How to write expect script which executes command and prints just the command's output?
I've tried various things but none works, e.g.
#!/usr/bin/expect
log_user 0
spawn bash
send "echo 1\r"
log_user 1
expect "1"
log_user 0
send "exit\r"
expect eof
Gives in output:
echo 1
While I need just "1" . I hope somebody knows simple solution how to fix my example
Capturing the output from sent commands is a bit of a pain in expect.
Here's a more general case that does not rely on the log_user setting, it captures the output with a regular expression:
#!/usr/bin/expect
log_user 0
spawn bash
# set the prompt to a known value
send "PS1='>'\r"
expect -re {>$}
# send a command: we don't know what the output is going to be
send "echo \$RANDOM\r"
# capture the portion of the output that occurs just before the prompt
expect -re "\r\n(.*?)\r\n>$"
puts "output is: $expect_out(1,string)"
send "exit\r"
expect eof
A thought just occurred to me: if the command does not require any interaction, then expect is overkill: just use exec
set output [exec bash -c {echo $RANDOM}]
Ok, it looks following script does (at least similar to) what I need:
log_user 0
spawn bash
expect "#" {} "\\\$" {}
send -- "echo AA\r"
expect -- "echo AA\r" {}
log_user 1
expect -- "AA"
log_user 0
send -- "exit\r"
expect eof
Related
I'm still pretty new to batch but I wanted to try out some different functionalities within the same batch program. Essentially, I am attempting to parse through a .txt file of hostnames and ping them 1 time each, and return a pass or fail by utilizing a find query for the specific ping result. The second find query is redundant but left in. A string of "timed out" indicates that the host is down. I have been able to achieve this with this bulky code, but my actual ping statistics are no longer being written to this log file.
I'm not too familiar with the command pipeline character but it may just be incompatible for this specific use case. If I remove the piped command for the find query, it is able to write the output of the ping command just fine, but then I lose the utility of a return pass or fail value. Is this just a simple syntax error? I've tried moving the location of the actual write to file argument on the line itself but it hasn't worked. Also, why are the errorlevel values inverted (line 21)? I can't seem to return what I'm looking for.
Is there anyway I can get all these parts to play nice together? I apologize if the answer is quite obvious...
#echo off
setlocal
set hosts=temp.txt
set count=0
echo.
echo Parsing File: %hosts%
echo.
echo Start > C:\Users\___\Downloads\pingLOG.txt
for /F "tokens=*" %%i in (%hosts%) do (
set /A count=count+1
echo+
echo.
echo [+] Pinging: %%i
echo [+] Pinging: %%i >> C:\Users\___\Downloads\pingLOG.txt
echo.
ping -n 1 "%%i" | find /I "timed out" >> C:\Users\___\Downloads\pingLOG.txt
if errorlevel == 1 (
echo Pass
) else (
echo Fail
)
echo.
echo %TIME% >> C:\Users\___\Downloads\pingLOG.txt
)
echo.
echo %count% Hosts Scanned
find /c "timed out" C:\Users\___\Downloads\pingLOG.txt
echo.
pause
You can't filter the output for a certain string and expect the complete output (eliminating unwanted parts is the very reason for filtering).
To accomplish your goal, you need a temporary file (the complete output) and filter that file, so you have both variants (filtered and unfiltered) (suboptimal, but well...):
ping -n 1 "%%i" >"%temp%\temp.tmp"
find /I "timed out" "%temp%\temp.tmp" >nul && set "status=Fail" || set "status=Pass"
type "%temp%\temp.tmp" >> "C:\Users\___\Downloads\pingLOG.txt"
echo %status% >> "C:\Users\___\Downloads\pingLOG.txt"
Hi I am trying to store the output of a command run through a spawn ssh remote window into my local host, I am new to expect and am not able to figure out where I am wrong.
My Code:
#!/bin/bash
while read line
do
/usr/bin/expect <<EOD
spawn ssh mininet#$line
expect "assword:"
send -- "mininet\r"
set output [open "outputfile.txt" "a+"]
expect "mininet#mininet-vm:*"
send -- "ls\r"
set outcome $expect_out(buffer)
send "\r"
puts $output "$outcome"
close $output
expect "mininet#mininet-vm:*"
send -- "exit\r"
interact
expect eof
EOD
done <read_ip.txt
I am getting the error
expect: spawn id exp6 not open
while executing
"expect "mininet#mininet-vm:*""
Please can any body help me on this code.
You have your expect program in a shell heredoc. The shell will expand variables in the heredoc before launching expect. You have to protect expect's variables from the shell.
