Shell Programming and Scripting

I'm using Expect to execute a command on a router and return the output to a file. The output is a list. At the end of the list there's a statement that reads, "Found 165 active connections" (Where "165" could be any number between 0 and 2000.) I'm familiar with using Expect to return data from a router, but I'm a noob when it comes to using regex to parse that information. In this case, I only want to return the number to a file. I figured I could do that with an expression like

Code:

(?<=Found\s)\d+

. However, this doesn't work when rolled into the Expect script. Here's what I have less the router login stuff:

Code:

set log [open "/var/tmp/file.txt" "w"]
set accum {}
expect {
        -regexp {(?<=Found\s)\d+} {
            set accum "${accum}$expect_out(0,string)"
            exp_continue
        }
    }
puts $log $accum
expect "$prompt"

I'm getting: "couldn't compile regular expression pattern: quantifier operand invalid" I assume that's from the braces, but I've messed with the syntax several different ways and can't seem to nail it. Hopefully someone with more Expect/Regex experience can assist. Thanks.

I am not familiar with expect, but are you sure that the quantifier mentioned in the error message isn't the question mark? If it is, the solution may be to escape it, "\?". Again, just a shot in the dark from someone not familiar with expect.

Good luck,
Alister

Thanks for the help. Escaping the ? seems to have stopped the error. The script now completes, but no data is returned. Hmmm.