[SOLVED] Help with awk and accessing variable values within the syntax

meridionaljet · 07-09-2011, 04:32 AM

Hello,

I'm writing a bash script, and I am having trouble with this line:

awk '{print "url = \"http://www.example.com/directory/$type/"$1"\""}' input.file > output.file

Within the URL being printed, I wish the value of the variable "$type" to be printed. However, the quotation marks associated with the awk command do not allow the value of the variable to be called (syntax highlighting doesn't recognize $type as a variable, but rather as a literal part of the URL). Thus, the URL that ends up being printed literally has "$type" printed as a part of it. I would like to know how to change this behavior so that the actual value of the variable $type is printed within the URL instead. Any help would be appreciated.

markush · 07-09-2011, 04:59 AM

Hello meridionaljet,

I can't help you with your question. But your question has much more to do with "Programming" than with "Software". I'd recommend that you use the "Report"-button at your post and ask a Moderator to move your thread to the Programming-forum of LQ, http://www.linuxquestions.org/questions/programming-9/

You will find more people there who can help you with this question.

Markus

catkin · 07-09-2011, 05:16 AM

Code:

awk '{print "url = \"http://www.example.com/directory/'"$type"'/"$1"\""}' input.file > output.file

grail · 07-09-2011, 07:28 AM

Or use awk's ability to create variables at calling:

Code:

awk -vtype=$type '{print "url = \"http://www.example.com/directory/"type"/"$1"\""}' input.file > output.file

meridionaljet · 07-09-2011, 01:08 PM

Quote:

Originally Posted by grail

Or use awk's ability to create variables at calling:

Code:

awk -vtype=$type '{print "url = \"http://www.example.com/directory/"type"/"$1"\""}' input.file > output.file

Thank you very much. That worked. I apologize for putting this in the wrong forum section.