How to pass argument in Expect through the command line in a shell script

Question

I am passing argument in Expect through the command line in a shell script.

I tried this

#!/usr/bin/expect -f
    
set arg1 [lindex $argv 0]
    
spawn lockdis -p
expect "password:" {send "$arg1\r"}
expect "password:" {send "$arg1\r"}
expect "$ "

But it's not working. How can I fix it?

bartimar · Accepted Answer · 2013-06-12T08:15:45.327

132

If you want to read from arguments, you can achieve this simply by

set username [lindex $argv 0];
set password [lindex $argv 1];

And print it

send_user "$username $password"

That script will print

$ ./test.exp user1 pass1
user1 pass1

You can use Debug mode

$ ./test.exp -d user1 pass1

edited Jun 12 '13 at 08:15

answered Jun 12 '13 at 07:43

bartimar

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While this works, be careful using it as your process will appear with the arguments, username and password, when doing something like "ps aux" – joebeeson Oct 30 '15 at 13:53
can we loop the argument list and put them in array? – Vivek Sep 07 '17 at 07:34
That's why you should pass it a file where your username and password are kept privately. – Han Zhang May 14 '22 at 05:50

score 10 · Answer 2 · answered Jun 12 '13 at 07:59

10

A better way might be this:

lassign $argv arg1 arg2 arg3

However, your method should work as well. Check that arg1 is retrieved. For example, with send_user "arg1: $arg1\n".

answered Jun 12 '13 at 07:59

spbnick

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The code above is for TCL which `expect` is using, and doesn't make sense for Bash. – spbnick Dec 08 '18 at 09:31

score 8 · Answer 3 · edited Dec 26 '17 at 22:16

8

#!/usr/bin/expect
set username [lindex $argv 0]
set password [lindex $argv 1]
log_file -a "/tmp/expect.log"
set timeout 600
spawn /anyscript.sh
expect "username: " { send "$username\r" }
expect "password: " { send "$password\r" }
interact

edited Dec 26 '17 at 22:16

doug65536

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answered Dec 12 '17 at 06:24

user128364

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score 7 · Answer 4 · edited Jul 25 '20 at 08:36

I like the answer provided with this guide.

It creates a parse argument process.

#process to parse command line arguments into OPTS array
proc parseargs {argc argv} {
    global OPTS
    foreach {key val} $argv {
        switch -exact -- $key {
            "-username"   { set OPTS(username)   $val }
            "-password"   { set OPTS(password)   $val }
        }
    }
}
parseargs $argc $argv
#print out parsed username and password arguments
puts -nonewline "username: $OPTS(username) password: $OPTS(password)"

The above is just a snippet. It's important to read through the guide in full and add sufficient user argument checks.

I like argument checking all the time – Sohail Si Apr 15 '22 at 09:12 — Sohail Si, Apr 15 '22 at 09:12

score -2 · Answer 5 · edited Jun 23 '20 at 21:16

-2

Note, sometimes argv 0 is the name of the script you are calling. So if you run it that way, argv 0 doesn't work.

For me I run

expect script.exp  password

That makes argv 1 = password and argv 0 = script.exp.

edited Jun 23 '20 at 21:16

Peter Mortensen

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answered Nov 30 '16 at 02:28

Sid Dakota

1

score -4 · Answer 6 · edited Jun 23 '20 at 21:18

-4

Arguments with spaces are fine, assuming the argument you want is the first after the script name ($0 is script name, $1 is the first argument, etc.)

Make sure you use "$ARG", not $ARG as it will not include the whitespace, but break them up into individual arguments. Do this in your Bash script:

#!/bin/bash

ARG="$1"
echo WORD FROM BASH IS: "$ARG" #test for debugging

expect -d exp.expect "$ARG"

exit 0

Also, as the first answer states, use debug mode, (the -d flag). It will output your argv variables as Expect sees them and should show you what is going on.

edited Jun 23 '20 at 21:18

Peter Mortensen

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answered May 02 '17 at 03:06

eulerworks

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The expect script is missing – Timo Mar 27 '18 at 18:57
1

It isn’t, “missing” it works the same way for any expect script that would be called via bash. Though in this example it would need to be called, “exp.expect.” There was no need for a downvote, my answer explains how to handle what was in question. – eulerworks Aug 13 '18 at 14:08
2

ok, I `expect`ed the content of your script which references the parameters from outside as the others showed in their examples. – Timo Aug 15 '18 at 06:40

How to pass argument in Expect through the command line in a shell script

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