I am running a application form the shell script. Now I like to know the exit status of the application to know whether it exit normally or abnormally ( crash etc). How I know it?
Example:
./mytestApp
Bash stores the last process' exit value in the special variable $?.
You can use special variable $? which contains an exit status of the last comand.
$? contains the exit status of the last command executed. So, if the last command was ./mytestapp, $? would contain its exit status immediately after (note that you can only retrieve this value once and it must be retrieved immediately after the command whose exit status you want to know). You may want to capture it in a variable, e.g.
#!/bin/bash
./mytestapp
APPSUCCESS=$?
# Continue doing whatever it is you're doing
This all assumes that you're using bash (sh and zsh will work as well, IIRC).
The special variable $? will contain the exit status of the last command in bash.
Related
What is the difference between the return and exit statement in Bash functions with respect to exit codes?
From man bash on return [n];
Causes a function to stop executing and return the value specified by n to its caller. If n is omitted, the return status is that of the last command executed in the function body.
... on exit [n]:
Cause the shell to exit with a status of n. If n is omitted, the exit status is that of the last command executed. A trap on EXIT is executed before the shell terminates.
EDIT:
As per your edit of the question, regarding exit codes, return has nothing to do with exit codes. Exit codes are intended for applications/scripts, not functions. So in this regard, the only keyword that sets the exit code of the script (the one that can be caught by the calling program using the $? shell variable) is exit.
EDIT 2:
My last statement referring exit is causing some comments. It was made to differentiate return and exit for the understanding of the OP, and in fact, at any given point of a program/shell script, exit is the only way of ending the script with an exit code to the calling process.
Every command executed in the shell produces a local "exit code": it sets the $? variable to that code, and can be used with if, && and other operators to conditionally execute other commands.
These exit codes (and the value of the $? variable) are reset by each command execution.
Incidentally, the exit code of the last command executed by the script is used as the exit code of the script itself as seen by the calling process.
Finally, functions, when called, act as shell commands with respect to exit codes. The exit code of the function (within the function) is set by using return. So when in a function return 0 is run, the function execution terminates, giving an exit code of 0.
return will cause the current function to go out of scope, while exit will cause the script to end at the point where it is called. Here is a sample program to help explain this:
#!/bin/bash
retfunc()
{
echo "this is retfunc()"
return 1
}
exitfunc()
{
echo "this is exitfunc()"
exit 1
}
retfunc
echo "We are still here"
exitfunc
echo "We will never see this"
Output
$ ./test.sh
this is retfunc()
We are still here
this is exitfunc()
I don't think anyone has really fully answered the question because they don't describe how the two are used. OK, I think we know that exit kills the script, wherever it is called and you can assign a status to it as well such as exit or exit 0 or exit 7 and so forth. This can be used to determine how the script was forced to stop if called by another script, etc. Enough on exit.
return, when called, will return the value specified to indicate the function's behavior, usually a 1 or a 0. For example:
#!/bin/bash
isdirectory() {
if [ -d "$1" ]
then
return 0
else
return 1
fi
echo "you will not see anything after the return like this text"
}
Check like this:
if isdirectory $1; then echo "is directory"; else echo "not a directory"; fi
Or like this:
isdirectory || echo "not a directory"
In this example, the test can be used to indicate if the directory was found. Notice that anything after the return will not be executed in the function. 0 is true, but false is 1 in the shell, different from other programming languages.
For more information on functions: Returning Values from Bash Functions
Note: The isdirectory function is for instructional purposes only. This should not be how you perform such an option in a real script.*
Remember, functions are internal to a script and normally return from whence they were called by using the return statement. Calling an external script is another matter entirely, and scripts usually terminate with an exit statement.
The difference "between the return and exit statement in Bash functions with respect to exit codes" is very small. Both return a status, not values per se. A status of zero indicates success, while any other status (1 to 255) indicates a failure. The return statement will return to the script from where it was called, while the exit statement will end the entire script from wherever it is encountered.
return 0 # Returns to where the function was called. $? contains 0 (success).
return 1 # Returns to where the function was called. $? contains 1 (failure).
exit 0 # Exits the script completely. $? contains 0 (success).
exit 1 # Exits the script completely. $? contains 1 (failure).
If your function simply ends without a return statement, the status of the last command executed is returned as the status code (and will be placed in $?).
Remember, return and exit give back a status code from 0 to 255, available in $?. You cannot stuff anything else into a status code (e.g., return "cat"); it will not work. But, a script can pass back 255 different reasons for failure by using status codes.
