I currently have a bunch of installer scripts which log stderr/stdout to a log file that works well but I need to also redirect stdin for user responses to the same log file. The install scripts sometime call functions in a shared library (an include), which may also read user input. I thought about adding a custom read function but this will require altering the shared library and wondered if there's a way to do this from the calling script.
At the moment the scripts are similar to this:
#!/usr/bin/bash
. ./libInstall
INSTALL_LOG="./install.log"
( (
echo "INFO: Installing..."
# Run some arbitrary commands...
# Read some input...
read ANSWER1
read ANSWER2
# Call function in libInstall which will prompt the user...
funcWhichAsksAQuestion ANSWER3
echo "INFO: Installation Complete"
) 2>&1 ) | tee -a "${INSTALL_LOG}"
If I change "( (" to reflect the line below I can tee off stdin to the log file:
cat - 2> /dev/null | tee -a ${INSTALL_LOG} | ( (
This works but requires 2 carriage returns once the script ends, presumably because the pipe is broken.
It's almost there but I'd it to work without having to press enter twice at the end to get back to the shell prompt.
These scripts have to be fairly portable to work on RHEL >=5, AIX >=5.1, Solaris >=9 with the lowest bash version being v2.05 I believe.
Any ideas how I can achieve this?
Thanks
Why not just add 'echo "\n\n"' after your "installation complete" line? Granted, you'll have two extra lines in your log file, but those seem relatively harmless.
I believe you have to return twice because of how tee is implemented. It "uses" one return by itself, and the other two come from the 'read' calls (well, one read, one funcWhichAsksAQuestion).
Related
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"
In an interactive bash script I use
exec > >(tee -ia logfile.log)
exec 2>&1
to write the scripts output to a logfile. However, if I ask the user to input something this is not written to this file:
read UserInput
Also, I issue commands with $UserInput as parameter. These command are also not written to the logfile.
The logfile should contain everything my script does, i.e. what the user entered interactively and also the resulting commands along with their output.
Of course I could use set -x and/or echo "user input: "$UserInput, but this would also be sent to the "screen". I dont want to read anything else on the screen except what my script or the commands echo.
How can this be done?
I am trying to modify a script someone created for unix in shell. This script is mostly used to run on backed servers with no human interaction, however I needed to make another script to allow users to input information. So, it is just modifying to old version for user input. But the biggest issue I am running into is trying to get both error logs and echos to be saved in a log file. The script has a lot of them, but I wanted to have those shown on the terminal as well as send them to the log file specified, to be looked into later.
What I have is this:
exec 1> ${LOG} 2>&1
This line is pretty much send everything to the log file. That is all good, but I also have people trying to enter in information in the script, and it is sending everything to the log file including the echo needed for the prompt. This line is also at the beginning of the script, but reading more into the stderr and stdout messages. I tried:
exec 2>&1 1>>${LOG}
exec 1 | tee ${LOG} But only getting error when running it this "./bash_pam.sh: line 39: exec: 1: not found"
I have went over site such as this to solve the issue, but I am not understanding why it does not print to both. The way I insert it, it either only sends it to the log location and not to the terminal, or it sends it to the terminal, but nothing is persevered in the log.
EDIT: Some of the solutions, for this have mentioned that certain fixes will work in bash, but not in /bin/sh.
If you would like all output to be printed onto the console, while also being printed to a logfile.txt you would run this command on your script:
bash your_script.sh 2>&1 | tee -a logfile.txt
Or calling it within the file:
<bash_command> 2>&1 | tee -a logfile.txt
To append to logfile.txt instead of overwriting, add the -a option to tee.
I am writing a script that among other things runs a shell command several times. This command doesn't handle exit codes very well and I need to know if the process ended successfully or not.
So what I was thinking is to analyze the stderr to find out the word error (using grep). I know this is not the best thing to do, I'm working on it....
Anyway, the only way I can imagine is to put the stderr of that program in a variable and then use grep to well, "grep" it and throw it to another variable. Then I can see if that variable is valorized, meaning that there was an error, and do my work.
The qustion is: how can I do this ?
I don't really want to run the program inside a variable, because it has got a lot of arguments (with special characters such as backslash, quotes, doublequotes...) and it's a memory and I/O intensive program.
Awaiting your reply, thanks.
Redirect the stderr of that command to a temporary file and check if the word "error" is present in that file.
mycommand 2> /tmp/temp.txt
grep error /tmp/temp.txt
Thanks #Jdamian, this was my answer too, in the end.
I asked my principal if I can write a temp file and it allowed, so this is the end result:
... script
command to be launched -argument -other "argument" -other "other" argument 2>&1 | tee $TEMPFILE
ERRORCODE=( `grep -i error "$TEMPFILE" `)
if [ -z $ERRORCODE ] ;
then
some actions ....
I didn't tested this yet because I got some other scripts involved that I need to write before.
What I'm trying to do is:
run the command, having its stderr redirected to stdout;
using tee, have the above result printed on screen and also to the temp file;
have grep to store the string error found on the temp file (if any) in a variable called ERRORCODE;
if that variable is populated (which mean if it has been created by grep), then the script stops, quitting with status 1, else it contiunes.
What do you think ?
If you don't need the standard output:
if mycommand 2>&1 >/dev/null | grep -q error; then
echo an error occurred
fi
I'm adding some custom logging functionality to a bash script, and can't figure out why it won't take the output from one named pipe and feed it back into another named pipe.
