hadi motamedi wrote:
On Thu, Dec 3, 2009 at 12:42 PM, mark m.roth@5-cent.us wrote:
John Doe wrote:
From: hadi motamedi motamedi24@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