script:

sed -i -e '/\.jp(e)?g/!d' myfile

But it will delete all lines, which means that it doesn't work as expected.

However, if I do sed -i -e '/\.jpg/!d' myfile or sed -i -e '/\.jpeg/!d' myfile. Both of them work well.

Answer 1

Captured group (()) and the quantifier ? (match the preceding token 0 or 1 time) comes (at least) with ERE (Extended RegEx), not BRE (Basic RegEx).

sed by default uses BRE, so the tokens are being treated literally.

To enable ERE, use -E (or -r if available) with sed:

sed -E '/\.jp(e)?g/!d' myfile

Capturing e is redundant here:

sed -E '/\.jpe?g/!d' myfile

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Note that, you can use ERE tokens from BRE by escaping them with \, so the following would work too:

sed '/\.jp\(e\)\?g/!d' myfile
sed '/\.jpe\?g/!d' myfile

Again, this does not look as clean as just using one option i.e. -E. Only case where you will want this is portability.