Awk - unterminated regex - bash

Question

I am writing a shell script which needs to pull values out of a text file which looks like this:
app.full.name /warfilelocation/ warfilename
My shell script will be iterating over a list of application names and pulling out either the location or name using AWK. I have tested doing this on the command line using the following:
awk "\$1 ~/app.full.name/ { print $2 }" applications.txt
which returns what I would expect however when i put this in a shell script I start having issues.
I have a function that looks like this:
function get_location() {
local application=$1
awk "\$1 ~/^$application/ { print \$2 }" applications.txt
}
But when i call this function i get the following error:
awk: $1 ~/^app.full.name
awk: ^ unterminated regexp
awk: cmd. line:1: app.full.name
awk: cmd. line:1: ^ syntax error
awk: cmd. line:2: app.full.name/ { print $2 }
awk: cmd. line:2: ^ syntax error
Does anyone have any ideas what I am doing wrong here. I presume I am not escaping the variable correct but no matter what i try it doesnt seem to work.
Thanks in advance

Answer 1

Use this approach to make awk recognize shell variables:
awk -v "v1=$VAR1" -v "v2=$VAR2" '{print v1, v2}' input_file
Update
$ cat input
tinky-winky
dipsy
laa-laa
noo-noo
po
$ teletubby='po'
$ awk -v "regexp=$teletubby" '$0 ~ regexp' input
po
Note that anything could go into the shell-variable,
even a full-blown regexp, e.g ^d.*y. Just make sure to use single-quotes
to prevent the shell from doing any expansion.

Answer 2

The error messages seem to indicate that there is a stray newline at the end of $application, which gives the "line 2" error messages.

Answer 3

see this: using awk match() function
kent$ app=app.ful
kent$ echo "app.full.name /warfilelocation/ warfilename"|awk -v a=$app '{if(match($1,a))print $2}'
/warfilelocation/

Answer 4

It's hard to tell without knowing exactly the value of $application, but it seems like you have a strange character in $application, such as a " or a / or something like that.
$ export application=foo/bar
$ awk "\$1 ~/^$application/ { print \$1 }"
gawk: cmd. line:1: $1 ~/^foo/bar/ { print $1 }
gawk: cmd. line:1: ^ parse error
I would look at the exact value that you have in $application, and if it contains a /, escape it.
One way to do this would be to use:
$ export application=`echo foo/bar | sed -e 's;/;\\\\/;g'`
$ awk "\$1 ~/^$application/ { print \$1 }"