find command from shell scipt (ksh) problem

UNIX for Advanced & Expert Users
1

Hi experts,

I have a simple shell script as follows.

#!/bin/ksh
FIND_STRING="\( -name 'testfile*.Z' -o -name 'DUMMY_*' \) "
find /tmp -type f $FIND_STRING -print

When I run this with ksh -x testscript, I get the following output.

+ FIND_STRING=\( -name 'testfile*.Z' -o -name 'DUMMY_' \)
+ find /tmp -type f \( -name 'testfile.Z' -o -name 'DUMMY_*' \) -print
find: 0652-009 There is a missing conjunction

If I cut and paste the second line of the output into the dollar prompt, it works fine. Donot know what is happening . I had posted a similar error few weeks before, but now I could pinpoint to a 2 liner.

Thanks in advance for your help.

2

The variable FIND_STRING is not being expanded by the shell when the find is run and that's why when you replace it with its contents the find command runs alright.

3

Well, "$FIND_STRING" is being evaluated in that the variable name is being replaced with the variable contents. However, the contents themselves are not being evaluated...that would be double-evaluation, a 15-yard penalty.

You could either use the command 'eval' in front of the 'find' command, or just remove the backslashes when you set the string.

4

Here is the ouput when you do set -x on the script

+ FIND_STRING=$'\\( -name \'testfile*.Z\' -o -name \'DUMMY_*\' \\)'
+ echo '\(' -name $'\'testfile*.Z\'' -o -name $'\'DUMMY_*\'' '\)'
\( -name 'testfile*.Z' -o -name 'DUMMY_*' \)
+ find /tmp -type f '\(' -name $'\'testfile*.Z\'' -o -name $'\'DUMMY_*\'' '\)' -print
find: \(: unknown option

If you change the find command line to

eval find /tmp -type f $FIND_STRING -print

things work as expected.

5

Thanks a lot for your explanation and the eval works fine. I had not used eval before and didn't know it is so vital. I haven't fully digested the solution but will spend some time on it.

Thanks again for your kind help.

6

pfmurphy, just wondered how did you get the output you had posted with set -x? In my system (ksh on aix) the output is same as I posted in the beginning.

7

Basically eval causes a shell to parse a command line twice. The following simple example should help clarify things

name=PATH
echo $name
eval echo $name
eval echo $$name
eval echo \$$name