I have this command sequence that I'm having trouble understanding:
[me#mine ~]$ (echo 'test'; cat) | bash
echo $?
1
echo 'this is the new shell'
this is the new shell
exit
[me#mine ~]$
As far as I can understand, here is what happens:
A pipe is created.
stdout of echo 'test' is sent to the pipe.
bash receives 'test' on stdin.
echo $? returns 1, which is what happens when you run test without args.
cat runs.
It is copying stdin to stdout.
stdout is sent to the pipe.
bash will execute whatever you type in, but stderr won't get printed to the screen (we used |, not |&).
I have three questions:
It looks like, even though we run two commands, we use the same pipe and bash process for both commands. Is that the case?
Where do the prompts go?
When something like cat uses stdin, does it take exclusive ownership of stdin as long as the shell runs, or can other things use it?
I suspect I'm missing some detail with ttys, but I'm not sure. Any help or details or man excerpt appreciated!
So...
Yes, there's a single pipe sending commands to a single instance of bash. Note:
$ echo 'date "+%T hello $$"; sleep 1; date "+%T world $$"' | bash
22:18:52 hello 72628
22:18:53 world 72628
There are no prompts. From the man page:
An interactive shell is one started without non-option arguments (unless -s is specified) and without the -c option whose standard input and error are both connected to terminals. PS1 is set and $- includes i if bash is interactive.
So a pipe is not an interactive shell, and therefore has no prompt.
Stdin and stdout can only connect to one thing at a time. cat will take stdin from the process that ran it (for example, your interactive shell) and send its stdout through the pipe to bash. If you need multiple things to be able to submit to the stdin of that cat, consider using a named pipe.
Does that cover it?
Related
I began with playing ctfs challenges, and I encountered a problem where I needed to send an exploit into a binary and then interact with the spawned shell.
I found a solution to this problem which looks something like this:
(echo -ne "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx\xbe\xba\xfe\xca" && cat) | nc pwnable.kr 9000
Meaning:
without the "cat" sub-command, I couldn't interact with the shell, but with it, i now able to send commands into the spawned shell and get the returned output to my console stdout.
What exactly happens there? this command line confuses me
If you just type in cat at the command line, you'll be able to see that this command simply copies stdin to stdout one line at a time. It will carry on doing this until you either quit with Ctrl-C or send an EOF with Ctrl-D.
In this example you're running cat immediately after successfully printing the payload (the concatenator && tells the shell to run the second command only if the first command has an exit code of zero; i.e., no error). As a result, the remote terminal won't see an EOF until you terminate it as described above. When this is piped to nc, everything you type in is sent via cat to the remote server, and everything it sends back appears on your stdout.
So yes, in effect you end up with an interactive shell. You can get pretty much the same effect on your own machine by running cat | sh.
I have a shell script which writes all output to logfile
and terminal, this part works fine, but if I execute the script
a new shell prompt only appear if I press enter. Why is that and how do I fix it?
#!/bin/bash
exec > >(tee logfile)
echo "output"
First, when I'm testing this, there always is a new shell prompt, it's just that sometimes the string output comes after it, so the prompt isn't last. Did you happen to overlook it? If so, there seems to be a race where the shell prints the prompt before the tee in the background completes.
Unfortunately, that cannot fixed by waiting in the shell for tee, see this question on unix.stackexchange. Fragile workarounds aside, the easiest way to solve this that I see is to put your whole script inside a list:
{
your-code-here
} | tee logfile
If I run the following script (suppressing the newline from the echo), I see the prompt, but not "output". The string is still written to the file.
#!/bin/bash
exec > >(tee logfile)
echo -n "output"
What I suspect is this: you have three different file descriptors trying to write to the same file (that is, the terminal): standard output of the shell, standard error of the shell, and the standard output of tee. The shell writes synchronously: first the echo to standard output, then the prompt to standard error, so the terminal is able to sequence them correctly. However, the third file descriptor is written to asynchronously by tee, so there is a race condition. I don't quite understand how my modification affects the race, but it appears to upset some balance, allowing the prompt to be written at a different time and appear on the screen. (I expect output buffering to play a part in this).