You can set variables contained in the calling script, or echo results in the function and use command substitution in the calling script; but the purpose of return and exit are to pass status codes, not values or computation results as one might expect in a programming language like C.
Sometimes, you run a script using . or source.
. a.sh
If you include an exit in the a.sh, it will not just terminate the script, but end your shell session.
If you include a return in the a.sh, it simply stops processing the script.
exit terminates the current process; with or without an exit code, consider this a system more than a program function. Note that when sourcing, exit will end the shell. However, when running, it will just exit the script.
return from a function go back to the instruction after the call, with or without a return code. return is optional and it's implicit at the end of the function. return can only be used inside a function.
I want to add that while being sourced, it's not easy to exit the script from within a function without killing the shell. I think, an example is better on a 'test' script:
#!/bin/bash
function die(){
echo ${1:=Something terrible wrong happen}
#... clean your trash
exit 1
}
[ -f /whatever/ ] || die "whatever is not available"
# Now we can proceed
echo "continue"
doing the following:
user$ ./test
Whatever is not available
user$
test -and- the shell will close.
user$ . ./test
Whatever is not available
Only test will finish and the prompt will show.
The solution is to enclose the potentially procedure in ( and ):
#!/bin/bash
function die(){
echo $(1:=Something terrible wrong happen)
#... Clean your trash
exit 1
}
( # Added
[ -f /whatever/ ] || die "whatever is not available"
# Now we can proceed
echo "continue"
) # Added
Now, in both cases only test will exit.
The OP's question:
What is the difference between the return and exit statement in BASH functions with respect to exit codes?
Firstly, some clarification is required:
A (return|exit) statement is not required to terminate execution of a (function|shell). A (function|shell) will terminate when it reaches the end of its code list, even with no (return|exit) statement.
A (return|exit) statement is not required to pass a value back from a terminated (function|shell). Every process has a built-in variable $? which always has a numeric value. It is a special variable that cannot be set like "?=1", but it is set only in special ways (see below *).
The value of $? after the last command to be executed in the (called function | sub shell) is the value that is passed back to the (function caller | parent shell). That is true whether the last command executed is ("return [n]"| "exit [n]") or plain ("return" or something else which happens to be the last command in the called function's code.
In the above bullet list, choose from "(x|y)" either always the first item or always the second item to get statements about functions and return, or shells and exit, respectively.
What is clear is that they both share common usage of the special variable $? to pass values upwards after they terminate.
* Now for the special ways that $? can be set:
When a called function terminates and returns to its caller then $? in the caller will be equal to the final value of $? in the terminated function.
When a parent shell implicitly or explicitly waits on a single sub shell and is released by termination of that sub shell, then $? in the parent shell will be equal to the final value of $? in the terminated sub shell.
Some built-in functions can modify $? depending upon their result. But some don't.
Built-in functions "return" and "exit", when followed by a numerical argument both set $? with their argument, and terminate execution.
It is worth noting that $? can be assigned a value by calling exit in a sub shell, like this:
# (exit 259)
# echo $?
3
In simple words (mainly for newbie in coding), we can say,
`return`: exits the function,
`exit()`: exits the program (called as process while running)
Also if you observed, this is very basic, but...,
`return`: is the keyword
`exit()`: is the function
If you convert a Bash script into a function, you typically replace exit N with return N. The code that calls the function will treat the return value the same as it would an exit code from a subprocess.
Using exit inside the function will force the entire script to end.
Adding an actionable aspect to a few of the other answers:
Both can give exit codes - default or defined by the function, and the only 'default' is zero for success for both exit and return. Any status can have a custom number 0-255, including for success.
Return is used often for interactive scripts that run in the current shell, called with . script.sh for example, and just returns you to your calling shell. The return code is then accessible to the calling shell - $? gives you the defined return status.
Exit in this case also closes your shell (including SSH connections, if that's how you're working).
Exit is necessary if the script is executable and called from another script or shell and runs in a subshell. The exit codes then are accessible to the calling shell - return would give an error in this case.
First of all, return is a keyword and exit is a function.
That said, here's a simplest of explanations.
return
It returns a value from a function.
exit
It exits out of or abandons the current shell.
I am using the bash script wait for it to check if a container is up and running before I launch another container.
https://github.com/vishnubob/wait-for-it
It seems that if it's successful it returns 0 and if not 124. Is it possible to check this value and if it's not successful exit the script?
I've tried
./wait-for-it.sh $BROKER_ADDRESS
echo $?
if ($?==124)
then
echo "exiting as broker service never became available"
exit
fi
And I can see it echos 124 in my terminal but then the check fails and it carries on launching the container. I'm assuming my conditional check is wrong but I can't seem to figure out why
$? Always returns the exit status of the previous command
so in this case in your 3rd line in if loop you are comparing exit status of the echo command in your second line to 124. So either remove the second line or store the exit status to a variable & use that variable in your if loop.
