I understand that to check for error level in Linux can be done by using $?. The thing is that the $? value is reset if one of the commands is successfully performed, even if the previous command failed. The code I use for testing as below:
cd /vobs/test2/test3
if [ $2 = "R" ]; then
mv missing ~/missing2
echo "Created"
Assuming that mv missing ~/missing2 failed the $? should be equal to 1 but due to the last command echo "Created" is performed the $? will be equal to 0. How to perform a scan for the code above so that the moment that $?=1 it will execute exit 1 command. I can perform if else for every command execute but it is not the best way to perform, is it? I need some advice on this, please.
You can exit on the first failing command using set -e. This is part of an unofficial "strict bash mode":
set -e
To exit on all errors.
set -o pipefail
To fail on the first issue in a pipeline rather than only last one.
set -u
To fail on access to undefined variables.
I am trying to create a linux shell script that compiles a program in java and runs it only if there are no errors from the compiler. This is my attempt, which doesn't work, and runs the program every time even if there are errors:
#!/bin/bash
cd ~/Desktop/Foo/src
compilerOutput=$(javac Foo.java)
if ["$compilerOutput" = ""]; then
java Foo
fi
Like any decent Unix tool, javac should set its exit code to non-zero if there is an error.
javac Foo.java && java Foo
Behind the scenes, the shell examines the exit code (which is available as $? if you want to examine it explicitly) and takes the next action only if it is zero.
If you want more complex logic, you can say
if javac Foo.java; then
date +"%c success" >>logfile.log
java Foo
else
date +"%c fail (exit code $?)" >>logfile.log
echo "Oh dear" | wall
fi
The root cause of problem is
compilerOutput=$(javac Foo.java)
When you use command substitution, only the standard output get assigned to compilerOutput
For javac there is no standard output, that is the standard output is always empty, What you see on the screen in standard error
There for even when there is no errors in the program or not
if ["$compilerOutput" = ""]; then
will always be true as the standard output is always empty
Solution
The proper way of doing this is to get the exit status of the previous command through the shell variable $?
Code
#!/bin/bash
cd ~/Desktop/Foo/src
javac Foo.java 2> errors
if [ $? = "0" ]; then
java Foo
fi
Changes made
javac Foo.java 2> errors
This statement redirects bstandard error to errors file
if [ $? = "0" ]; then
Checks if the exit status of previous command, javac Foo.java is 0 (Successfull compilation)
I have an bash script "build.sh" like this:
# load Xilinx environment settings
source $XILINX/../settings32.sh
cp -r "../../../EDK/platform" "hw_platform"
if [ $# -ne 0 ]; then
cp $1/system.xml hw_platform/system.xml
fi
echo "Done"
Normally I run it as "./build.sh" and it execute the "source" statement to set environment variables correct. Sometimes I need to let the script to copy file from an alternative place, I run it as "./build.sh ~/alternative_path/"; My script check whether there is an cmd line argument by checking $# against 0.
When I do that, the "source" statement at the beginning of the script somehow get skipped, and build failed. I have put two "echo" before and after the "source", and I see echo statements get executed.
Currently I circumvent this issue by "source $XILINX/../settings32.sh; build.sh". However, please advise what I have done wrong in the script? Thanks.
Try storing the values of your positional paramaters first on an array variable then reset them to 0. "$XILINX/../settings32.sh" may be acting differently when it detects some arguments.
# Store arguments.
ARGS=("$#")
# Reset to 0 arguments.
set --
# load Xilinx environment settings
source "$XILINX/../settings32.sh"
cp -r "../../../EDK/platform" "hw_platform"
if [[ ${#ARGS[#]} -ne 0 ]]; then
cp "${ARGS[0]}/system.xml" hw_platform/system.xml
fi
echo "Done"
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.
This question already has answers here:
Aborting a shell script if any command returns a non-zero value
(10 answers)
Closed 1 year ago.
I am a noob in shell-scripting. I want to print a message and exit my script if a command fails. I've tried:
my_command && (echo 'my_command failed; exit)
but it does not work. It keeps executing the instructions following this line in the script. I'm using Ubuntu and bash.
Try:
my_command || { echo 'my_command failed' ; exit 1; }
Four changes:
Change && to ||
Use { } in place of ( )
Introduce ; after exit and
spaces after { and before }
Since you want to print the message and exit only when the command fails ( exits with non-zero value) you need a || not an &&.
cmd1 && cmd2
will run cmd2 when cmd1 succeeds(exit value 0). Where as
cmd1 || cmd2
will run cmd2 when cmd1 fails(exit value non-zero).
Using ( ) makes the command inside them run in a sub-shell and calling a exit from there causes you to exit the sub-shell and not your original shell, hence execution continues in your original shell.
To overcome this use { }
The last two changes are required by bash.
The other answers have covered the direct question well, but you may also be interested in using set -e. With that, any command that fails (outside of specific contexts like if tests) will cause the script to abort. For certain scripts, it's very useful.
If you want that behavior for all commands in your script, just add
set -e
set -o pipefail
at the beginning of the script. This pair of options tell the bash interpreter to exit whenever a command returns with a non-zero exit code. (For more details about why pipefail is needed, see http://petereisentraut.blogspot.com/2010/11/pipefail.html)
This does not allow you to print an exit message, though.
Note also, each command's exit status is stored in the shell variable $?, which you can check immediately after running the command. A non-zero status indicates failure:
my_command
if [ $? -eq 0 ]
then
echo "it worked"
else
echo "it failed"
fi
I've hacked up the following idiom:
echo "Generating from IDL..."
idlj -fclient -td java/src echo.idl
if [ $? -ne 0 ]; then { echo "Failed, aborting." ; exit 1; } fi
echo "Compiling classes..."
javac *java
if [ $? -ne 0 ]; then { echo "Failed, aborting." ; exit 1; } fi
echo "Done."
Precede each command with an informative echo, and follow each command with that same
if [ $? -ne 0 ];... line. (Of course, you can edit that error message if you want to.)
Provided my_command is canonically designed, ie returns 0 when succeeds, then && is exactly the opposite of what you want. You want ||.
Also note that ( does not seem right to me in bash, but I cannot try from where I am. Tell me.
my_command || {
echo 'my_command failed' ;
exit 1;
}
You can also use, if you want to preserve exit error status, and have a readable file with one command per line:
my_command1 || exit
my_command2 || exit
This, however will not print any additional error message. But in some cases, the error will be printed by the failed command anyway.
The trap shell builtin allows catching signals, and other useful conditions, including failed command execution (i.e., a non-zero return status). So if you don't want to explicitly test return status of every single command you can say trap "your shell code" ERR and the shell code will be executed any time a command returns a non-zero status. For example:
trap "echo script failed; exit 1" ERR
Note that as with other cases of catching failed commands, pipelines need special treatment; the above won't catch false | true.
Using exit directly may be tricky as the script may be sourced from other places (e.g. from terminal). I prefer instead using subshell with set -e (plus errors should go into cerr, not cout) :
set -e
ERRCODE=0
my_command || ERRCODE=$?
test $ERRCODE == 0 ||
(>&2 echo "My command failed ($ERRCODE)"; exit $ERRCODE)