If I want to match DEF_23 using the following regexp:
expect {
-re "DEF_\[0-9]*"
set result $expect_out(1,string)
}
why does it say no such element in array?
How does $expect_out work, and how can I capture the DEF using a regexp and assign it to the variable result?
You're looking for expect_out(0,string) -- the array element 1,string would be populated if you had capturing parentheses in your regular expression.
The expect manpage documents the use of expect_out in the documentation of the expect command:
Upon matching a pattern (or eof or full_buffer), any matching and previously unmatched output is saved in the variable expect_out(buffer). Up to 9 regexp substring matches are saved in the variables expect_out(1,string) through expect_out(9,string). If the -indices flag is used before a pattern, the starting and ending indices (in a form suitable for lrange) of the 10 strings are stored in the variables expect_out(X,start) and expect_out(X,end) where X is a digit, corresponds to the substring position in the buffer. 0 refers to strings which matched the entire pattern and is generated for glob patterns as well as regexp patterns.
There is an illustrative example in the manpage.
It seems that the above explication is not precise! Check this example:
$ cat test.exp
#!/usr/bin/expect
set timeout 5
log_user 0
spawn bash
send "ls -1 db*\r"
expect {
-re "^db.*$" {
set bkpfile $expect_out(0,string)
}
}
send_user "The filename is: $bkpfile\n"
close
$ ls -1 db*
dbupgrade.log
$ ./test.exp
can't read "bkpfile": no such variable
while executing
"send_user "The filename is: $bkpfile\n""
(file "./test.exp" line 15)
$
The test result is the same when $expect_out(1,string) or $expect_out(buffer)is used. Am I missing something or this is the expected behavior?
Aleksandar - it should work if you change the match to "\ndb.*$".
If you turn on exp_internal 1, you will see the buffer contains something like this: "ls -1 db*\r\ndbupgrade.log\r\n08:46:09"
So, the caret (^) will throw your pattern match off.
