find command in shell script

Shell Programming and Scripting
1

Hi,
dirs.conf fine contains below data

/a/b/c/dir1|50
/a/b/c/dir2|50
/a/b/c/dir3|50

In a shell script I do as below

while read file_rec
do
        dir_name=`echo "${file_rec}" | cut -d "|" -f 1`
        purge_days=`echo "${file_rec}" | cut -d "|" -f 2`

        if [ -d $dir_name ]
        then
        echo "#### The purge days for the directory ${dir_name} is ${purge_days} ####" >> $log
        # To list files with the format YY-MM-DD
	find ${dir_name} -type f -name \"*[1-2][09][0-9][0-9]-[01][0-9]-[0-3][0-9]*\" -mtime +${purge_days} -exec ls -ltr {} \\; >> $log
        else
        echo "*** The directory $dir_name does not exists ***" >> $log
        fi
        dir_name=""
        purge_days=""
done < dirs.conf

My problem is, the find command is not running and listing the old files in log file

Can someone help...

Regards,

2

You dont need to call external commands to split strings. Either use parameter expansion:

dir_name=${file_rec%|*}
purge_days=${file_rec#*|}

Or use multiple variables in your read command:

IFS='|' read dir_name purge_days

Quote your variable references:

  if [ -d "$dir_name" ]

Don't escape the quotes or the backslash:

find "$dir_name" -type f -name "*[1-2][09][0-9][0-9]-[01][0-9]-[0-3][0-9]*" \
                                -mtime +"$purge_days" -exec ls -ltr {} \; >> $log

The -t and -r options to ls will not do anything as you are giving ls a single filename

3

Two things: a) if you run this with set -x does it appear to interpolate all of your variables correctly?; and, b) if it does, and you take this same interpolated find query and run it manually, does it work then?

I also notice the extra ;-char at the end of the find query. Why do you need to include this? The find shouldn't need additional prompting.

4

Thanks Johnson the double quote substitution worked.