Hello,

I am calling a line of awk from a bash script.

MYSUBSTRING="<factories xmi:type=\"resources.url:URL\" xmi:id=\"URL"
POSITION=`awk -v MYSUBSTRING=$MYSUBSTRING '{ print index ($0, MYSUBSTRING)}' <<< ${line}`
echo $POSITION

awk does not like the colon character in the string. How do I escape it?

Error encountered:
awk: cmd. line:1: xmi:type="resources.url:URL"
awk: cmd. line:1 ^ syntax error


Cheers,
q

You gotta buncha stuff to look at here.

First, this is my cut-and-paste of what you said you did, with some echo statements to show what's going on:

Did I copy it right?

Second, the output I got from running that was this:

Note: no error message at all. Maybe I didn't copy what you did correctly. Please recheck.

Third, what's that $0 doing in there? In a bash command line, $0 is expanded to be the name of the running script if it's in a double quoted or back quoted string; it's left alone if it's in a single quoted string. I think that's what you want here, but you're not getting it. You're getting the name of the program, because this is really part of the back-quoted string; the single quotes within the back-quoted string don't count for squat when bash is interpreting the line; they just get sent through as they are, and bash doesn't even care if it finds an even number of them.

I discovered that by modifying this script somewhat (and saving it as e2, obviously):

... I got this output:

See that ./e2 in there?

Anyway, either come back here with comments, or just break it down into simpler pieces so you can see what's going on.

Hope this helps.

Quote shell variables when you use them: