Hi all

I have a script that run fine ,Actually if find 777 directory and take its count and report,There is no problem with the script.But our reporting system have some limitation that dont allow more then 1000 directory to report,Now i want some way i can break this up and then report to the reporting system,My code is below,At the moment there are like 5000 777 directories.Please help i am blanked.

#!/bin/bash

check=/var/www/html


res=$(find $check -type d -perm 777 2>/dev/null )
count=$(find $check -type d -perm 777 | wc -l)

echo $count
#echo $res
                   
Reporting system command.

#! /bin/bash

TD=$(mktemp -d /tmp/xxxxxxxxx)
ALL="${TD}/allfiles"
REP=<Reporting system command>
trap "rm -Rf ${TD}" 0 1 2 3 9 13 15

cd ${TD}
find /etc -type f > ${ALL}
split -l 100 ${ALL} "prfx-"
find . -name "prfx-*" -exec ${REP} {} \;

exit 0

HI

I think this will work,Its good if you explain how and what is going on

#! /bin/bash

TD=$(mktemp -d /tmp/xxxxxxxxx)         # create a temp-directory
ALL="${TD}/allfiles"                               # a filename for the output of find
REP=<Reporting system command>    # your command
trap "rm -Rf ${TD}" 0 1 2 3 9 13 15       # erase the temp directory at signal 0,1 etc.

cd ${TD}
find /etc -type f > ${ALL}                        # find and put all output in one file
split -l 100 ${ALL} "prfx-"                       # split this file into smaller files with 100 
                                                              # lines each (filenames start with "prfx-")
find . -name "prfx-*" -exec ${REP} {} \;  # find the "prfx-"-files and call your 
                                                              # command with them (I don't know
                                                              # how your command works)

exit 0

HI

This is not working Dear,Can anyone post simpler and nicer solution.