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Print lines after the search string until blank line is found


 
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# 1  
Print lines after the search string until blank line is found

All I want is to look for the pattern in the file...If I found it at # places... I want print lines after those pattern(line) until I find a blank line.



Code:
Log EXAMPLE :


MT:Exception caught 
The following Numbers were affected:
1234
2345
2346

Error
java.lang.InternalError: Can't connect to X11 
      Exception caught 

The following numbers were affected:
11111
22222
33333
44444
55555

java.lang.InternalError: Can't connect to


This is the command I used



Code:
nawk 'c-->0;$0~s{if(b)for(c=b+1;c>1;c--)print r[(NR-c+1)%b];print;c=a}b{r[NR%b]=$0}' b=0 a=10 s="The following numbers were affected:" Process.log

Above command gives me 10 lines after the search string.

But I just want the 3 and 5 lines in this case. I want to break the print if it hits the blank line.
# 2  
Code:
awk '$0~s{p=1;next}/^$/{p=0}p' s="The following numbers were affected:" Process.log

# 3  
Thanks...

Quote:
Originally Posted by bartus11
Code:
awk '$0~s{p=1;next}/^$/{p=0}p' s="The following numbers were affected:" Process.log

Thanks for your reply ..

I am getggin this error when I run the command ..
Code:
awk: syntax error near line 1
awk: bailing out near line 1

# 4  

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