Here is a basic version of the script (http://pastebin.com/RMt1FYPc):
#!/bin/bash
PROGNAME=$(basename $(readlink -f $0))
LOG="$PROGNAME.log"
PIPE_LOG="$PROGNAME-$$-log"
PIPE_ECHO="$PROGNAME-$$-echo"
# program output to log file and optionally echo to screen (if $1 is "-e")
log () {
if [ "$1" = '-e' ]; then
shift
$# > $PIPE_ECHO 2>&1
else
$# > $PIPE_LOG 2>&1
fi
}
# create named pipes if not exist
if [[ ! -p $PIPE_LOG ]]; then
mkfifo -m 600 $PIPE_LOG
fi
if [[ ! -p $PIPE_ECHO ]]; then
mkfifo -m 600 $PIPE_ECHO
fi
# cat pipe data to log file
while read data; do
echo -e "$PROGNAME: $data" >> $LOG
done < $PIPE_LOG &
# cat pipe data to log file & echo output to screen
while read data; do
echo -e "$PROGNAME: $data"
log echo $data # this doesn't work
echo -e $data > $PIPE_LOG 2>&1 # and neither does this
echo -e "$PROGNAME: $data" >> $LOG # so I have to do this
done < $PIPE_ECHO &
# clean up temp files & pipes
clean_up () {
# remove named pipes
rm -f $PIPE_LOG
rm -f $PIPE_ECHO
}
#execute "clean_up" on exit
trap "clean_up" EXIT
log echo "Log File Only"
log -e echo "Echo & Log File"
I thought the commands on line 34 & 35 would take the $data from $PIPE_ECHO and output it to the $PIPE_LOG. But, it doesn't work. Instead I have to send that output directly to the log file, without going through the $PIPE_LOG.
Why is this not working as I expect?
EDIT: I changed the shebang to "bash". The problem is the same, though.
SOLUTION: A.H.'s answer helped me understand that I wasn't using named pipes correctly. I have since solved my problem by not even using named pipes. That solution is here: http://pastebin.com/VFLjZpC3
it seems to me, you do not understand what a named pipe really is. A named pipe is not one stream like normal pipes. It is a series of normal pipes, because a named pipe can be closed and a close on the producer side is might be shown as a close on the consumer side.
The might be part is this: The consumer will read data until there is no more data. No more data means, that at the time of the read call no producer has the named pipe open. This means that multiple producer can feed one consumer only when there is no point in time without at least one producer. Think of it of door which closes automatically: If there is a steady stream of people keeping the door always open either by handing the doorknob to the next one or by squeezing multiple people through it at the same time, the door is open. But once the door is closed it stays closed.
A little demonstration should make the difference a little clearer:
Open three shells. First shell:
1> mkfifo xxx
1> cat xxx
no output is shown because cat has opened the named pipe and is waiting for data.
Second shell:
2> cat > xxx
no output, because this cat is a producer which keeps the named pipe open until we tell him to close it explicitly.
Third shell:
3> echo Hello > xxx
3>
This producer immediately returns.
First shell:
Hello
The consumer received data, wrote it and - since one more consumer keeps the door open, continues to wait.
Third shell
3> echo World > xxx
3>
First shell:
World
The consumer received data, wrote it and - since one more consumer keeps the door open, continues to wait.
Second Shell: write into the cat > xxx window:
And good bye!
(control-d key)
2>
First shell
And good bye!
1>
The ^D key closed the last producer, the cat > xxx, and hence the consumer exits also.
In your case which means:
Your log function will try to open and close the pipes multiple times. Not a good idea.
Both your while loops exit earlier than you think. (check this with (while ... done < $PIPE_X; echo FINISHED; ) &
Depending on the scheduling of your various producers and consumers the door might by slam shut sometimes and sometimes not - you have a race condition built in. (For testing you can add a sleep 1 at the end of the log function.)
You "testcases" only tries each possibility once - try to use them multiple times (you will block, especially with the sleeps ), because your producer might not find any consumer.
So I can explain the problems in your code but I cannot tell you a solution because it is unclear what the edges of your requirements are.
It seems the problem is in the "cat pipe data to log file" part.
Let's see: you use a "&" to put the loop in the background, I guess you mean it must run in parallel with the second loop.
But the problem is you don't even need the "&", because as soon as no more data is available in the fifo, the while..read stops. (still you've got to have some at first for the first read to work). The next read doesn't hang if no more data is available (which would pose another problem: how does your program stops ?).
I guess the while read checks if more data is available in the file before doing the read and stops if it's not the case.
You can check with this sample:
mkfifo foo
while read data; do echo $data; done < foo
This script will hang, until you write anything from another shell (or bg the first one). But it ends as soon as a read works.
Edit:
I've tested on RHEL 6.2 and it works as you say (eg : bad!).
The problem is that, after running the script (let's say script "a"), you've got an "a" process remaining. So, yes, in some way the script hangs as I wrote before (not that stupid answer as I thought then :) ). Except if you write only one log (be it log file only or echo,in this case it works).
(It's the read loop from PIPE_ECHO that hangs when writing to PIPE_LOG and leaves a process running each time).
I've added a few debug messages, and here is what I see:
only one line is read from PIPE_LOG and after that, the loop ends
then a second message is sent to the PIPE_LOG (after been received from the PIPE_ECHO), but the process no longer reads from PIPE_LOG => the write hangs.
When you ls -l /proc/[pid]/fd, you can see that the fifo is still open (but deleted).
If fact, the script exits and removes the fifos, but there is still one process using it.
If you don't remove the log fifo at the cleanup and cat it, it will free the hanging process.
Hope it will help...