Bash script, trying pass variables to "find"

theCzar

Hi,

I'm looking for log files that may or may not exist, and I don't know exact file names besides. I know they have a 3 digit identifier ($seq) followed by an '_ 'so I'm looking to write a script that will read through my 30+ directorys (5000+ log files) and do a bit of cleaning up.

I'm trying to use find, as I would normally, but having it search in a directory defined by a variable for files defined by variables, return the value to form an array $LISTS(). (directory names are in an array $JOB[], referenced with variable $x)

LISTS=(`find ./${JOB[${x}]}/ -name \'*${seq}_*\'`)

It doesn't work.

If I use constants instead of variables, it does. I've echoed the variables and the values are good but "find" seems to try and evaluate the command before the variables are converted to real values. I tried \ escaping the quotes.

Send "you should just use perl" > /dev/null

AndrewMc

theCzar wrote: »

LISTS=(`find ./${JOB[${x}]}/ -name \'*${seq}_*\'`)

I'd have to see more of the script to see how you're manipulating $x (are you running this in a loop, updating $x each time?). However, it looks like ${seq} is inside single quotes — as far as I can remember this means no variables inside will be expanded. Can you try " instead, or just leave the quotes out altogether?

Khannie

AndrewMc wrote: »

Can you try " instead

That's your answer. I think that's safer than not using quotes because that can result in unwanted shell substitution.

theCzar

Thanks for the answer, the double quotes work.

Sort of annoyed I didn't think to try that before coming on here!

pickarooney

Don't be. Think of all the people who will find the same answer now that you posted the question.