I trying to write simple output logger. And it's just refuse to work. I can swear, it worked once and it was beautiful.
It's practice, so I don't want to use pre-build bash tools. (like script)
Code:
#!/bin/bash
# create_log.sh
exec 6>&1
exec &> log
s
a=0
while true
do
sleep 1
echo love
((a++))
if [ "$a" -eq 1000 ]
then
break
fi
done
exec 1>&6 6>&-
echo "Stopped doing love"
I run this script in console . /create_log.sh &
And as long as the cycle turns, stdout and stderr should be redirected to log file. But they simply doesn't.
Log file full of love, but I simply can not get date. (or any other output from console)
P.S. If I just type exec > log in console it's work perfectly.
An approach needs to be run natively in the shell for which you intend to redirect output, not in any subprocess of that shell. Running anything with a & as the command separating it from the next command puts it in a subprocess, rather than running in the shell itself.
Consider this pair of functions (for bash 4.1 or newer):
# for this example, consider this content to belong to file-with-functions.bash
start_redir() {
exec {orig_stdout}>&1
exec > >(tee log >&$orig_stdout)
}
end_redir() {
[[ $orig_stdout ]] || {
echo "Not redirected with start_redir previously" >&2
return 1
}
exec 1>&$orig_stdout
exec {orig_stdout}>&-
}
...this can be used as follows:
. ./file-with-functions.bash # source these functions into the current shell; no &
start_redir
ls
end_redir
You can put these functions in a file that you source, but that sourcing needs to be done in the foreground, as putting anything in the background makes it happen in a subprocess, not the shell you're using itself.
I am new to bash scripting and learning through some examples. One of the examples that I saw is using an if-statement to test if a previously assigned output file is valid, like this:
if [ -n "$outputFile" ] && ! 2>/dev/null : >> $outputFile ; then
exit 1
fi
I understand what [ -n "$outputFile" ] does but not the rest of the conditional. Can someone explain what ! 2>/dev/null : >> $outputFile mean/does?
I have googled for answers but most links found were explanations on I/O redirection, which are definitely relevant but still unclear about the ! : >> structure.
That's some oddly written code!
The : command is built into bash. It's equivalent to true; it does nothing, successfully.
: >> $outputFile
attempts to do nothing, and appends the (empty) output to $outputFile -- which has already been confirmed to be a non-empty string. The >> redirection operator will create the file if it doesn't already exist.
I/O redirections such as 2>/dev/null can appear anywhere in a simple command; they don't have to be at the end. So the stdout of the : command (which is empty) is appended to $outputFile, and any stderr output is redirected to /dev/null. Any such stderr output would be the result of a failure in the redirection, since the : command itself does nothing and won't fail to do so. I don't know why the redirection of stdout (onto the end of $outputFile and the redirection of stderr (to /dev/null) are on opposite sides of the : command.
The ! operator is a logical "not"; it checks whether the following command succeeded, and inverts the result.
The net result, written in English-ish text is:
if "$outputFile" is set and is not an empty string, and if we don't have permission to write to it, then terminate the script with a status of 1.
In short, it tests whether we're able to write to $outputFile, and bails out if we don't.
The script is attempting to make sure $outputFile is writable in a not-so-obvious way.
: is the null command in bash, it does nothing. The fact that stderr is redirected to /dev/null is simply to suppress the permission denied error, should one occur.
If the file is not writable, then the command fails, which makes the condition true since it's negated with ! and the script exits.
I am getting a non-deterministic crash in a library I am using which occurs a lot less frequently when the library's full debugging is turned on. I want to run it repeatedly until the program crashes, and then look at the detailed debug (let's assume that unit-test-command with args calls the code I am interested in)
This is the code I have in my script:
#!/bin/bash
while [[ $(unit-test-command with args) == 0 ]]
do
echo ""
done
However, not only does it only go through the loop once irrespective of the return value of the command (which is non-zero when it crashes), but it also only displays the output of my program, and not the output of the library debugging.
What I am doing wrong?
$(command) expands to the console output, not to return code. For example, uname returns 0 and $(uname) returns "linux".
Try so:
while unit-test-command with args ; do : ; done
#!/bin/bash
while [ true ];do
unit-test-command with args
if [ $? != 0 ];then
echo "failed"
break
fi
echo "didn't faile
sleep 10
done
I dont know if it is weird that read is not taking the input from the terminal.
The configure script, which is used in source code making process, should ask the user to give the input to select the type of Database either MYSQL or ORACLE(below is the code).