I have a main script. Inside it I call other three shell scripts, A,B and C. All were successful. Exit codes are all equal to zero. However, when I looked into the output file of the first script which is A, it contains an error message. Now I want to exit the main script and not to continue running the other scripts after the script that has output error. Can anyone help me on this? Thanks!
Even if some command in your first bash script results in an error, the script as a whole may complete with exit code 0.
You can check the exit code of any individual command in your script by using the $? variable. This variable stores the exit code of the previous command. This will allow you to check for errors within the script.
The easiest way is to append || exit 1 to the statement which is throwing the error. That will cause the script to exit if the exit code of the command is 1 (i.e. an error).
So assuming you had a command sqlscript and you wanted the entire script to exit if sqlscript exited with a non-zero exit code you would do
sqlscript || exit 1
As a point of trivia, the 1 in exit 1 is not needed. A plain exit command would also exit with the exit status of the last executed command.
Which would be false (code=1) if the sqlscript command fails. If the sqlscript command succeeds, the exit code is the exit code of sqlscript. In that case, the || does not trigger and the exit command is not executed.
I have a main script. Inside it I call other three shell scripts, A,B
and C. All were successful. Exit codes are all equal to zero. However,
when I looked into the output file of the first script which is A, it
contains an error message. Now I want to exit the main script and not
to continue running the other scripts after the script that has output
error.
Since script A doesn't return an error exit code, you have to inspect its output. This is quite easy with grep provided that you have a search string which clearly identifies an error message, e. g.:
# this echo command simulates script A - it outputs "error" and exits with 0:
echo "contains an error message" >StoreKey_All.csv # assumed this output file
grep error StoreKey_All.csv && exit 1 # exit if output has error
# continue with scripts B and C
echo B
I'm writing a shell script with #!/bin/sh as the first line so that the script exits on the first error. There are a few lines in the file that are in the form of command || true so that the script doesn't exit right there if the command fails. However, I still want to know know the exit code of the command. How would I get the exit code without having to use set +e to temporarily disable that behavior?
Your question appears to imply set -e.
Assuming set -e:
Instead of command || true you can use command || exitCode=$?. The script will continue and the exit status of command is captured in exitCode.
$? is an internal variable that keeps the exit code of the last command.
Since || short-circuits if command succeeds, set exitCode=0 between tests or instead use: command && exitCode=0 || exitCode=$?.
But prefer to avoid set -e style scripting altogether, and instead add explicit error handling to each command in your script.
If you want to know the status of the command, then presumably you take different actions depending on its value. In which case your code should look something like:
if command; then
# do something when command succeeds
else
# do something when command fails
fi
In that case you don't need to do anything special, since the shell will not abort when command fails.
The only reasons set -e would give you any problems is if you write your code as:
command
if test $? = 1; ...
So don't do that.
What is the difference between the return and exit statement in Bash functions with respect to exit codes?
From man bash on return [n];
Causes a function to stop executing and return the value specified by n to its caller. If n is omitted, the return status is that of the last command executed in the function body.
... on exit [n]:
Cause the shell to exit with a status of n. If n is omitted, the exit status is that of the last command executed. A trap on EXIT is executed before the shell terminates.
EDIT:
As per your edit of the question, regarding exit codes, return has nothing to do with exit codes. Exit codes are intended for applications/scripts, not functions. So in this regard, the only keyword that sets the exit code of the script (the one that can be caught by the calling program using the $? shell variable) is exit.
EDIT 2:
My last statement referring exit is causing some comments. It was made to differentiate return and exit for the understanding of the OP, and in fact, at any given point of a program/shell script, exit is the only way of ending the script with an exit code to the calling process.
Every command executed in the shell produces a local "exit code": it sets the $? variable to that code, and can be used with if, && and other operators to conditionally execute other commands.
These exit codes (and the value of the $? variable) are reset by each command execution.
Incidentally, the exit code of the last command executed by the script is used as the exit code of the script itself as seen by the calling process.
