UNIX for Dummies Questions & Answers

Hi !

I am a newbie with Unix and I try to remove the "|" FS in this file:

Code:

1|2|3|4|5|6

to get:

Code:

123456

by using:

Code:

gawk 'BEGIN{FS=""} {gsub("|","",$0); print $0}' file

but nothing changes.

Or even when I try to change "|" by "%"with this command:

Code:

gawk 'BEGIN{FS=""} {gsub("|","%",$0) ;print $0}' file

I get:

Code:

%1%|%2%|%3%|%4%|%5%

I don't get it !
If someone could explain me, it would be great !

Vertical bar is a special character in a regular expression. Use:

Code:

 gsub( "\\|", "", $0 );

This will escape it. Two escapes are necessary to ensure that one escape reaches the regex parser.

Thanks agama !

I didn't know 2 escapes was necessary.

Two levels of quoting are required only if a string literal is eventually used as a regular expression. The first argument to gsub is actually a regular expression, so there is never any reason to use a string literal in the function call. Do yourself a favor and spare yourself the hassle of two levels of quoting different sets of special characters.

Code:

 gsub( /\|/, "", $0 );

However, there are times when it's necessary to use a dynamically-assigned, string-valued variable as the first argument, but even then, the first argument to gsub isn't itself a string literal.

Regards,
Alister