You might also try running your script after running the script command, which will log everything written to the terminal; if you wade through all the control characters in the file, you may notice the prompt in the file just prior to the output written by tee. In support of my race condition theory, I'll note that after running the script a few times, it was no longer displaying "abnormal" behavior; my shell prompt was displayed as expected after the string "output", so there is definitely some non-deterministic element to this situation.
#chepner's answer provides great background information.
Here's a workaround - works on Ubuntu 12.04 (Linux 3.2.0) and on OS X 10.9.1:
#!/bin/bash
exec > >(tee logfile)
echo "output"
# WORKAROUND - place LAST in your script.
# Execute an executable (as opposed to a builtin) that outputs *something*
# to make the prompt reappear normally.
# In this case we use the printf *executable* to output an *empty string*.
# Use of `$ec` is to ensure that the script's actual exit code is passed through.
ec=$?; $(which printf) ''; exit $ec
Alternatives:
#user2719058's answer shows a simple alternative: wrapping the entire script body in a group command ({ ... }) and piping it to tee logfile.
An external solution, as #chepner has already hinted at, is to use the script utility to create a "transcript" of your script's output in addition to displaying it:
script -qc yourScript /dev/null > logfile # Linux syntax
This, however, will also capture stderr output; if you wanted to avoid that, use:
script -qc 'yourScript 2>/dev/null' /dev/null > logfile
Note, however, that this will suppress stderr output altogether.
As others have noted, it's not that there's no prompt printed -- it's that the last of the output written by tee can come after the prompt, making the prompt no longer visible.
If you have bash 4.4 or newer, you can wait for your tee process to exit, like so:
#!/usr/bin/env bash
case $BASH_VERSION in ''|[0-3].*|4.[0-3]) echo "ERROR: Bash 4.4+ needed" >&2; exit 1;; esac
exec {orig_stdout}>&1 {orig_stderr}>&2 # make a backup of original stdout
exec > >(tee -a "_install_log"); tee_pid=$! # track PID of tee after starting it
cleanup() { # define a function we'll call during shutdown
retval=$?
exec >&$orig_stdout # Copy your original stdout back to FD 1, overwriting the pipe to tee
exec 2>&$orig_stderr # If something overwrites stderr to also go through tee, fix that too
wait "$tee_pid" # Now, wait until tee exits
exit "$retval" # and complete exit with our original exit status
}
trap cleanup EXIT # configure the function above to be called during cleanup
echo "Writing something to stdout here"
I'm looking at https://stackoverflow.com/a/10225050/1737158
And in same Q there is an answer with timeout command but it's not in all OSes, so I want to avoid it.
What I try to do is:
demo="$(top)" &
TASK_PID=$!
sleep 3
echo "TASK_PID: $TASK_PID"
echo "demo: $demo"
And I expect to have nothing in $demo variable while top command never ends.
Now I get an empty result. Which is "acceptable" but when i re-use the same thing with the command which should return value, I still get an empty result, which is not ok. E.g.:
demo="$(uptime)" &
TASK_PID=$!
sleep 3
echo "TASK_PID: $TASK_PID"
echo "demo: $demo"
This should return uptime result but it doesn't. I also tried to kill the process by TASK_PID but I always get. If a command fails, I expect to have stderr captures somehow. It can be in different variable but it has to be captured and not leaked out.
What happens when you execute var=$(cmd) &
Let's start by noting that the simple command in bash has the form:
[variable assignments] [command] [redirections]
for example
$ demo=$(echo 313) declare -p demo
declare -x demo="313"
According to the manual:
[..] the text after the = in each variable assignment undergoes tilde expansion, parameter expansion, command substitution, arithmetic expansion, and quote removal before being assigned to the variable.
