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Print ALL lines except if field is 999

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# 1  
Old 11-05-2014
Print ALL lines except if field is 999
Hi All!!! :-)

I need a command that will print each line of a text file UNLESS the 3rd field of that line is equal to the number 999. (space seperated fields)

Solaris10/BASH SHELL:

INPUT.TXT
Code:
aaa bbb 111 222
ccc ddd 333 444
eee fff 999 555
ggg hhh 666 777
aaa bbb 999 222
ccc ddd 333 444

OUTPUT.TXT
Code:
aaa bbb 111 222
ccc ddd 333 444
ggg hhh 666 777
ccc ddd 333 444

The INPUT.TXT file might have 5 or 10 fields, but it is field 3=999 that I dont want. I'm hoping there is a bash shell command (awk?) that could do this.

Thank you so very much!
Take cares!
-aj
Last edited by Scrutinizer; 11-06-2014 at 03:16 PM.. Reason: Added code tags
# 2  
Old 11-05-2014
Please use code tags as required by forum rules!

Any attempts from your side?
# 3  
Old 11-05-2014
Code:
awk '{if $3=999 print $0}' INPUT.TXT > OUTPUT.TXT

I know that probably wont work...but it helps get the idea out there of what we need (hope)
# 4  
Old 11-05-2014
Code:
awk '$3 !~ "999"' input.file > output.file

Fixing yours:

Code:
awk '{if ($3 != "999") print $0}' INPUT.TXT > OUTPUT.TXT

Last edited by Aia; 11-05-2014 at 03:37 PM..
This User Gave Thanks to Aia For This Post:
ajp7701
# 5  
Old 11-05-2014
That has many an syntax error plus a logical one. Correcting the syntaxes awk '{if ($3==999) print $0}' fileyieldseee fff 999 555
aaa bbb 999 222 and shows the logical one. Try
Code:
 awk '$3!=999' file
aaa bbb 111 222
ccc ddd 333 444
ggg hhh 666 777
ccc ddd 333 444

This User Gave Thanks to RudiC For This Post:
ajp7701
# 6  
Old 11-05-2014
NICE!! thanks so much!
# 7  
Old 11-05-2014
@Aia: That would suppress lines with 1999 or 99921234 as well... well, the first, unedited version ...
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