Shell Programming and Scripting

How to pass a file path to open a file?
I am using cygwin.
I want to open a file from a particular path, say C:\Test\File1,. This file path is stored in a variable.
I am able to cat a file like this :
cat "c:\Test\File1"
but i want the same thing to happen in my script file through a variable and it is not working this way.

Eg:
v1="C:\Test\Ad.A"
v2="C:\Test\Ar.A"
gawk -v file2=$v1 -f prog.awk $v2

Please help in this regard

I got the solution
it can be done like this
v1="C:/\Test/\Ad.A"


Thanks
Sandeep

But i still have 2 doubts.
1.) How to access folders or path which contains spaces in their names like this:

C:\Documents and Settings\san deep\Desktop\File1



2.) Suppose i want to access a file whose name is stored in a variable(var) and file path is fixed(C:\Test\)
So how to add the file name after a fixed file path

Eg:
C:/\Test/\<some expresiooon to put var in filepath>



Please help
Regards
Sandeep

In Cygwin can you get away with using the /cygdrive link and some quotes:

Code:

ls -l "/cygdrive/c/Program Files/Whatever the long file name is.txt"