One way is to use a 'quoted' heredoc, and pass the shell variable to expect through the environment:
#!/bin/bash
export host ## an environment variable
while read host
do
/usr/bin/expect <<'EOD' ## note the quotes here
spawn ssh mininet#$env(host) ## get the value from the environment
expect "assword:"
send -- "mininet\r"
set output [open "outputfile.txt" "a+"]
expect "mininet#mininet-vm:*"
send -- "ls\r"
set outcome $expect_out(buffer)
send "\r"
puts $output "$outcome"
close $output
expect "mininet#mininet-vm:*"
send -- "exit\r"
expect eof ## don't want both "interact" and "expect eof"
EOD
done <read_ip.txt
Putting single quotes around the heredoc terminator means the whole heredoc acts like a single quoted string, and expect's variables are left for expect to handle.
You might also investigate the expect log_file command: you can enable and disable logging at will, much as you are doing manually here.
I was following this simple tutorial to try out a simple lua script
http://www.redisgreen.net/blog/2013/03/18/intro-to-lua-for-redis-programmers/
I created a simple hello.lua file with these lines
local msg = "Hello, world!"
return msg
And i tried running simple command
EVAL "$(cat /Users/rsingh/Downloads/hello.lua)" 0
And i am getting this error
(error) ERR Error compiling script (new function): user_script:1: unexpected symbol near '$'
I can't find what is wrong here and i haven't been able to find someone who has come across this.
Any help would be deeply appreciated.
Your problem comes from the fact you are executing this command from an interactive Redis session:
$ redis-cli
127.0.0.1:6379> EVAL "$(cat /path/to/hello.lua)" 0
(error) ERR Error compiling script (new function): user_script:1: unexpected symbol near '$'
Within such a session you cannot use common command-line tools like cat et al. (here cat is used as a convenient way to get the content of your script in-place). In other words: you send "$(cat /path/to/hello.lua)" as a plain string to Redis, which is not Lua code (of course), and Redis complains.
To execute this sample you must stay in the shell:
$ redis-cli EVAL "$(cat /path/to/hello.lua)" 0
"Hello, world!"
If you are coming from windows and trying to run a lua script you should use this format:
redis-cli --eval script.lua
Run this from the folder where your script is located and it will load a multi line file and execute it.
On the off chance that anyone's come to this from Windows instead, I found I had to do a lot of juggling to achieve the same effect. I had to do this:
echo “local msg = 'Hello, world!'; return msg” > hello.lua
for /F "delims=" %i in ('type hello.lua') do #set cmd=%i
redis-cli eval "%cmd%" 0
.. if you want it saved as a file, although you'll have to have all the content on one line. If you don’t just roll the content into a set command
set cmd=“local msg = 'Hello, world!'; return msg”
redis-cli eval "%cmd%" 0
I've got a small script called "onewhich". Its purpose is to behave like which, except that it will only give the FIRST occurrence of any executables specified as options, as found in the order they'd appear in the path.
So for example, if my path is /opt/bin:/usr/bin:/bin, and I have both /opt/bin/runme and /usr/bin/runme, then the command onewhich runme would return /opt/bin/runme.
But if I also have a /usr/bin/doit, then the command onewhich doit runme would return /usr/bin/doit instead.
The idea is to walk through the path, check for each executable specified, and if it exists, show it and exit.
Here's the script so far.
#!/bin/sh
for what in "$#"; do
for loc in `echo "${PATH}" | awk -vRS=: 1`; do
if [ -f "${loc}/${what}" ]; then
echo "${loc}/${what}"
exit 0
fi
done
done
exit 1
The problem is, I want to be better about PATH directories with special characters. Every second shell question here on StackOverflow talks about how bad it is to parse paths with tools like awk and sed. There's even a bash faq entry about it. (Proviso: I'm not using bash for this, but the recommendation is still valid.)
So I tried rewriting the script to separate paths in a pipe, like this"
#!/bin/sh
for what in "$#"; do
echo "${PATH}" | awk -vRS=: 1 | while read loc ; do
if [ -f "${loc}/${what}" ]; then
echo "${loc}/${what}"
exit 0
fi
done
done
exit 1
I'm not sure if this gives me any real advantage (since $loc is still inside quotes), but it also doesn't work because for some reason, the exit 0 seems to be ignored. Or ... it exits something (the sub-shell with the while loop that terminates the pipe, maybe), but the script exits with a value of 1 every time.
What's a better way to step through directories in ${PATH} without the risk that special characters will confuse things?
