Find command - Performance

I'm trying write a script to find certain files in a Unix system. I have 3 large directories called /Archive1 /Archive2 and Archive3. I have to find a file in each directory. Here is my script:

#!/bin/bash


tt=/usr/tmp/zznineteenzz
arch=/usr/tmp/foo.tomk
dirs="/Archives /Archives2 /Archives3"

for i in `cat $tt`; do

for dd in $dirs; do

for ff in `.dosu find $dd -name $i*Z -print`; do
if [ -f $arch ]
then
echo $ff >> $arch
fi
done
done
done
====

I'm worried about performance issues. Is there a better way? Thanks everyone!

This way you read thru each directory only once with find.

Not totally sure, but I think you can do find /dir1 /dir2 /dir3 -type f ... etc., so do all 3 dirs at once - man find will ngive a yea/nay.

Some find implementations may do that. You mentioned "a UNIX system", this will work anywhere.

Do not confuse the number of lines in a script with performance. The critical step here is just reading thru a directory and then grepping against a list of files contained in /usr/tmp/zznineteenzz

A little quicker (just one command not in an external loop, o shaving off those all important microseconds! ) and also easier/less to type ...

Try using find and xargs.....

find /path/to/ <anyother find options> | xargs grep 'search value'

ignore this I didn't read it right...thought you were looking to find something in a file.

My bad!

Since we're beating performance to death here... POSIX find supports a + operator
which improves performance of the -exec clause by having the subprocess command work on a large set of data:

So, this is another way to approach the problem.