Shell Programming and Scripting

Hi
I need to parse a file in the following format

Code:

.PCI\Image test\Batch::I0Main

Status code = 1

.Test\Batch::Tgg0Main

Status code = 1

and need the output like

Code:

PCI\Image test
Test

I tried using simple awk command

Code:

awk -F"\\" '{print $1}'

but it just gives

Code:

PCI
Test

and doesnt get rid of the empty lines and lines with status code .
Any help appriciated

Thanks
Arif

I'm not sure if this helps. I did a little command line sample for you if I understood correctly.

Duplicates your data?

Code:

jnetnix@nix:~$ cat /tmp/test.txt 
.PCI\Image test\Batch::I0Main

Status code = 1

.Test\Batch::Tgg0Main

Status code = 1
jnetnix@nix:~$

Modified correctly? Of course this would may work depending on your full data file...if Batch is not there.

Code:

jnetnix@nix:~$ cat /tmp/test.txt | grep "^\." | sed -r 's/(^\.|\\Batch:.*)//g'
PCI\Image test
Test
jnetnix@nix:~$

Or...maybe better I snagged some nice nawk code from another post. This will split on the \ and remove the last field.

Code:

jnetnix@nix:~$ cat /tmp/test.txt | grep "^\." | sed -r 's/^\.//' | nawk -v OFS='\' -F '\' {'NF=NF-1; print;'}
PCI\Image test
Test

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Thanks for the reply , the sed -r options doesnt works on my machine (soalris) . ALso there is no . in teh begining of teh text , the file is actually like below

Code:

Transaction\Operating Fund\Batch::OF0Main

Status code = 1

and the out put needs to be like

Code:

Transaction\Operating Fund

.Sorry for the confusion.

Thanks
Arif

Code:

sed -n 's#\(.*\)\\.*#\1#p' myFile

Code:

sed -n 's#\(.*\)\\.*#\1#p' myFile

The above code gives me PCI\Image test\ out of the string PCI\Image test\Batch::I0Main ,
How can I ge the last part in Batch::I0Main.

Sorry not pretty good at regular expression.

Thanks
Arif

Code:

sed -n 's#.*\\\(.*\)#\1#p' myFile