Alternately, am I reinventing the wheel? Is there maybe a way to do this that's built in to existing shell tools?
This needs to run in both Linux and FreeBSD, which is why I'm writing it in Bourne instead of bash.
Thanks.
This doesn't directly answer your question, but does eliminate the need to parse PATH at all:
onewhich () {
for what in "$#"; do
which "$what" 2>/dev/null && break
done
}
This just calls which on each command on the input list until it finds a match.
To parse PATH, you can simply set `IFS=':'.
if [ "${IFS:-x}" = "${IFS-x}" ]; then
# Only preserve the value of IFS if it is currently set
OLDIFS=$IFS
fi
IFS=":"
for f in $PATH; do # Do not quote $PATH, to allow word splitting
echo $f
done
if [ "${OLDIFS:-x}" = "${OLDIFS-x}" ]; then
IFS=$OLDIFS
fi
The above will fail if any of the directories in PATH actually contain colons.
Your first method looks to me as if it should work. In practical terms, if it's really the $PATH you'll be searching, it's unlikely you'll have spaces and newlines embedded in directories there. If you do, it's probably time to refactor.
But still, I don't think you're at risk from the possibility of bad names clobbering your loop, since you're wrapping variables in quotes. At worst, I suspect you might miss the odd valid executable, but I can't see how the script would generate errors. (I don't see how the script would miss valid executables, and I haven't tested - I'm just saying I don't see problems at first glance.)
As for your second question, about the loop, I think you've hit the nail on the head. When you run a pipe like this | that | while condition; do things; done, the while loop runs in its own shell at the end of the pipe. Exiting that shell may terminate the actions of the pipe, but that only brings you back to the parent shell, which has its own thread of execution that terminates with exit 1.
As for a better way to do this, I would consider which.
#!/bin/sh
for what in "$#"; do
which "$what"
done | head -1
And if you really want the exit values as well:
#!/bin/sh
for what in "$#"; do
which "$what" && exit 0
done
exit 1
The second might even be fewer resources, as it doesn't have to open a file handle and pipe through head.
You can also split your path using IFS. For example, if you wanted to wrap your loops the other way around, you could do this:
#!/bin/sh
IFS=":"
for loc in $PATH; do
for what in "$#"; do
if [ -x "$loc"/"$what" ]; then
echo "$loc"/"$what"
exit 0
fi
done
done
exit 1
Note that under normal circumstances, you might want to save the old value of $IFS, but you seem to be doing things in a stand-alone script, so the "new" value gets thrown out when the script exits.
All the above code is untested. YMMV.
Another way to get around the need to parse PATH at all is to run the builtin type command in new shell with a stripped environment (i. e. there simply are no functions or aliases to look up; cf. env -i sh -c 'type cmd 2>/dev/null).
# using `cmd` instead of $(cmd) for portability
onewhich() {
ec=0 # exit code
for cmd in "$#"; do
command -p env -i PATH="$PATH" sh -c '
export LC_ALL=C LANG=C
cmd="$1"
path="`type "$cmd" 2>/dev/null`"
if [ X"$path" = "X" ]; then
printf "%s\n" "error: command \"${cmd}\" not found in PATH" 1>&2
exit 1
else
case "$path" in
*\ /*)
path="/${path#*/}"
printf "%s\n" "$path";;
*)
printf "%s\n" "error: no disk file: $path" 1>&2
exit 1;;
esac
exit 0
fi
' _ "$cmd"
[ $? != 0 ] && ec=1
done
[ $ec != 0 ] && return 1
}
onewhich awk ls sed
onewhich builtin
onewhich if
Since which on success returns two full command paths if two commands are specified as arguments, exit 0 in the first onewhich script above aborts the program prematurely. In addition, if two commands are specified as arguments to which, the exit code of which is set to 1 even if only one command lookup failed (cf. which awk sedxyz ls; echo $?). To mimic this behaviour of the which command it is necessary to toggle on/off two variables (cnt and nomatches below).
onewhich() (
IFS=":"
nomatches=0
for cmd in "$#"; do
cnt=0
for loc in $PATH ; do
if [ $cnt = 0 ] && [ -x "$loc"/"$cmd" ]; then
echo "$loc"/"$cmd"
cnt=1
fi
done
[ $cnt = 0 ] && nomatches=1
done
[ $nomatches = 1 ] && exit 1 || exit 0 # exit 1: at least one cmd was not in PATH
)
onewhich awk ls sed
onewhich awk lsxyz sed
onewhich builtin
onewhich if
I'm trying to set up a shell script that will start a screen session (or rejoin an existing one) only if it is invoked from an interactive shell. The solution I have seen is to check if $- contains the letter "i":
#!/bin/sh -e
echo "Testing interactivity..."
echo 'Current value of $- = '"$-"
if [ `echo \$- | grep -qs i` ]; then
echo interactive;
else
echo noninteractive;
fi
However, this fails, because the script is run by a new noninteractive shell, invoked as a result of the #!/bin/sh at the top. If I source the script instead of running it, it works as desired, but that's an ugly hack. I'd rather have it work when I run it.
