Question

bash guru 😉 I’m trying to improve some string in bash which grep specific keyword’s matches in specific files. It looks like that:

find /<path>/hp -iname '*.ppd' -print0 | xargs -0 grep "\*ModelName\:"

which works very fast for me! In 20 times faster than this one:

find /<path>/hp -iname '*.ppd' -print0 | xargs -0 -I {} bash -c 'grep "\*ModelName\:" {}'

But the problem is that in the first script I’m getting the following lines:

/<path>/hp/hp-laserjet_m9040_mfp-ps.ppd:*ModelName: "HP LaserJet M9040 M9050 MFP"

but desired result is just

*ModelName: "HP LaserJet M9040 M9050 MFP"  

(as in the second script). How can I achieve it?

P.S.: I’m using find for flexibility and future improvements of the script.

Answer 1

The -h option to grep suppress filenames from the output.

find /<path>/hp -iname '*.ppd' -print0 | xargs -0 grep -h "\*ModelName\:"

If your grep does not provide -h the use cat:

find /<path>/hp -iname '*.ppd' -print0 | xargs -0 cat | grep "\*ModelName\:"

Also, for your information, find provides the -exec option which would render xargs unnecessary had you wanted to pursue your second option:

find /<path>/hp -iname '*.ppd' -exec grep grep "\*ModelName\:" '{}' \;