replacing the `'` char using awk

Question

I have lines with a single : and a\' in them that I want to get rid of. I want to use awk for this. I\'ve tried using:

Answer 1

You could use:

1. Octal code for the single quote:

   [:\47]
   

2. The single quote inside double quotes, but in that case special characters will be expanded by the shell:

   % print a\': | awk "sub(/[:']/, x)"        
   a
   

3. Use a dynamic regexp, but there are performance implications related to this approach:

   % print a\': | awk -vrx="[:\\\']" 'sub(rx, x)'  
   a
   

Answer 2

With bash you cannot insert a single quote inside a literal surrounded with single quotes. Use '"'"' for example.

First ' closes the current literal, then "'" concatenates it with a literal containing only a single quote, and ' reopens a string literal, which will be also concatenated.

What you want is:

awk '{gsub ( "[:'"'"']","" ) ; print $0; }'

ssapkota's alternative is also good ('\'').

Answer 3

This works:

awk '{gsub( "[:'\'']","" ); print}'

Answer 4

This also works:

awk '{gsub("\x27",""); print}'

Answer 5

I don't know why you are restricting yourself to using awk, anyways you've got many answers from other users. You can also use sed to get rid of " :' "

sed 's/:\'//g'

This will also serve your purpose. Simple and less complex.

Answer 6

simplest awk '{gsub(/\047|:/,"")};1'