MYSQLLIBPATH="/usr/lib/mysql"
echo "Enter DataBase-Type 1-ORACLE, 2-MySQL (default MySQL):"
read in
echo $? >> /tmp/error.log
if test -z "$in" -o "$in" = "2"
then
DATABASE=-DDB_MYSQL
if true; then
MYSQL_TRUE=
MYSQL_FALSE='#'
else
MYSQL_TRUE='#'
MYSQL_FALSE=
fi
echo "Enter Mysql Library Path: (eg: $MYSQLLIBPATH (default))"
read in
echo $? >> /tmp/error.log
if test -n "$in"
then
MYSQLLIBPATH=`echo $in`
fi
echo "Mysql Lib path is $MYSQLLIBPATH"
else
if false; then
MYSQL_TRUE=
MYSQL_FALSE='#'
else
MYSQL_TRUE='#'
MYSQL_FALSE=
fi
DATABASE=-DDB_ORACLE
LD_PATH=
fi
But, the read command is not asking for the user input. Its failing to take the input from the stdin.
When I checked the status of the command in the error.log it was showing
1
1
Could anyone tell why read is failing to take the input from the stdin.
Are there any builtin variable which can block read taking the input?
Most likely read executes with standard input redirected from a file that has reached EOF. If the above is not the whole of your configure code, check that there are no input redirections. Could the code above be a part of a function which was invoked with some input from a pipe or a file? Otherwise check how configure is executed - are there any redirections?
Otherwise, the universal advice applies: try simplifying and stripping down your code until it is obvious what's happening.
BTW, it is not a good idea to make configure interactive, if you want to have your program packaged for a distribution - it's not easy to control execution of interactive programs. Consider adding support for supplying parameters through command line options.
When I use exit command in a shell script, the script will terminate the terminal (the prompt). Is there any way to terminate a script and then staying in the terminal?
My script run.sh is expected to execute by directly being sourced, or sourced from another script.
EDIT:
To be more specific, there are two scripts run2.sh as
...
. run.sh
echo "place A"
...
and run.sh as
...
exit
...
when I run it by . run2.sh, and if it hit exit codeline in run.sh, I want it to stop to the terminal and stay there. But using exit, the whole terminal gets closed.
PS: I have tried to use return, but echo codeline will still gets executed....
The "problem" really is that you're sourcing and not executing the script. When you source a file, its contents will be executed in the current shell, instead of spawning a subshell. So everything, including exit, will affect the current shell.
Instead of using exit, you will want to use return.
Yes; you can use return instead of exit. Its main purpose is to return from a shell function, but if you use it within a source-d script, it returns from that script.
As §4.1 "Bourne Shell Builtins" of the Bash Reference Manual puts it:
return [n]
Cause a shell function to exit with the return value n.
If n is not supplied, the return value is the exit status of the
last command executed in the function.
This may also be used to terminate execution of a script being executed
with the . (or source) builtin, returning either n or
the exit status of the last command executed within the script as the exit
status of the script.
Any command associated with the RETURN trap is executed
before execution resumes after the function or script.
The return status is non-zero if return is used outside a function
and not during the execution of a script by . or source.
You can add an extra exit command after the return statement/command so that it works for both, executing the script from the command line and sourcing from the terminal.
Example exit code in the script:
if [ $# -lt 2 ]; then
echo "Needs at least two arguments"
return 1 2>/dev/null
exit 1
fi
The line with the exit command will not be called when you source the script after the return command.
When you execute the script, return command gives an error. So, we suppress the error message by forwarding it to /dev/null.
Instead of running the script using . run2.sh, you can run it using sh run2.sh or bash run2.sh
A new sub-shell will be started, to run the script then, it will be closed at the end of the script leaving the other shell opened.
Actually, I think you might be confused by how you should run a script.
If you use sh to run a script, say, sh ./run2.sh, even if the embedded script ends with exit, your terminal window will still remain.
However if you use . or source, your terminal window will exit/close as well when subscript ends.
for more detail, please refer to What is the difference between using sh and source?
This is just like you put a run function inside your script run2.sh.
You use exit code inside run while source your run2.sh file in the bash tty.
If the give the run function its power to exit your script and give the run2.sh
its power to exit the terminator.
Then of cuz the run function has power to exit your teminator.
#! /bin/sh
# use . run2.sh
run()
{
echo "this is run"
#return 0
exit 0
}
echo "this is begin"
run
echo "this is end"
Anyway, I approve with Kaz it's a design problem.
I had the same problem and from the answers above and from what I understood what worked for me ultimately was:
Have a shebang line that invokes the intended script, for example,
#!/bin/bash uses bash to execute the script
I have scripts with both kinds of shebang's. Because of this, using sh or . was not reliable, as it lead to a mis-execution (like when the script bails out having run incompletely)
The answer therefore, was
Make sure the script has a shebang, so that there is no doubt about its intended handler.
chmod the .sh file so that it can be executed. (chmod +x file.sh)
Invoke it directly without any sh or .
