# variable inside a variable **URL:** \ **Category:** Shell Programming and Scripting\ **Created:** [November 2, 2011, 7:44am UTC](https://community.unix.com/t/variable-inside-a-variable/298440 "2011-11-02T07:44:37Z")\ **Posts on this page:** 3\ **Page:** 1
**Author:** ![anilcliff](https://community.unix.com/letter_avatar/anilcliff/32/5_5575768a8748004e209b776fc1b2916d.png) [@anilcliff](https://community.unix.com/u/anilcliff)\ **Post date:** [November 2, 2011, 7:44am UTC](https://community.unix.com/t/variable-inside-a-variable/298440/1 "2011-11-02T07:44:37Z")
I am trying manipulate variables inside a variable ```plaintext root@server [~]# hour=`date | awk {'print $4'} | cut -d: -f 1`;lasthour=`expr $hour - 01`; lasthour=`printf %02d $lasthour`;grep "`date -I` $lasthour" /var/log/exim_mainlog |egrep -o 'dovecot_login[^]+' | sort|uniq -c|sort -nk 1 8 dovecot_login:user1@domain.com 4 dovecot_login:user2@domain.com ``` The above prints output in two lines. When I take that to a variable and analze each line, it becomes 4 lines. What I need is to check the count of each entry and output if count is more than a limit, say 5. So the below script should only display "8 [dovecot\_login:user1@domain.com](mailto:dovecot_login:user1@domain.com)" ```plaintext root@server [~]# for i in $(hour=`date | awk {'print $4'} | cut -d: -f 1`;lasthour=`expr $hour - 01`; lasthour=`printf %02d $lasthour`;grep "`date -I` $lasthour" /var/log/exim_mainlog |egrep -o 'dovecot_login[^]+' | sort|uniq -c|sort -nk 1); do echo $i; done 8 dovecot_login:user1@domain.com 4 dovecot_login:user2@domain.com ``` ---
**Author:** ![vgersh99](https://community.unix.com/user_avatar/community.unix.com/vgersh99/32/14851_2.png) [@vgersh99](https://community.unix.com/u/vgersh99)\ **Post date:** [November 2, 2011, 8:01am UTC](https://community.unix.com/t/variable-inside-a-variable/298440/2 "2011-11-02T08:01:35Z")
I believe you're over-complicating the task with the use of so many tools - there could be a more graceful way to do all of it. Please provide a sample of your /var/log/exim\_mainlog file using [code tags](http://www.unix.com/misc.php?do=bbcode&#code) and a desired output with the explanation. Also what happens for the mid-night hour - hour 0? ---
**Author:** ![anilcliff](https://community.unix.com/letter_avatar/anilcliff/32/5_5575768a8748004e209b776fc1b2916d.png) [@anilcliff](https://community.unix.com/u/anilcliff)\ **Post date:** [November 2, 2011, 10:18am UTC](https://community.unix.com/t/variable-inside-a-variable/298440/3 "2011-11-02T10:18:15Z")
I got it corrected. If the count is more than 3, it will display it. ```nohighlight hour=`date | awk {'print $4'} | cut -d: -f 1`;lasthour=`expr $hour - 01`; lasthour=`printf %02d $lasthour`;grep "`date -I` $lasthour" /var/log/exim_mainlog |egrep -o 'dovecot_login[^]+' | sort|uniq -c|sort -nk 1| while read login; do num=`echo $login | awk {'print $1'}`; if [$num -gt 3]; then echo $login;fi ; done 4 dovecot_login:user1@domain.net 4 dovecot_login:user2@domain.com 11 dovecot_login:user3@domain.com 27 dovecot_login:user4@domain.ca ``` ---------- Post updated at 07:48 PM ---------- Previous update was at 07:36 PM ---------- I have corrected the midnight 00 issue. Here is the complete script ```nohighlight #!/bin/sh hour=`date | awk {'print $4'} | cut -d: -f 1` if [$hour= '00']; then hour = 24 fi lasthour=`expr $hour - 01` lasthour=`printf %02d $lasthour` grep "`date -I` $lasthour" /var/log/exim_mainlog |egrep -o 'dovecot_login[^]+' | sort|uniq -c|sort -nk 1| while read login do num=`echo $login | awk {'print $1'}` if [$num -gt 3]; then echo $login fi done ```