Using Regular Expressions to get Data Between Two Newlines in Expect-CodePudding

I'm writing this expect script that sshs into a device and runs a cat command to view the mac address of the device. I wan't to assign this mac address to a variable. However, I'm having trouble parsing the data.

This is the contents of my expect_out(buffer):

expect: set expect_out(buffer) "cat /sys/class/net/eth0/address\r\nd0:21:f9:4f:c1:68\r\nU6-Lite-BZ.6.0.21# "

The d0:21:f9:4f:c1:68 is what I actually want.

This is what commands I have been running:

expect -re {(?n)^d0.*}
set test "$expect_out(0,string)"
puts "$test"

And this does store the correct output to a variable. However, normally I wouldn't know the first 2 chars of the mac address (d0), so I need to edit the line such that it just reads after the newline to the next newline. Or reads a number of characters after the first newline ending. Does anyone know how to do this? Thanks for the help.

CodePudding user response：

Since you already know the file content is a MAC addr which is 17 chars ^{(: and HEX digits)} long --

expect -re {[:[:xdigit:]]{17}}
set test "$expect_out(0,string)"
puts "$test"

CodePudding user response：

You could do

set lines [split $expect_out(buffer) \n]
set macaddr [string trim [lindex $lines end-1] \r]