Shell Programming and Scripting

How to write in awk to remove lines starting with "#" and then process the file:

This is not working:

Code:

cat file|awk '{if ($0 ~ /^#/) $0="";print NF>0}'

When I just give

Code:

cat file|awk '{if ($0 ~ /^#/) $0="";print }'

it prints the blank lines . I don't wnat the blank lines along with the data lines.

Thanks,

Why don't you simply do:

Code:

awk '$0 !~ /^#/ && NF' file

If you want to perform any action:

Code:

awk '$0 !~ /^#/ && NF{ action here }' file

Try:

Code:

awk '!/^#/' file

Code:

grep -v '^#' file

To suppress printing of blank lines and remove lines starting with #

Code:

awk '!/^#|^$/' file

PS: no need to cat data to awk

Thanks Yoda,
I was looking for something like this

Code:

awk '$0 !~ /^#/ && NF{ action here }' file

, it worked great, but it is also removing exiting empty lines, but that is fine in this case,

- && NF : worked and seems it removed empty lines.


Scrutinizer ,
I dont wanted to use grep -v, as wanted to pass in one awk statement. Thanks again...

Jotne, This worked great as well,

Code:

awk '!/^#|^$/' file

,
But surprised it leaves only one blank line on the data file, after it removed all starting hash and blank lines.
Thanks all. .

You may have blanks or tabs in the remaining line.

Try this:

Code:

awk 'NF && !/#/ {action}' file

If you don't want to remove existing empty lines, then I guess what Scrutinizer posted is the right approach.

Just add your action part to it:

Code:

awk '!/^#/ { action here } ' file