Re: Another fgrep query

From: Tapani Tarvainen (tt+gn20030730T130537_at_it.jyu.fi)
Date: 07/30/03 

Date: 30 Jul 2003 14:43:41 +0300

"Dave Hedgehog" <x@x.com> writes:

> I have a directory which contains loads (probably thousands) of different
> files.
> I want to search for a string but when I try an fgrep it bombs out because
> the 'arg list is too long'.
> I know roughly what date the data I am looking for was saved so is there a
> way I can limit my fgrep to only search files which were saved on a certain
> day.
> I'm sure there probably is but I don't have the knowledge to do it.

First, you can avoid the 'arg list too long' problem with xargs:

ls | xargs fgrep 'string' /dev/null

That will fail if there are spaces in filenames, though.
Alternatively you can do

ls | while read f; do fgrep 'string' /dev/null "$f" ;done

That can handle spaces in filenames (not newlines though),
but it is very slow. The /dev/null is needed to force
printing the filename (not needed if you use -l option).

Also, both of those will check all files and thus waste time
compared to selecting files by timestamp first.

Probably easiest way to select files by age is find.
E.g., assuming you want to search only files that are
older than 7 days but younger than 14 days, you can do

find . \( -type f -mtime +14 -print \) -o ! -name . -prune |
 xargs fgrep 'string' /dev/null

If there are no subdirectories you can simplify that:

find . -type f -mtime +7 -mtime -14 | xargs fgrep 'string' /dev/null

The "-type f" is just to prevent grepping ".".

These also fail if there are spaces in filenames; if that
is possible, try

find . -type f -mtime +7 -mtime -14 -exec fgrep 'string' /dev/null {} \;

slower though that is, or if you have a Gnu system,

find . -type f -mtime +7 -mtime -14 -print0 |
 xargs -0 fgrep 'string' /dev/null 

-- 
Tapani Tarvainen

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