change the all files which contains xxxxx to yyyyyy

grep -r -l xxxxx . | xargs perl -i -pe "s/xxxxx/yyyyy/g"
This script first find all files which contains word xxxxx recursively. Then replace the word xxxxx to yyyyy of the files. Use case: - Web site domain change - Function name change of the program

By: hassylin
2009-02-06 08:18:50

What Others Think

The perl command is wrong, shoud be perl -i -pe "s/xxxx/yyyy/g" Also, this breaks on files with spaces. echo "This contains xxxxx, see?" > "filename" echo "This also contains xxxxx, see?" > "filename two" cat "filename" This contains xxxxx, see? cat "filename two" This also contains xxxxx, see? grep -r -l xxxxx . | xargs perl -pi "s/xxxxx/yyyyy/g" Can't open perl script "s/xxxxx/yyyyy/g": No such file or directory grep -r -l xxxxx . | xargs perl -i -pe "s/xxxxx/yyyyy/g" Can't open two: No such file or directory, line 1. cat "filename" This contains yyyyy, see? cat "filename two" This also contains xxxxx, see? This works better: grep -r -l xxxxx . | while read f ; do perl -i -pe 's/xxxxx/yyyyy/g' "$f" ; done cat "filename two" This also contains yyyyy, see?
zjuul · 611 weeks and 2 days ago
Thanks. I fixed the typo.
hassylin · 611 weeks and 2 days ago
Here's a find/sed equivalent: find . -exec sed -ir 's/xxxxx/yyyyy/g' {} \;
porges · 611 weeks and 1 day ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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