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Shell Programming and Scripting

need to print lines between repeated pattern

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Top Forums Shell Programming and Scripting need to print lines between repeated pattern
# 8  
Old 09-04-2012
Quote:
Originally Posted by Eman_in_forum
Code:
-bash-3.00$ awk '/uid=/{p=0;}/uid=nijss/{p=1;}p'  shortfile
awk: syntax error near line 1
awk: bailing out near line 1

Use nawk or /usr/xpg4/bin/awk on Solaris.
# 9  
Old 09-04-2012
Try something like this..

Code:
Var_a="bessemsj"
awk -v var="$Var_a" '/^uid=/{(s=0);if($2 == var){s=1}}{if(s == 1){ print;next}}' FS='=' file

# 10  
Old 09-04-2012
using

Code:
/usr/xpg4/bin/awk -v var="$Var_a" '/^uid=/{(s=0);if($2 == var){s=1}}{if(s == 1){ print;next}}' FS='=' file

works pretty well !
Many thanks


Moderator's Comments:
Mod Comment Please use code tags next time for your code and data.
Last edited by zaxxon; 09-04-2012 at 02:21 PM.. Reason: code tags
# 11  
Old 09-06-2012
It's working finally Smilie

Code:
************************************************
DIFFERENCE FOUND FOR UID = cremersy :
************************************************
a17
 leg 1 is missing following line(s):
dn: cn=Desk ESSD,ou=Desks,dc=WEO,c=nl,o=Vodafone
-bash-3.00$ cat diffreport


************************************************
DIFFERENCE FOUND FOR UID = wasbauej :
************************************************
8
 leg 1 is missing following line(s):
dn: cn=WEO Advisors, ou=Groups, dc=WEO,c=nl,o=Vodafone


but something annoying me , there is a set of strings that appear above the word 'leg 1', becoz of an extra printing in a while loop

can u tell me please how to remove a line above matched word 'leg 1'
Last edited by Scott; 08-12-2013 at 09:33 AM.. Reason: Code tags
# 12  
Old 09-06-2012
Quote:
Originally Posted by Eman_in_forum
but something annoying me , there is a set of strings that appear above the word 'leg 1', becoz of an extra printing in a while loop

can u tell me please how to remove a line above matched word 'leg 1'
Must be having in the input..

Please check input first.

Script doesn't add any extra thing...Smilie
# 13  
Old 09-06-2012
yes you are correct
i used the command you sent me-which was perfect-inside a loop
i need to remove one line above pattern 'leg1' plz advise
# 14  
Old 09-06-2012
Try with this...Smilie

Code:
/usr/xpg4/bin/awk -v var="$Var_a" '/^uid=/{(s=0);if($2 ~ var){s=1;p=NR}}{if(s == 1){if((p+2) != NR){ print;next}}}' FS='=' file
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