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Full Discussion: looking for help on script to capture file permission.
Top Forums Shell Programming and Scripting looking for help on script to capture file permission. Post 302471786 by anurag.singh on Monday 15th of November 2010 06:11:09 AM
Old 11-15-2010
Code:
 
#!/bin/ksh
find /opt/ora10g/oracle/ -name "*" | while read file
do
       if [ "$file != "." -a -f $file ]
             echo $file >> temp.txt
             ls -l $file | awk '{ print $1 }' | cut -c 2-10 | sed -e 's/[rwx]/1,/g' -e 's/-/0,/g' | awk '{ split($1, arr, ","); a=4*arr[1] + 2*arr[2] + 1*arr[3]; b=4*arr[4] + 2*arr[5] + 1*arr[6]; c=4*arr[7] + 2*arr[8] + 1*arr[9]; printf("chmod %d%d%d \n",a,b,c); }' >> temp1.txt
       fi
done
paste -d" " temp1.txt temp.txt >> restore_prem.ksh
rm -f temp.txt
rm -f temp1.txt

 

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SYNOPSIS

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DESCRIPTION

       This program enables you to generate a bug report and send it to Oracle Corporation. It is a shell script and runs on Unix.

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       a bug report by using the mysqlbug script.

       mysqlbug helps you generate a report by determining much of the following information automatically, but if something important is missing,
       please include it with your message.  mysqlbug can be found in the scripts directory (source distribution) and in the bin directory under
       your MySQL installation directory (binary distribution).

       Invoke mysqlbug without arguments:

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       The script will place you in an editor with a copy of the report to be sent. Edit the lines near the beginning that indicate the nature of
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