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ksh : need to get the 4 th line above and 2 nd below the matched pattern in the log file

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Top Forums Shell Programming and Scripting ksh : need to get the 4 th line above and 2 nd below the matched pattern in the log file
# 8  
Old 10-09-2013
Quote:
Originally Posted by shamrock
Does it have to be awk as ex can do it all...
Code:
ex -s +"/Jobnumber '23_369468343464564' was successful/-4 p | /Jobnumber '23_369468343464564' was successful/+2 p | q!" file


I really liked this ed-based approach, as I often find myself looking for an N-th line above/below a given line in logs and I usually write an ugly awk which remembers lines before/after.

I tried it and somehow it did not work please help me to find what do I do wrong here:
Code:
$ cat xyz                    
line 1
line to show
line 2
line 3
line key
line 4
line 5
$ ed -s +"/key/-3 p | q!" xyz
ed: illegal commands -- /
usage: ed [- | -s] [-p string] [-x] [-C] [file]

sorry for offtop
# 9  
Old 10-09-2013
Lightbulb
Quote:
Originally Posted by Akshay Hegde
can you post your log file ? Its difficult for me understand pattern with 9 line
Here is the log with more lines
Code:

012/01/21 10:29:02  (111111)  Processing Job '23_369468343464564'
     2012/01/21 10:29:02  (111111)
     Making Job '23_369468343464564.0'...
     2012/01/21 10:29:04 (111111)
     Jobnumber '23_369468343464564' was successful 
     2012/01/21 10:29:04  (111111) 
Job has 123 Docs 234 Pgs 456 Lines 4 Errors
     2012/01/21 10:29:04  (111111)
     Total in Job: 123 documents using 29 KBytes
012/01/21 10:29:02  (111111)  Processing Job '34_794862826912345'
     2012/01/21 10:29:02  (111111)
     Making Job '34_794862826912345.0'...
     2012/01/21 10:29:04 (111111)
     Jobnumber '34_794862826912345' was successful 
     2012/01/21 10:29:04  (111111) 
Job has 123 Docs 234 Pgs 456 Lines 4 Errors
     2012/01/21 10:29:04  (111111)
     Total in Job: 123 documents using 29 KBytes
012/01/21 10:29:02  (111111)  Processing Job '98_973986746290898'
     2012/01/21 10:29:02  (111111)
     Making Job '98_973986746290898.0'...
     2012/01/21 10:29:04 (111111)
     Jobnumber '98_973986746290898' was successful 
     2012/01/21 10:29:04  (111111) 
Job has 123 Docs 234 Pgs 456 Lines 4 Errors
     2012/01/21 10:29:04  (111111)
     Total in Job: 123 documents using 29 KBytes

# 10  
Old 10-09-2013
try

Code:
$ awk '{j=e=""}/Processing/{j=$0}/^Job.*Errors$/{e=$0}(j){print j}(e){print e}' log

Code:
012/01/21 10:29:02  (111111)  Processing Job '23_369468343464564'
Job has 123 Docs 234 Pgs 456 Lines 4 Errors
012/01/21 10:29:02  (111111)  Processing Job '34_794862826912345'
Job has 123 Docs 234 Pgs 456 Lines 4 Errors
012/01/21 10:29:02  (111111)  Processing Job '98_973986746290898'
Job has 123 Docs 234 Pgs 456 Lines 4 Errors

# 11  
Old 10-10-2013
Quote:
Originally Posted by migurus
I really liked this ed-based approach, as I often find myself looking for an N-th line above/below a given line in logs and I usually write an ugly awk which remembers lines before/after.

I tried it and somehow it did not work please help me to find what do I do wrong here:
Code:
$ cat xyz                    
line 1
line to show
line 2
line 3
line key
line 4
line 5
$ ed -s +"/key/-3 p | q!" xyz
ed: illegal commands -- /
usage: ed [- | -s] [-p string] [-x] [-C] [file]

sorry for offtop
I posted an ex one liner...which is not doable in ed as it is far more primitive than ex...but you can do the same thing in ed using the here document facility...
Code:
ed -s file <<eof
/key/-3 p
q
eof

Hope this helps...
This User Gave Thanks to shamrock For This Post:
migurus
# 12  
Old 10-10-2013
I guess the requestor's request has not yet been satisfied. You may want to try
Code:
awk     '                               {Arr[NR%5]=$0}
         /34_794862826912345.*succ/     {print Arr[(NR+1)%5]; L=NR+2}
         NR==L
        ' file
012/01/21 10:29:02  (111111)  Processing Job '34_794862826912345'
Job has 123 Docs 234 Pgs 456 Lines 4 Errors

Last edited by RudiC; 10-10-2013 at 01:39 PM.. Reason: pretty print
This User Gave Thanks to RudiC For This Post:
rpm120
# 13  
Old 10-10-2013
Quote:
Originally Posted by RudiC
I guess the requestor's request has not yet been satisfied. You may want to try
Code:
awk     '                               {Arr[NR%5]=$0}
         /34_794862826912345.*succ/     {print Arr[(NR+1)%5]; L=NR+2}
         NR==L
        ' file
012/01/21 10:29:02  (111111)  Processing Job '34_794862826912345'
Job has 123 Docs 234 Pgs 456 Lines 4 Errors


worked perfectly...Smilie
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