Shell Programming and Scripting

Hi Gurus, i need your help to create a script the will print a characters after the pattern was found.
Sample lines are below:

Code:

My birthday:"1977-16-07", My birthday:"1975-16-07"
My birthday:"1970-16-07"

.

My patter should be "birthday:", then i want to print the following characters which is the complete date: Note( some lines has multiple occurence of the patters)
So my desired output would be:

Code:

1977-16-07
1975-16-07
1970-16-07

appreciate your help. Thanks

Moderator's Comments:

I added CODE tags, but I am not convinced that they are in the right spot.

Perl way...there are many ways to do it with awk and grep as well...

Code:

$ perl -nle '$,="\n";@array = /birthday:\"(\d{4,4}-\d{2,2}-\d{2,2})\"/g; print @array' file
1977-16-07
1975-16-07
1970-16-07

please provide some back ground information.Seems the query is not pretty clear to create a script.there can be multiple ways to do as per above statements but not yet clear what is the specific requirement

you can try with awk and regex.

In your sample, the pattern doesn't matter, and your statement specifying your requirements doesn't match the output you say you want (the double quotes are missing in the output).

To get the output you requested from the input you gave (if I guessed correctly about where the CODE tags should have been), you could try something simple like:

Code:

awk -F'"' '{for(i = 2; i < NF; i += 2) print $i}' file

If file contains your sample input, the output produced matches your desired output.

As always, if you're using a Solaris/SunOS system, you need to use /usr/xpg4/bin/awk, /usr/xpg6/bin/awk, or nawk instead of /bin/awk or /usr/bin/awk.

Thanks.. Can you show me how it is done in awk and grep? Thanks a lot

scripter123,

Check this out:

Code:

awk '{A=split($0,a,/"/);for(i=1;i<=A;i++) if(a[i]~/birthday/) print a[i+1]}' file
1977-16-07
1975-16-07
1970-16-07

Enjoy ,Have fun!.