|
|
|
google unix.com
|
|||||||
| Forums | Register | Forum Rules | Links | Albums | FAQ | Members List | Calendar | Search | Today's Posts | Mark Forums Read |
| Shell Programming and Scripting Post questions about KSH, CSH, SH, BASH, PERL, PHP, SED, AWK and OTHER shell scripts and shell scripting languages here. |
More UNIX and Linux Forum Topics You Might Find Helpful
|
||||
| Thread | Thread Starter | Forum | Replies | Last Post |
| Network Solution | disturbe_d | IP Networking | 2 | 11-04-2008 03:42 PM |
| Solution accepted | sudhamacs | Post Here to Contact Site Administrators and Moderators | 1 | 08-19-2008 10:44 AM |
| need the solution | paniruddha | Shell Programming and Scripting | 3 | 07-09-2008 05:39 PM |
| Is there a awk solution for this?? | timj123 | Shell Programming and Scripting | 7 | 03-14-2008 06:28 AM |
| Is There a Sed Solution for This? | racbern | Shell Programming and Scripting | 1 | 03-13-2008 11:31 AM |
Powered by 
|
|
|
LinkBack | Thread Tools | Search this Thread | Rate Thread | Display Modes |
11-20-2008
|
||||
|
||||
|
Need good solution
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. Code:
#!/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.
|
| Bookmarks |
| Thread Tools | Search this Thread |
| Display Modes | Rate This Thread |
|
|
Threaded Mode