Also, after the [command] above is expanded, the first word is taken to be the name of the command, but:
If no command name results, the variable assignments affect the current shell environment. Otherwise, the variables are added to the environment of the executed command and do not affect the current shell environment.
So, as expected, when demo=$(cmd) is run, the result of $(..) command substitution is assigned to the demo variable in the current shell.
Another point to note is related to the background operator &. It operates on the so called lists, which are sequences of one or more pipelines. Also:
If a command is terminated by the control operator &, the shell executes the command asynchronously in a subshell. This is known as executing the command in the background.
Finally, when you say:
$ demo=$(top) &
# ^^^^^^^^^^^ simple command, consisting ONLY of variable assignment
that simple command is executed in a subshell (call it s1), inside which $(top) is executed in another subshell (call it s2), the result of this command substitution is assigned to variable demo inside the shell s1. Since no commands are given, after variable assignment, s1 terminates, but the parent shell never receives the variables set in child (s1).
Communicating with a background process
If you're looking for a reliable way to communicate with the process run asynchronously, you might consider coprocesses in bash, or named pipes (FIFO) in other POSIX environments.
Coprocess setup is simpler, since coproc will setup pipes for you, but note you might not reliably read them if process is terminated before writing any output.
#!/bin/bash
coproc top -b -n3
cat <&${COPROC[0]}
FIFO setup would look something like this:
#!/bin/bash
# fifo setup/clean-up
tmp=$(mktemp -td)
mkfifo "$tmp/out"
trap 'rm -rf "$tmp"' EXIT
# bg job, terminates after 3s
top -b >"$tmp/out" -n3 &
# read the output
cat "$tmp/out"
but note, if a FIFO is opened in blocking mode, the writer won't be able to write to it until someone opens it for reading (and starts reading).
Killing after timeout
How you'll kill the background process depends on what setup you've used, but for a simple coproc case above:
#!/bin/bash
coproc top -b
sleep 3
kill -INT "$COPROC_PID"
cat <&${COPROC[0]}
I have a problem with the nohup command.
When I run my job, I have a lot of data. The output nohup.out becomes too large and my process slows down. How can I run this command without getting nohup.out?
The nohup command only writes to nohup.out if the output would otherwise go to the terminal. If you have redirected the output of the command somewhere else - including /dev/null - that's where it goes instead.
nohup command >/dev/null 2>&1 # doesn't create nohup.out
Note that the >/dev/null 2>&1 sequence can be abbreviated to just >&/dev/null in most (but not all) shells.
If you're using nohup, that probably means you want to run the command in the background by putting another & on the end of the whole thing:
nohup command >/dev/null 2>&1 & # runs in background, still doesn't create nohup.out
On Linux, running a job with nohup automatically closes its input as well. On other systems, notably BSD and macOS, that is not the case, so when running in the background, you might want to close input manually. While closing input has no effect on the creation or not of nohup.out, it avoids another problem: if a background process tries to read anything from standard input, it will pause, waiting for you to bring it back to the foreground and type something. So the extra-safe version looks like this:
nohup command </dev/null >/dev/null 2>&1 & # completely detached from terminal
Note, however, that this does not prevent the command from accessing the terminal directly, nor does it remove it from your shell's process group. If you want to do the latter, and you are running bash, ksh, or zsh, you can do so by running disown with no argument as the next command. That will mean the background process is no longer associated with a shell "job" and will not have any signals forwarded to it from the shell. (A disowned process gets no signals forwarded to it automatically by its parent shell - but without nohup, it will still receive a HUP signal sent via other means, such as a manual kill command. A nohup'ed process ignores any and all HUP signals, no matter how they are sent.)
