bash script variable passing to find

Question

I have a simple bash script where I generate some temporary files using split, do some processing and then try to track down all the files at the end and merge them

rand_int=$RANDOM
split -d -l $n_lines_split $1 $rand_int   #works fine

for f in $(find . -amin -200 -regex '.*$rand_int.*' ); do 
    (some processing here) ; 
done

My problem is that in the find command $rand_int is interpreted literally, whereas I want to use the variable's value.

Dan Fego · Accepted Answer · 2011-12-22 19:49:38Z

6

In the shell, single-quotes (') cause what's inside to be interpreted literally. What you want to do is use double-quotes (") around the expression with $rand_int.

So for the find expression:

find . -amin -200 -regex ".*$rand_int.*"

answered Dec 22, 2011 at 19:49

Dan Fego

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1 Comment

lonestar21 Over a year ago

beautiful! Funny, I used to use Perl (same convention), and that hadn't even occurred to me

user237419 · Accepted Answer · 2011-12-22 19:49:13Z

2

use " " instead of ''

for f in $(find . -amin -200 -regex ".*$rand_int.*" ); do

answered Dec 22, 2011 at 19:49

user237419

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