Unix redirection:
Recently I faced an issue where one of the script was using the command below to execute it in background. The issue was that the script was executing twice when it is started.
For example:
In the script I put an echo "Hello" to print to log file. When the script executed I saw in the log file that it printed twice at the same time. Can any one tell me what caused here to execute the script twice.
nohup <runScript> </dev/null 1>&- 2>&- &
The original version of your question was slightly confusing. The subject line asks about (with command and argument inferred):
somecmd arg1 </dev/null 1>&- 2>&- &
The body of the question appeared to ask about:
nohup &- 2>&- &
which could reasonably be inferred to mean:
nohup somecmd arg1 &- 2>&- &
The edited version of your question is also confusing — though the change was just to indent code fragment. The notation <runscript> is ill-chosen when you are asking about I/O redirections. I'm guessing that what you wrote as <runscript> is equivalent to me writing somecmd, rather than redirecting standard input from runscript and an ill-formed output redirection. However, the revised
code does at least match the subject line:
nohup runScript </dev/null 1>&- 2>&- &
So, I'll ignore the &- notation (a previous version of this answer did not).
Notation </dev/null 1>&- 2>&- &
The first command line redirects standard input from /dev/null, and closes both standard output and standard error and executes the command in background. Redirecting from /dev/null is good; closing standard output and standard error is not so good — programs are entitled to have those three file descriptors open, and that can be done by redirecting to /dev/null too:
somecmd arg1 </dev/null >/dev/null 2>&1 &
or:
somecmd arg1 </dev/null >/dev/null 2>/dev/null &
There is not much difference between these two.
Double running
There is nothing in any of the code that would account for the script being run twice, or the output appearing in a log file twice. Since you have not shown the script that was run, we cannot deduce any cause from that. On the whole, the charge would be 'operator error' — you managed to run the command twice. If you want us to look into that, you'll have to provide a reproducible script that:
Shows the script to be run.
Empties the log file.
Runs the script once with your chosen notation.
Shows that the log file contains two entries.
Without such a reproducible script, there's nothing anyone can do to help you.
Related
I have a simple bash script that launches an executable in the background and redirects stdout + stderr to a log file:
#!/usr/bin/bash
myexec >& logfile &
It works. However, output from myexec isn't the only thing that gets redirected: any messages that bash emits while attempting to invoke myexec are also going to logfile. To wit, if bash doesn't find myexec, I don't get to see the myexec: No such file or directory error because it went straight to logfile instead of to the terminal. This behavior annoys me because I end up not knowing whether the script succeeded in starting up myexec.
It occurs to me that the script could just test for the existence of myexec before trying to invoke it, but I'm wondering whether there isn't a way to do the redirection itself in such a way that only myexec's output, and not the shell's, gets redirected.
It's not possible to separate the outputs in the way the OP describes. As Charles Duffy explains in his comment, the system call that opens (or fails to open) the executable myexec takes place after Bash has forked a new process, at which point all of the I/O redirection has already been set up. There is, however, a workaround that suffices for the purpose stated in the OP, namely, "knowing whether the script succeeded in starting up myexec":
myexec > logfile 2>&1 && echo "ok" >&2 || echo "nope." >&2
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 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
I frequently execute from a shell (in my case Bash) commands that I want to fork immediately and whose output I want to ignore. So frequently in fact that I created a script (silent) to do it:
#!/bin/bash
$# &> /dev/null &
I can then run, e.g.
silent inkscape myfile.svg
and my terminal will not be polluted by the debug output of the process I just forked.
I have two questions:
Is there an "official" way of doing this?, i.e. something shorter but equivalent to &> /dev/null & ?
If not, is there a way I can make tab-completion work after my silent command as if it weren't there ? To give an example, after I've typed silent inksc, I'd like bash to auto-complete my command to silent inkscape when I press [tab].
aside: probably want to exec "$#" &> /dev/null & in your silent script, to cause it to discard the sub-shell, and the quotes around "$#" will keep spaces from getting in the way.
