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FAST Search and Replace for Strings in all Files in Directory

Terminal - FAST Search and Replace for Strings in all Files in Directory
sh -c 'S=askapache R=htaccess; find . -mount -type f|xargs -P5 -iFF grep -l -m1 "$S" FF|xargs -P5 -iFF sed -i -e "s%${S}%${R}%g" FF'
2009-10-02 05:03:10
User: AskApache
Functions: find grep sed sh xargs
9
FAST Search and Replace for Strings in all Files in Directory

I needed a way to search all files in a web directory that contained a certain string, and replace that string with another string. In the example, I am searching for "askapache" and replacing that string with "htaccess". I wanted this to happen as a cron job, and it was important that this happened as fast as possible while at the same time not hogging the CPU since the machine is a server.

So this script uses the nice command to run the sh shell with the command, which makes the whole thing run with priority 19, meaning it won't hog CPU processing. And the -P5 option to the xargs command means it will run 5 separate grep and sed processes simultaneously, so this is much much faster than running a single grep or sed. You may want to do -P0 which is unlimited if you aren't worried about too many processes or if you don't have to deal with process killers in the bg.

Also, the -m1 command to grep means stop grepping this file for matches after the first match, which also saves time.

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