Explanation:
In Unixy systems, every source of input or target of output has a number associated with it called a "file descriptor", or "fd" for short. Every running program ("process") has its own set of these, and when a new process starts up it has three of them already open: "standard input", which is fd 0, is open for the process to read from, while "standard output" (fd 1) and "standard error" (fd 2) are open for it to write to. If you just run a command in a terminal window, then by default, anything you type goes to its standard input, while both its standard output and standard error get sent to that window.
But you can ask the shell to change where any or all of those file descriptors point before launching the command; that's what the redirection (<, <<, >, >>) and pipe (|) operators do.
The pipe is the simplest of these... command1 | command2 arranges for the standard output of command1 to feed directly into the standard input of command2. This is a very handy arrangement that has led to a particular design pattern in UNIX tools (and explains the existence of standard error, which allows a program to send messages to the user even though its output is going into the next program in the pipeline). But you can only pipe standard output to standard input; you can't send any other file descriptors to a pipe without some juggling.
The redirection operators are friendlier in that they let you specify which file descriptor to redirect. So 0<infile reads standard input from the file named infile, while 2>>logfile appends standard error to the end of the file named logfile. If you don't specify a number, then input redirection defaults to fd 0 (< is the same as 0<), while output redirection defaults to fd 1 (> is the same as 1>).
Also, you can combine file descriptors together: 2>&1 means "send standard error wherever standard output is going". That means that you get a single stream of output that includes both standard out and standard error intermixed with no way to separate them anymore, but it also means that you can include standard error in a pipe.
So the sequence >/dev/null 2>&1 means "send standard output to /dev/null" (which is a special device that just throws away whatever you write to it) "and then send standard error to wherever standard output is going" (which we just made sure was /dev/null). Basically, "throw away whatever this command writes to either file descriptor".
When nohup detects that neither its standard error nor output is attached to a terminal, it doesn't bother to create nohup.out, but assumes that the output is already redirected where the user wants it to go.
The /dev/null device works for input, too; if you run a command with </dev/null, then any attempt by that command to read from standard input will instantly encounter end-of-file. Note that the merge syntax won't have the same effect here; it only works to point a file descriptor to another one that's open in the same direction (input or output). The shell will let you do >/dev/null <&1, but that winds up creating a process with an input file descriptor open on an output stream, so instead of just hitting end-of-file, any read attempt will trigger a fatal "invalid file descriptor" error.
nohup some_command > /dev/null 2>&1&
That's all you need to do!
Have you tried redirecting all three I/O streams:
nohup ./yourprogram > foo.out 2> foo.err < /dev/null &
You might want to use the detach program. You use it like nohup but it doesn't produce an output log unless you tell it to. Here is the man page:
NAME
detach - run a command after detaching from the terminal
SYNOPSIS
detach [options] [--] command [args]
Forks a new process, detaches is from the terminal, and executes com‐
mand with the specified arguments.
OPTIONS
detach recognizes a couple of options, which are discussed below. The
special option -- is used to signal that the rest of the arguments are
the command and args to be passed to it.
-e file
Connect file to the standard error of the command.
-f Run in the foreground (do not fork).
-i file
Connect file to the standard input of the command.
-o file
Connect file to the standard output of the command.
-p file
Write the pid of the detached process to file.
EXAMPLE
detach xterm
Start an xterm that will not be closed when the current shell exits.
AUTHOR
detach was written by Robbert Haarman. See http://inglorion.net/ for
contact information.
Note I have no affiliation with the author of the program. I'm only a satisfied user of the program.
Following command will let you run something in the background without getting nohup.out:
nohup command |tee &
In this way, you will be able to get console output while running script on the remote server:
sudo bash -c "nohup /opt/viptel/viptel_bin/log.sh $* &> /dev/null" &
Redirecting the output of sudo causes sudo to reask for the password, thus an awkward mechanism is needed to do this variant.
If you have a BASH shell on your mac/linux in-front of you, you try out the below steps to understand the redirection practically :
Create a 2 line script called zz.sh
#!/bin/bash
echo "Hello. This is a proper command"
junk_errorcommand
The echo command's output goes into STDOUT filestream (file descriptor 1).