(./myscript.sh)
Hope this helps someone with similar question or problem.
To write a script that is secure to be run as either a shell script or sourced as an rc file, the script can check and compare $0 and $BASH_SOURCE and determine if exit can be safely used.
Here is a short code snippet for that
[ "X$(basename $0)" = "X$(basename $BASH_SOURCE)" ] && \
echo "***** executing $name_src as a shell script *****" || \
echo "..... sourcing $name_src ....."
I think that this happens because you are running it on source mode
with the dot
. myscript.sh
You should run that in a subshell:
/full/path/to/script/myscript.sh
'source' http://ss64.com/bash/source.html
It's correct that sourced vs. executed scripts use return vs. exit to keep the same session open, as others have noted.
Here's a related tip, if you ever want a script that should keep the session open, regardless of whether or not it's sourced.
The following example can be run directly like foo.sh or sourced like . foo.sh/source foo.sh. Either way it will keep the session open after "exiting". The $# string is passed so that the function has access to the outer script's arguments.
#!/bin/sh
foo(){
read -p "Would you like to XYZ? (Y/N): " response;
[ $response != 'y' ] && return 1;
echo "XYZ complete (args $#).";
return 0;
echo "This line will never execute.";
}
foo "$#";
Terminal result:
$ foo.sh
$ Would you like to XYZ? (Y/N): n
$ . foo.sh
$ Would you like to XYZ? (Y/N): n
$ |
(terminal window stays open and accepts additional input)
This can be useful for quickly testing script changes in a single terminal while keeping a bunch of scrap code underneath the main exit/return while you work. It could also make code more portable in a sense (if you have tons of scripts that may or may not be called in different ways), though it's much less clunky to just use return and exit where appropriate.
Also make sure to return with expected return value. Else if you use exit when you will encounter an exit it will exit from your base shell since source does not create another process (instance).
Improved the answer of Tzunghsing, with more clear results and error re-direction, for silent usage:
#!/usr/bin/env bash
echo -e "Testing..."
if [ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ]; then
echo "***** You are Executing $0 in a sub-shell."
exit 0
else
echo "..... You are Sourcing $BASH_SOURCE in this terminal shell."
return 0
fi
echo "This should never be seen!"
Or if you want to put this into a silent function:
function sExit() {
# Safe Exit from script, not closing shell.
[ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ] && exit 0 || return 0
}
...
# ..it have to be called with an error check, like this:
sExit && return 0
echo "This should never be seen!"
Please note that:
if you have enabled errexit in your script (set -e) and you return N with N != 0, your entire script will exit instantly. To see all your shell settings, use, set -o.
when used in a function, the 1st return 0 is exiting the function, and the 2nd return 0 is exiting the script.
if your terminal emulator doesn't have -hold you can sanitize a sourced script and hold the terminal with:
#!/bin/sh
sed "s/exit/return/g" script >/tmp/script
. /tmp/script
read
otherwise you can use $TERM -hold -e script
If a command succeeded successfully, the return value will be 0. We can check its return value afterwards.
Is there a “goto” statement in bash?
Here is some dirty workaround using trap which jumps only backwards.
#!/bin/bash
set -eu
trap 'echo "E: failed with exitcode $?" 1>&2' ERR
my_function () {
if git rev-parse --is-inside-work-tree > /dev/null 2>&1; then
echo "this is run"
return 0
else
echo "fatal: not a git repository (or any of the parent directories): .git"
goto trap 2> /dev/null
fi
}
my_function
echo "Command succeeded" # If my_function failed this line is not printed
Related:
https://stackoverflow.com/a/19091823/2402577
How to use $? and test to check function?
I couldn't find solution so for those who want to leave the nested script without leaving terminal window:
# this is just script which goes to directory if path satisfies regex
wpr(){
leave=false
pwd=$(pwd)
if [[ "$pwd" =~ ddev.*web ]]; then
# echo "your in wordpress instalation"
wpDir=$(echo "$pwd" | grep -o '.*\/web')
cd $wpDir
return
fi
echo 'please be in wordpress directory'
# to leave from outside the scope
leave=true
return
}
wpt(){
# nested function which returns $leave variable
wpr
# interupts the script if $leave is true
if $leave; then
return;
fi
echo 'here is the rest of the script, executes if leave is not defined'
}
I have no idea whether this is useful for you or not, but in zsh, you can exit a script, but only to the prompt if there is one, by using parameter expansion on a variable that does not exist, as follows.
${missing_variable_ejector:?}
Though this does create an error message in your script, you can prevent it with something like the following.
{ ${missing_variable_ejector:?} } 2>/dev/null
1) exit 0 will come out of the script if it is successful.
2) exit 1 will come out of the script if it is a failure.
You can try these above two based on ur req.