Printing single quotes with awk
 

Hi,

I have the following command, which works fine:

ll /backup/PROD |grep PROD_ |awk -F" " '{print "catalog backuppiece /backup/PROD/"$9";"}'

The output is:

catalog backuppiece /backup/PROD/PROD_29mi42vn_1_1;

BUT... I want single quotes in the result. One before /back.. and one before the ;
Like this:

catalog backuppiece '/backup/PROD/PROD_29mi42vn_1_1';

How do I do that? I tried using \' but it does not work.
The following command works, but I get an error message, which prevents me from using the command in a bash file:

ll /backup/PROD |grep PROD_ |awk -F" " '{print "catalog backuppiece \'\''/backup/PROD/"$9"\'\'';"}'
awk: warning: escape sequence `\'' treated as plain `''

Regards,

Casper

Hi,

I found a solution to this question - No need for any replies then.

I used awk -v

ls -l /backup/PROD |grep PROD_ |awk -F" " -v xx="'" '{print "catalog backuppiece " xx "/backup/PROD/"$9 xx";"}'

Output is:

catalog backuppiece '/backup/PROD/PROD_29mi42vn_1_1';


Regards,

Casper

There is a much better way:

Or an even better idea:
The usual story here would also be not to parse ls as on my machine I only have 8 fields total so your $9 would fail for me.