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Full Discussion: find command in shell script
Top Forums Shell Programming and Scripting find command in shell script Post 302422700 by cfajohnson on Wednesday 19th of May 2010 07:11:06 AM
Old 05-19-2010
Quote:
Originally Posted by icefish
Hi,
dirs.conf fine contains below data
Code:
/a/b/c/dir1|50
/a/b/c/dir2|50
/a/b/c/dir3|50

In a shell script I do as below
Code:
while read file_rec
do
        dir_name=`echo "${file_rec}" | cut -d "|" -f 1`
        purge_days=`echo "${file_rec}" | cut -d "|" -f 2`


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

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

Or use multiple variables in your read command:
Code:
IFS='|' read dir_name purge_days

Quote:
Code:
  if [ -d $dir_name ]


Quote your variable references:
Code:
  if [ -d "$dir_name" ]

Quote:
Code:
  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


Don't escape the quotes or the backslash:
Code:
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
Quote:
Code:
  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
 

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