So how can I test for interactivity within a script?
Give this a try and see if it does what you're looking for:
#!/bin/sh
if [ $_ != $0 ]
then
echo interactive;
else
echo noninteractive;
fi
The underscore ($_) expands to the absolute pathname used to invoke the script. The zero ($0) expands to the name of the script. If they're different then the script was invoked from an interactive shell. In Bash, subsequent expansion of $_ gives the expanded argument to the previous command (it might be a good idea to save the value of $_ in another variable in order to preserve it).
From man bash:
0 Expands to the name of the shell or shell script. This is set
at shell initialization. If bash is invoked with a file of com‐
mands, $0 is set to the name of that file. If bash is started
with the -c option, then $0 is set to the first argument after
the string to be executed, if one is present. Otherwise, it is
set to the file name used to invoke bash, as given by argument
zero.
_ At shell startup, set to the absolute pathname used to invoke
the shell or shell script being executed as passed in the envi‐
ronment or argument list. Subsequently, expands to the last
argument to the previous command, after expansion. Also set to
the full pathname used to invoke each command executed and
placed in the environment exported to that command. When check‐
ing mail, this parameter holds the name of the mail file cur‐
rently being checked.
$_ may not work in every POSIX compatible sh, although it probably works in must.
$PS1 will only be set if the shell is interactive. So this should work:
if [ -z "$PS1" ]; then
echo noninteractive
else
echo interactive
fi
try tty
if tty 2>&1 |grep not ; then echo "Not a tty"; else echo "a tty"; fi
man tty :
The tty utility writes the name of the terminal attached to standard
input to standard output. The name that is written is the string
returned by ttyname(3). If the standard input is not a terminal, the
message ``not a tty'' is written.
You could try using something like...
if [[ -t 0 ]]
then
echo "Interactive...say something!"
read line
echo $line
else
echo "Not Interactive"
fi
The "-t" switch in the test field checks if the file descriptor given matches a terminal (you could also do this to stop the program if the output was going to be printed to a terminal, for example). Here it checks if the standard in of the program matches a terminal.
Simple answer: don't run those commands inside ` ` or [ ].
There is no need for either of those constructs here.
Obviously I can't be sure what you expected
[ `echo \$- | grep -qs i` ]
to be testing, but I don't think it's testing what you think it's testing.
That code will do the following:
Run echo \$- | grep -qs i inside a subshell (due to the ` `).
Capture the subshell's standard output.
Replace the original ` ` expression with a string containing that output.
Pass that string as an argument to the [ command or built-in (depending on your shell).
Produce a successful return code from [ only if that string was nonempty (assuming the string didn't look like an option to [).
Some possible problems:
The -qs options to grep should cause it to produce no output, so I'd expect [ to be testing an empty string regardless of what $- looks like.
It's also possible that the backslash is escaping the dollar sign and causing a literal 'dollar minus' (rather than the contents of a variable) to be sent to grep.
On the other hand, if you removed the [ and backticks and instead said
if echo "$-" | grep -qs i ; then
then:
your current shell would expand "$-" with the value you want to test,
echo ... | would send that to grep on its standard input,
grep would return a successful return code when that input contained the letter i,
grep would print no output, due to the -qs flags, and
the if statement would use grep's return code to decide which branch to take.
Also:
no backticks would replace any commands with the output produced when they were run, and
no [ command would try to replace the return code of grep with some return code that it had tried to reconstruct by itself from the output produced by grep.
For more on how to use the if command, see this section of the excellent BashGuide.
If you want to test the value of $- without forking an external process (e.g. grep) then you can use the following technique:
if [ "${-%i*}" != "$-" ]
then
echo Interactive shell
else
echo Not an interactive shell
fi
This deletes any match for i* from the value of $- then checks to see if this made any difference.
(The ${parameter/from/to} construct (e.g. [ "${-//[!i]/}" = "i" ] is true iff interactive) can be used in Bash scripts but is not present in Dash, which is /bin/sh on Debian and Ubuntu systems.)