10-07-2010
Simple AWK question
Hi,
let's assume i have an output below:
orgauser 23826 :E:Validity
senerse 2096
senerse 2111
senerse 21585
senerse 21596
root 12653 -bash
root 17262
root 17278
Some lines have not any string in their third column. I don't want to see those lines. i just want to see the lines which have strings in the third column.
So, output should be as below:
orgauser 23826 :E:Validity
root 12653 -bash
How can i do that? Any help would be greatly appreciated
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COMM(1) BSD General Commands Manual COMM(1)
NAME
comm -- select or reject lines common to two files
SYNOPSIS
comm [-123f] file1 file2
DESCRIPTION
The comm utility reads file1 and file2, which should be sorted lexically, and produces three text columns as output: lines only in file1;
lines only in file2; and lines in both files.
The filename ``-'' means the standard input.
The following options are available:
-1 Suppress printing of column 1.
-2 Suppress printing of column 2.
-3 Suppress printing of column 3.
-f Fold case in line comparisons.
Each column will have a number of tab characters prepended to it equal to the number of lower numbered columns that are being printed. For
example, if column number two is being suppressed, lines printed in column number one will not have any tabs preceding them, and lines
printed in column number three will have one.
comm assumes that the files are lexically sorted; all characters participate in line comparisons.
EXIT STATUS
comm exits 0 on success, >0 if an error occurred.
SEE ALSO
cmp(1), diff(1), sort(1), uniq(1)
STANDARDS
The comm utility conforms to IEEE Std 1003.2-1992 (``POSIX.2'').
BSD
June 6, 1993 BSD