hadi motamedi wrote: > On Thu, Dec 3, 2009 at 12:42 PM, mark <m.roth at 5-cent.us> wrote: > >> John Doe wrote: >>> From: hadi motamedi <motamedi24 at gmail.com> >>>> Can you please do me favor and let me know if I can go further and try >> for >>>> advanced search like finding how many rows inside a file have data that >>>> does not start with a zero after the third comma ? >>> Something like: awk -F, ' { print $4 } ' | grep -v "^0" | wc -l Use one >>> command at a time to see how they work with each other (you might have to >>> modify the grep a bit)... >> *sigh* >> >> Drive me crazy, why use multiple commands? >> >> awk -F 'BEGIN { FS = ","; }{if ( $3 !~ /^0 ) { count++; }} END { print >> count }' >> filename > > Sorry . I tried for your proposed procedure , as the followings : > #awk -F 'BEGIN { FS = ","; }{if ( $3 !~ /^0 ) { count++; }} END { print > count }' HLRSubscriber-20091111173349.csv > But my CentOS server didn't return to the prompt . Can you please let me > know why it is in an end-less iterated loop ? > Thank you in advance Syntax error. You wrote if ( $3 !~ /^0 not if ( $3 !~ /^0/ PLEASE: if you ask for help, and someone gives you examples, READ THE MAN PAGES SO THAT YOU KNOW WHAT YOU'RE DOING. I could have just as well have given you something that would have wiped your system (like system("rm -rf /"). mark -- America is the only country that went from barbarism to decadence without civilization in between. - Oscar Wilde