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Find a string and place two blank lines


 
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# 1  
Old 09-15-2009
Find a string and place two blank lines

Hi friends,

I am looking for a line to find a particular string in my file and once found then replace with 2-3 blank lines before the string

Example:
Code:
 
aaa 11 bbb
1
2
3
 
aaa 22 bbb
4
5
6

Output
Code:
 
 
aaa 11 bbb
1
2
3
 
 
 
aaa 22 bbb
4
5
6

Thanks in advance
# 2  
Old 09-15-2009
Is this what you are after?

Code:
$ cat file
aaa 11 bbb
1
2
3

aaa 22 bbb
4
5
6

aaa 33 bbb
1
2
3

Code:
$ awk -v RS="" -v ORS="\n\n" '/aaa 22 bbb/{print "\n\n" $0;next}{print}' file
aaa 11 bbb
1
2
3



aaa 22 bbb
4
5
6

aaa 33 bbb
1
2
3

# 3  
Old 09-15-2009
I am getting below error:
Code:
omcadmin>awk -v RS="" -v ORS="\n\n" '/aaa 22 bbb/{print "\n\n" $0;next}{print}' file
awk: syntax error near line 1
awk: bailing out near line 1



---------- Post updated at 05:59 PM ---------- Previous update was at 05:55 PM ----------

Sorry that worked..
But i need for all lines starting with aaa and ending with bbb
I tried this but not working
Code:
nawk -v RS="" -v ORS="\n\n" '/^aaa.*bbb$/{print "\n\n" $0;next}{print}' file

# 4  
Old 09-15-2009
Tested with GNUawk, original-awk and mawk. All successfully.

Try:
Code:
awk 'BEGIN{RS="";ORS="\n\n"} /aaa 22 bbb/{print "\n\n" $0;next}{print}' file



---------- Post updated at 02:37 PM ---------- Previous update was at 02:33 PM ----------
Quote:
Originally Posted by shaliniyadav
But i need for all lines starting with aaa and ending with bbb
Ok. Then, simply try

Code:
awk '/^aaa.+bbb$/{print "\n\n" $0;next}{print}' file

# 5  
Old 09-15-2009
Thanks a LOT Smilie

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