As for #2: complete -F _command silent should do something like what you want. (I call my version of that script launch and have complete -F launch in my .bash_profile)
It looks like nohup does more or less what you want. The tab-completion problem is because bash thinks that you are trying to complete a filename as an argument to the script, whereas its completion rules know that nohup takes a command as its first argument.
Nohup redirects stout and stderr to nohup.out and will also leave the command running if your shell exits.
Here's a little script I use for launching interactive (and chatty) X apps from e.g. an xterm
#!/bin/bash
exe="$1"
shift
"$exe" "$#" 2>/tmp/$$."$exe".err 1>&2 & disown $!
No output, won't die if the terminal exits, but in case something goes wrong there's a log of all output in /tmp
If you don't want the log just use /dev/null instead.
Also will work from a function if you're script-alergic.
Perhaps if you could 'rebind' the tab key? An example on superuser Stackoverflow with the enter key is shown. Is this the right idea?
I want to start a script remotely via ssh like this:
ssh user#remote.org -t 'cd my/dir && ./myscript data my#email.com'
The script does various things which work fine until it comes to a line with nohup:
nohup time ./myprog $1 >my.log && mutt -a ${1%.*}/`basename $1` -a ${1%.*}/`basename ${1%.*}`.plt $2 < my.log 2>&1 &
it is supposed to do start the program myprog, pipe its output to mylog and send an email with some datafiles created by myprog as attachment and the log as body. Though when the script reaches this line, ssh outputs:
Connection to remote.org closed.
What is the problem here?
Thanks for any help
Your command runs a pipeline of processes in the background, so the calling script will exit straight away (or very soon afterwards). This will cause ssh to close the connection. That in turn will cause a SIGHUP to be sent to any process attached to the terminal that the -t option caused to be created.
Your time ./myprog process is protected by a nohup, so it should carry on running. But your mutt isn't, and that is likely to be the issue here. I suggest you change your command line to:
nohup sh -c "time ./myprog $1 >my.log && mutt -a ${1%.*}/`basename $1` -a ${1%.*}/`basename ${1%.*}`.plt $2 < my.log 2>&1 " &
so the entire pipeline gets protected. (If that doesn't fix it it may be necessary to do something with file descriptors - for instance mutt may have other issues with the terminal not being around - or the quoting may need tweaking depending on the parameters - but give that a try for now...)
This answer may be helpful. In summary, to achieve the desired effect, you have to do the following things:
Redirect all I/O on the remote nohup'ed command
Tell your local SSH command to exit as soon as it's done starting the remote process(es).
Quoting the answer I already mentioned, in turn quoting wikipedia:
Nohuping backgrounded jobs is for example useful when logged in via SSH, since backgrounded jobs can cause the shell to hang on logout due to a race condition [2]. This problem can also be overcome by redirecting all three I/O streams:
nohup myprogram > foo.out 2> foo.err < /dev/null &
UPDATE
I've just had success with this pattern:
ssh -f user#host 'sh -c "( (nohup command-to-nohup 2>&1 >output.file </dev/null) & )"'
Managed to solve this for a use case where I need to start backgrounded scripts remotely via ssh using a technique similar to other answers here, but in a way I feel is more simple and clean (at least, it makes my code shorter and -- I believe -- better-looking), by explicitly closing all three streams using the stream-close redirection syntax (as discussed at the following locations:
https://unix.stackexchange.com/questions/131801/closing-a-file-descriptor-vs
https://unix.stackexchange.com/questions/70963/difference-between-2-2-dev-null-dev-null-and-dev-null-21
http://www.tldp.org/LDP/abs/html/io-redirection.html#CFD
https://www.gnu.org/software/bash/manual/html_node/Redirections.html
Rather than the more widely used but (IMHO) hackier "redirect to/from /dev/null", resulting in the deceptively simple:
nohup script.sh >&- 2>&- <&-&
2>&1 works just as well as 2>&-, but I feel the latter is ever-so-slightly more clear. ;) Most people might have a space preceding the final "background job" ampersand, but since it is not required (as the ampersand itself functions like a semicolon in normal usage), I prefer to omit it. :)