The error command's output goes into STDERR filestream (file descriptor 2)
Currently, simply executing the script sends both STDOUT and STDERR to the screen.
./zz.sh
Now start with the standard redirection :
zz.sh > zfile.txt
In the above, "echo" (STDOUT) goes into the zfile.txt. Whereas "error" (STDERR) is displayed on the screen.
The above is the same as :
zz.sh 1> zfile.txt
Now you can try the opposite, and redirect "error" STDERR into the file. The STDOUT from "echo" command goes to the screen.
zz.sh 2> zfile.txt
Combining the above two, you get:
zz.sh 1> zfile.txt 2>&1
Explanation:
FIRST, send STDOUT 1 to zfile.txt
THEN, send STDERR 2 to STDOUT 1 itself (by using &1 pointer).
Therefore, both 1 and 2 goes into the same file (zfile.txt)
Eventually, you can pack the whole thing inside nohup command & to run it in the background:
nohup zz.sh 1> zfile.txt 2>&1&
You can run the below command.
nohup <your command> & > <outputfile> 2>&1 &
e.g.
I have a nohup command inside script
./Runjob.sh > sparkConcuurent.out 2>&1
In bash 1:
$ mkfifo /tmp/pipe
$ echo 'something' > /tmp/pipe
Now it hangs and waits that data to be read.
In bash 2:
$ </tmp/pipe
Now shell 1 goes away, it is closed, my terminal is gone.
Why is this happening?
In bash manual there is written
The command substitution $(cat file) can be replaced by the
equivalent but faster $(< file).
So I was experimenting if plain "< file" works in a similar way to cat file content to terminal.
$ bash --version | head -1
GNU bash, version 4.3.11(1)-release (x86_64-pc-linux-gnu)
$ cat /proc/version
Linux version 3.16.0-71-generic (buildd#lgw01-46) (gcc version 4.8.2 (Ubuntu 4.8.2-19ubuntu1) ) #92~14.04.1-Ubuntu SMP Thu May 12 23:31:46 UTC 2016
Edit
After seeing initial comments and answers I will add a bit of clarification.
I'm not concerned about different command line syntaxes.
But what I was really after was that in reader shell $ < /tmp/pipe scenario writer shell exits, but with $ cat /tmp/pipe in reader shell the writer shell does not exit. Why?
I see that I really did not phrase that in question and in body and should probably initiate another question?
From the pipe(7) manual page:
If all file descriptors referring to the read end of a pipe have been closed, then a write(2) will cause a SIGPIPE signal to be generated for the calling process.
What happens is that when the reading shell has finished reading and closes its end of the pipe, the writing shell will receive the SIGPIPE signal, and if it doesn't catch it then the shell will be terminated.
In manual sign $ is connected with variable not command prompt.
Try the following scripts:
1)
#!/bin/bash
echo $(< /tmp/pipe);
2)
#!/bin/bash
echo $(cat /tmp/pipe);
Both works correctly.
When you type < /tmp/pipe, you connect the standard input of the current shell to the named pipe instead. bash works by continuously reading from its input and executing what it reads as a command.
In shell 1, echo something > /tmp/pipe opens the pipe for writing, writes the string, then blocks until something reads it. As soon as echo completes, it will close its end of the pipe.
< /tmp/pipe opens the pipe for reading, and connects it to shell 2's standard input.
Shell 2 reads from the pipe (and tries to execute a command).
Back in shell 1, the echo, having unblocked after the 2nd shell read from the pipe, completes. The write end of the pipe closes.
With the write-end of the pipe closed, shell 2 will get a SIGPIPE when it tries to read another command, then exit.
(An alternate possibility is that shell 2 exits if the command it reads from the pipe and tries to execute causes an error.)
$(< file), on the other hand, is a special case of command substitution. When bash sees that, it simply reads from file itself, rather than spawning a cat process and capturing its output.