UNIX for Dummies Questions & Answers

Hi I would like to grep a text with pattern starting with $$WS_S. and ending with a delimiter space or comma.. please help me with the syntax
Eg:

Code:

aaaaaaaaasdjkghg $$WS_S.table1 sdef
dsf $$WS_S.table2,sdfdff
$$WS_S.table3 dasssssssssds

Moderator's Comments:

Use code tags please, see PM.

Show what you have tried so far, thanks.

Code:

grep '\$\$WS_S.' filename|awk -F" " ' { if($NF=="2"){print$0} } '

You have 37 posts in this forum now. You should be familiar with code tags and got a PM and warning before, but still post without them, that's why you got an infraction now.

awk can do everything what grep can do and much more.
So 2 commands are not necessary.
awk does not need a single space as field separator, since this is part of the default. Default is a blank or any number/combination of spaces including tabs.
Your awk code is missing a blank between the "print" and "$0".
From your description I can't see if you want the whole line printed or just the part of it with the pattern; though the example would not be good for the 1st case, as this will always a match.

Anyway, if you grep supports "-o":

Code:

$ grep "\$\$WS_S\.[^ ,]\+" infile
aaaaaaaaasdjkghg $$WS_S.table1 sdef
dsf $$WS_S.table2,sdfdff
$$WS_S.table3 dasssssssssds
$ grep -o "\$\$WS_S\.[^ ,]\+" infile
$$WS_S.table1
$$WS_S.table2
$$WS_S.table3