How to grep pattern


 
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# 1  
Old 05-05-2011
How to grep pattern

Hi All,

I need to grep pattern from text file. Also, I need to print above 3 lines from where the pattern starts. For expample, in the below texts, I need to grep "ORA-00600" and also the date, which will tell, on which date, the error belongs to? I hope, I am able to clarify my points..


Wed May 4 05:35:09 2011
Errors in file /pdsdev/ptmdev/oracle/admin/ptmdev/udump/ora_15615_ptmdev.trc:
ORA-00600: internal error code, arguments: [729], [96], [space leak], [], [], [], [], []
# 2  
Old 05-05-2011
With linux or GNU grep
Code:
grep -B 3 'pattern' inputfile

Otherwise try awk
Code:
awk '{arr[NF]=$0 
        if(index($0, "pattern") > 0 ) 
        {
            for(i=NF-3; i<=NF; i++) { print arr[i] }
        }
       } '   inputfile

PS: I think you really want 2 lines above not 3.
# 3  
Old 05-05-2011
Code:
awk 'NF==5{n=0}{A[n++]=$0}/ORA-00600/{for(i=0;i<=n;i++) print A[i]}' alertSID.log

---------- Post updated at 07:34 PM ---------- Previous update was at 07:26 PM ----------

My code will display from the date until (included) the ORA-00600 error (it could be 2 3 or more lines)

oops little fix should be i<n instead of i<=n
Code:
awk 'NF==5{n=0}{A[n++]=$0}/ORA-00600/{for(i=0;i<n;i++) print A[i]}' ora

 
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