How to use 'expect' to pass UID & Password to a "for loop" in shell script?

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Top Forums Shell Programming and Scripting How to use 'expect' to pass UID & Password to a "for loop" in shell script?
# 15  
Old 11-19-2013
Yes the output you have highlighted in red is the problem, as you pointed out the PS1 change is getting lost. Most likely because it is being sent too early and the login process is clearing input buffers.

Please read the first 3 paragraphs of post #13 again, here I give some clues about possible causes of this, and a propose increasing the timeout to assist debugging:

#!/usr/bin/expect -f
set timeout 25
set u "souvikm"
set p "pass^123"
set command "oslevel -s"

# Open server_list for read (handle hosts) and config_details for write (handle config)
set hosts [open server_list r]
set config [open config_details w]

# This like turns off local echo -
# for debugging leave it commented and all output will be echoed
#log_user 0

# Loop through each host in the $hosts file

while {[gets $hosts host] >= 0} {
    spawn -noecho ssh -t $u@$host
    expect "?assword:*"
    send -- "$p\r"

    # Wait for original server side prompt
    expect -re "(%|#|>|\$) *$"

    # Change server side prompt to make matching more reliable
    send "PS1='PROMPT# '\r"

    # Wait for new prompt
    expect -re "PROMPT# '(.*)PROMPT# "

    # Execute command
    send "$command\r"

    # Match command output and prompt
    expect -re "\r\n(.*)\r\n(.*)PROMPT# "

    # Load command output into variable output
    set output $expect_out(1,string)

    # Append output into (and host name) to local logfile
    puts $config "$host: $output\r"

    # Send exit to host and wait for connection to close
    send "exit\r"
    expect eof
close $hosts
close $config

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