Find files containing a given text

In bash I want to return file name (and the path to the file) for every file of type .php|.html|.js containing the case-insensitive string "document.cookie" | "setcookie"

How would I do that?

egrep -ir --include=*.{php,html,js} "(document.cookie|setcookie)" .

The r flag means to search recursively (search subdirectories). The i flag means case insensitive.

If you just want file names add the l (lowercase L) flag:

egrep -lir --include=*.{php,html,js} "(document.cookie|setcookie)" .

Try something like grep -r -n -i --include="*.html *.php *.js" searchstrinhere .

the -i makes it case insensitlve

the . at the end means you want to start from your current directory, this could be substituted with any directory.

the -r means do this recursively, right down the directory tree

the -n prints the line number for matches.

the --include lets you add file names, extensions. Wildcards accepted

For more info see: http://www.gnu.org/software/grep/

find them and grep for the string:

This will find all files of your 3 types in /starting/path and grep for the regular expression '(document\.cookie|setcookie)'. Split over 2 lines with the backslash just for readability...

find /starting/path -type f -name "*.php" -o -name "*.html" -o -name "*.js" | \
 xargs egrep -i '(document\.cookie|setcookie)'

Sounds like a perfect job for grep or perhaps ack

Or this wonderful construction:

find . -type f \( -name *.php -o -name *.html -o -name *.js \) -exec grep "document.cookie\|setcookie" /dev/null {} \;