在 Bash 脚本中期望：expect_out 不打印出缓冲区

Question

I am trying to capture the output of the "dir" command by logging into a switch, but I am unable to do so. I am using Expect within Bash. I am making use of expect_out to capture output of that command in a buffer and print it out. Actually I want to capture the output and perform some operations on it.

Script:

#!/bin/bash
expect -c "
spawn telnet 1.1.1.1 2000
sleep 1
send \"\r\"
send \"\r\"
expect {
Prompt> { send \"dir\r\"  }
}
set output $expect_out(buffer)
"
echo "$output"

Output:

spawn telnet 1.1.1.1 2000
Trying 1.1.1.1...
Connected to 1.1.1.1 (1.1.1.1).
Escape character is '^]'.




Prompt>

Prompt>

After these prompts are displayed, the scripts just exits. How can I fix this problem?

After I split it, I can use parameter substitution as well as single quotes. Now I am facing different error.

Script:

expect -c "
spawn telnet $IP $PORT1
sleep 1
send \"\r\"
send \"\r\"
"
expect -c '
expect {
Prompt> { send \"dir\r\" }
set output $expect_out(buffer)
puts "$output"
}
'

Output:

spawn telnet 172.23.149.139 2033
can't read "expect_out(buffer)": no such variable
while executing
"expect {
Prompt> { send \"dir\r\" }
set output $expect_out(buffer)
puts "$output"
}
"

---

I changed it to according to the suggestions. But I am still facing errors.

Script:

output=$(expect -c '
spawn telnet '"$IP $PORT1"'
sleep 1
send '"\r"'
send '"\r"'

expect Prompt> { send '"dir\r"'  }
expect '"\n"'
expect -indices Prompt>
puts '"[string range $expect_out(buffer) 0 [expr $expect_out(0,end) - 1]]"'

')

echo "======="
echo "$output"
echo "======="

Output:

syntax error in expression "(0,end) - 1"
    while executing
"expr (0,end) - 1"
    invoked from within
"string range (buffer) 0 [expr (0,end) - 1]"
    invoked from within
"puts [string range (buffer) 0 [expr (0,end) - 1]]"

=======
spawn telnet 1.1.1.1 2000
Trying 1.1.1.1...
Connected to 1.1.1.1 (1.1.1.1).
Escape character is '^]'.




Prompt>

Prompt>

=======

Hence to circumvent the error, I changed, the line

puts '"[string range $expect_out(buffer) 0 [expr $expect_out(0,end) - 1]]"'

to

puts '"$expect_out(buffer)"'

But then I am getting no error, but the output of dir is also not getting printed. Something like:

Prompt>

Prompt> (buffer)

Answer 1

The second of your “split” Expect programs does not have access to the spawned telnet process. When you split them like that, you made them independent (one can not access the variables or state of the other; actually by the time the second one has started the first one, and its telnet process, no longer exist).

---

The shell will automatically concatenate any strings (that are not separated by unquoted/unescaped whitespace) without regard to the kind of quotes (if any) they use. This means you can start put the first part of your Expect program in single quotes, switch to double quotes for the parameter substitution, and then go back to single quotes for the rest of the program (to avoid having to escape any of "$\` that occur in your Expect code).

expect -c '
spawn telnet '"$HOST $PORT"'
sleep 1
⋮ (rest of Expect program)
'

It uses single quotes to protect most of the program, but switches back to double quotes to let the shell substitute the its HOST and IP parameters into the text of the Expect program.

---

Next, the shell that started expect can not access variable set inside the Expect program. The normal way to collect output from a child process is to have it write to stdout or stderr and have the shell collect the output via a command substitution ( $() ).

In Tcl (and Expect), you can use puts to send a string to stdout. But, by default, Expect will also send to stdout the normal “interaction” output (what it receives from any spawned commands and what it sent to them; ie what you would see if you were running the spawned command manually). You can disable this default logging with log_user 0 .

You might write your program like this:

#!/bin/sh
output=$(expect -c '
# suppress the display of the process interaction
log_user 0

spawn telnet '"$HOST $PORT"'
sleep 1
send "\r"
send "\r"
# after a prompt, send the interesting command
expect Prompt> { send "dir\r"  }
# eat the \n the remote end sent after we sent our \r
expect "\n"
# wait for the next prompt, saving its position in expect_out(buffer)
expect -indices Prompt>

# output what came after the command and before the next prompt
# (i.e. the output of the "dir" command)
puts [string range $expect_out(buffer) \
                   0 [expr $expect_out(0,start) - 1]]
')
echo "======="
echo "$output"
echo "======="

Answer 2

Because your Expect script is enclosed in double quotes, the shell is expanding $expect_out to an empty string. Put the body of the expect script in single quotes.

When you set a variable in Expect, the shell will have no idea. When the Expect script completes, its variables vanish too. You have to puts the expect_out buffer.