Bash – List all files containing a specific string

bash

I need to recursively look for all files containing a specific string so that I can copy a new file over the old one. E.g.:

If a file has the string "replace me":

find the full path of the files:

/dir/of/file/filename.php
/dir/of/different/file/filename.php

replace the file with my new filename.php file.

It sounds simple, but I have wasted several hours on it getting nowhere.

Best Answer

find some/dir \
    -exec grep -q "replace me" {} \; \
    -exec cp some/new/filename.php {} \;

Related Solutions

Bash – Replace very long string in files recursively

I assume that the string can contain any character except newlines and null bytes. You can quote the string for use as a sed pattern. The characters $*./[\^ need to be preceded by a backslash. In the replacement text, you need to quote the characters \&/.

regexp=$(printf %s "$old" | sed 's:[$*./\[^]:\\&:g')
replacement=$(printf %s "$new" | sed 's:[\&/]:\\&:g')
sed -e "s/$regexp/$replacement/g"

If you have Perl available, it's simpler.

export old new
perl -pe 's/\Q$ENV{old}/$ENV{new}/'

To act on all the files in the current directory and its subdirectories recursively:

regexp=$(printf %s "$old" | sed 's:[$*./\[^]:\\&:g')
replacement=$(printf %s "$new" | sed 's:[\&/]:\\&:g')
export regexp replacement
find . -type f -exec sh -c 'for x; do sed -e "s/$regexp/$replacement/g" <"$x" >"$x.new" && mv "$x.new" "$x"; done' _ {} +

export old new
find . -type f -exec perl -i -pe 's/\Q$ENV{old}/$ENV{new}/' {} +

Best Answer

Related Solutions

Bash – Replace very long string in files recursively

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