I have a file, one of whose line contains:
number 8
how can i use sed, grep or whatever linux script to find out what integer is there in front of the line that starts with "number"?
Thanks...
+1
A:
use grep and cut, this will return only the number
cat ./file.txt | grep number | cut -d " " -f 2
martin.malek
2009-11-23 06:59:12
no need cat. grep number file.txt | .....
ghostdog74
2009-11-23 07:08:31
I prefer using cat, makes the command more readable, and portable. You can just replace the cat with whatever generates the output you want to parse.
Andre Miller
2009-11-23 09:12:23
with cat, you add 3 more characters. Plus, you create one extra pipe process. Portable?? in what sense? And providing the file to awk's input is not/or less portable?
ghostdog74
2009-11-23 14:14:22
+1
A:
Another way is to use awk:
awk '/number/ {print $2}' < ./file.txt
It's a single command, which some prefer. If it's a large file, you may prefer the cat | grep | cut-way, as the three programs run in separate processes.
Morten Siebuhr
2009-11-23 07:04:19
if its a large file, use just awk!. cat + grep + cut +whatever is expensive operation on large files. Also, there is no need for redirection. just pass the file directly to awk.
ghostdog74
2009-11-23 07:11:59
Thanks man. Does the command mean to print field 2 where field 1 == number?
baltusaj
2009-11-23 07:21:20