awk - do not contain string

Shell Programming and Scripting
1

Hi, I have file which contain following..

1>
(1 rows
(3 rows
(6 rows

I do not want to print these linse.

I came up with this but getting an error.

awk '(NR >13) && ($1!="1>") && if (! ($0 ~ /rows/)) {print $1}' frile1

Ho do I write command so i can skip this lines?

Thanks.

2

You can't put an if statement in a pattern and you don't use slashes with the ~ operator. Try the !~ operator...
$0 !~ "rows"

3

I am still getting an error:

awk '(NR >13) && ($1!='1') && ($0!~"rows") {print $1}' file.out
awk: syntax error near line 1
awk: bailing out near line 1

4

I don't get an error when I try it. You must be using Solaris. If so, switch to nawk. And please mention your OS when you post a question.

5

Yes I am using Solaris.

i was able to execute the command but still prints the lines that have "rows" in it.

6

got it. column 2 was printing ROW not ROWS. I change ROWS to ROW and it works.

Thanks for your help.

7

I can't reproduce that. It doesn't print lines with "rows" in it for me. But I can't test on Solaris right now. Maybe a Solaris user will check in.