[Solved] Sed/awk print between patterns the first occurrence

Shell Programming and Scripting
1

Guys,

I am trying the following:

i have a log file of a webbap which logs in the following pattern:

2011-08-14 21:10:04,535 blablabla ERROR blablabla
       bla
       bla
      bla
      bla
2011-08-14 21:10:04,535 blablabla ERROR blablabla
       bla
       bla
      bla
      bla
2011-08-14 21:10:04,535 blablabla ERROR blablabla
       bla
       bla
      bla
      bla

what i need is to print everything between the ERROR pattern but only the first occurrence :

2011-08-14 21:10:04,535 blablabla ERROR blablabla
       bla
       bla
      bla
      bla

Please help!

2

the example you made is not so clear. timestamps are the same, also those "bla"s.

try this one liner, if it gives your needs:

awk  '/ERROR/{if(b) exit; else b=1}1'  yourFile
3

thanks a lot, that's exactly what i needed!

4

Alternate solution..

awk '/ERROR/{++t}t==1' inputfile
5

this short one-liner looks nice. however it will go through the whole file anyway after getting the part of text we needed.

6

Hi,

Using 'sed':

$ cat infile
2011-08-14 21:10:04,535 blablabla ERROR Exception1 blablabla
bla
bla
bla
bla
2011-08-14 21:10:04,535 blablabla ERROR Exteption2 blablabla
bla
bla
bla
bla
2011-08-14 21:10:04,535 blablabla ERROR Exception1 blablabla
bla
bla
bla
bla
2011-08-14 21:10:04,535 blablabla ERROR Exception3 blablabla
bla
bla
bla
$ sed -ne '/ERROR/I,/ERROR/I { H; b }; x; s/^\n//; s/\(.*\)\n.*/\1/; p; q' infile
2011-08-14 21:10:04,535 blablabla ERROR Exception1 blablabla
bla
bla
bla
bla

Regards,
Birei

7

Thanks!

---------- Post updated at 05:19 PM ---------- Previous update was at 05:12 PM ----------

The above doesnt work as expected. The Exception1 is just an example. This command will run as a script and the pattern is a variable. What i need is to separate the errors and store an example of it in a file. The errors are different so i have to use the description of the error (i.e. Exception1) as a pattern.