Expect regex number pattern

Question

I'm having this poblem with a expect script. I want to capture when the output of a command is "0" and distinguish from other numbers or chains like "000" or "100"

my code is:

#!/usr/bin/expect -f
set timeout 3
spawn bash send "echo 0\n" 
expect {
-regexp {^0$} { send_user "\nzero\n" }  
-re {\d+} { send_user "\number\n"}
}
send "exit\n"

I have the follwing response:

spawn bash
echo 0

number

But the following regex don't work:

-regexp {^0$} { send_user "\nzero\n" }

if I change it to:

-regexp {0} { send_user "\nzero\n" }

works, but it also capture "00" "10" "1000" etc.

send "echo 100\n"
expect {
    -regexp {0} { send_user "\nzero\n" }
    -re {\d+} { send_user "\nnumber\n"}
}

Result:

spawn bash
echo 10

zero

I don't know what I'm doing wrong and I didn't found any from help on google. I also searched for similar problems resolved here I couldn't make any of it work on my code.

Update:

I tried the proposed fix code:

#!/usr/bin/expect -f
set timeout 3
spawn bash
send "echo 0\n"
expect {
    -regexp {\b0\b} { send_user "\nzero\n" }
    -re {\d+} { send_user "\nnumber\n"}
}
send "exit\n"

But I still having this response:

spawn bash
echo 0

number

Update 2

I also tried this line:

 -regexp {"^0\n$"} { send_user "\nzero\n" }

But I still having the same results.

Thanks in advance

Answer 1

I could resolve the problem:

#!/usr/bin/expect -f
set timeout 3

spawn bash
send "echo 0\n"
expect -re "(\\d+)" {
    set result $expect_out(1,string)
}
if { $result == 0 } {
    send_user "\nzero\n";
} else {
    send_user "\nnumber\n";
}

send "exit\n"