Finally, functions, when called, act as shell commands with respect to exit codes. The exit code of the function (within the function) is set by using return. So when in a function return 0 is run, the function execution terminates, giving an exit code of 0.
return will cause the current function to go out of scope, while exit will cause the script to end at the point where it is called. Here is a sample program to help explain this:
#!/bin/bash
retfunc()
{
echo "this is retfunc()"
return 1
}
exitfunc()
{
echo "this is exitfunc()"
exit 1
}
retfunc
echo "We are still here"
exitfunc
echo "We will never see this"
Output
$ ./test.sh
this is retfunc()
We are still here
this is exitfunc()
I don't think anyone has really fully answered the question because they don't describe how the two are used. OK, I think we know that exit kills the script, wherever it is called and you can assign a status to it as well such as exit or exit 0 or exit 7 and so forth. This can be used to determine how the script was forced to stop if called by another script, etc. Enough on exit.
return, when called, will return the value specified to indicate the function's behavior, usually a 1 or a 0. For example:
#!/bin/bash
isdirectory() {
if [ -d "$1" ]
then
return 0
else
return 1
fi
echo "you will not see anything after the return like this text"
}
Check like this:
if isdirectory $1; then echo "is directory"; else echo "not a directory"; fi
Or like this:
isdirectory || echo "not a directory"
In this example, the test can be used to indicate if the directory was found. Notice that anything after the return will not be executed in the function. 0 is true, but false is 1 in the shell, different from other programming languages.
For more information on functions: Returning Values from Bash Functions
Note: The isdirectory function is for instructional purposes only. This should not be how you perform such an option in a real script.*
Remember, functions are internal to a script and normally return from whence they were called by using the return statement. Calling an external script is another matter entirely, and scripts usually terminate with an exit statement.
The difference "between the return and exit statement in Bash functions with respect to exit codes" is very small. Both return a status, not values per se. A status of zero indicates success, while any other status (1 to 255) indicates a failure. The return statement will return to the script from where it was called, while the exit statement will end the entire script from wherever it is encountered.
return 0 # Returns to where the function was called. $? contains 0 (success).
return 1 # Returns to where the function was called. $? contains 1 (failure).
exit 0 # Exits the script completely. $? contains 0 (success).
exit 1 # Exits the script completely. $? contains 1 (failure).
If your function simply ends without a return statement, the status of the last command executed is returned as the status code (and will be placed in $?).
Remember, return and exit give back a status code from 0 to 255, available in $?. You cannot stuff anything else into a status code (e.g., return "cat"); it will not work. But, a script can pass back 255 different reasons for failure by using status codes.
You can set variables contained in the calling script, or echo results in the function and use command substitution in the calling script; but the purpose of return and exit are to pass status codes, not values or computation results as one might expect in a programming language like C.
Sometimes, you run a script using . or source.
. a.sh
If you include an exit in the a.sh, it will not just terminate the script, but end your shell session.
If you include a return in the a.sh, it simply stops processing the script.
exit terminates the current process; with or without an exit code, consider this a system more than a program function. Note that when sourcing, exit will end the shell. However, when running, it will just exit the script.
return from a function go back to the instruction after the call, with or without a return code. return is optional and it's implicit at the end of the function. return can only be used inside a function.
I want to add that while being sourced, it's not easy to exit the script from within a function without killing the shell. I think, an example is better on a 'test' script:
#!/bin/bash
function die(){
echo ${1:=Something terrible wrong happen}
#... clean your trash
exit 1
}
[ -f /whatever/ ] || die "whatever is not available"
# Now we can proceed
echo "continue"
doing the following:
user$ ./test
Whatever is not available
user$
test -and- the shell will close.
user$ . ./test
Whatever is not available
Only test will finish and the prompt will show.
The solution is to enclose the potentially procedure in ( and ):
#!/bin/bash
function die(){
echo $(1:=Something terrible wrong happen)
#... Clean your trash
exit 1
}
( # Added
[ -f /whatever/ ] || die "whatever is not available"
# Now we can proceed
echo "continue"
) # Added
Now, in both cases only test will exit.
The OP's question:
What is the difference between the return and exit statement in BASH functions with respect to exit codes?
Firstly, some clarification is required:
A (return|exit) statement is not required to terminate execution of a (function|shell). A (function|shell) will terminate when it reaches the end of its code list, even with no (return|exit) statement.
A (return|exit) statement is not required to pass a value back from a terminated (function|shell). Every process has a built-in variable $? which always has a numeric value. It is a special variable that cannot be set like "?=1", but it is set only in special ways (see below *).
The value of $? after the last command to be executed in the (called function | sub shell) is the value that is passed back to the (function caller | parent shell). That is true whether the last command executed is ("return [n]"| "exit [n]") or plain ("return" or something else which happens to be the last command in the called function's code.
In the above bullet list, choose from "(x|y)" either always the first item or always the second item to get statements about functions and return, or shells and exit, respectively.
What is clear is that they both share common usage of the special variable $? to pass values upwards after they terminate.
* Now for the special ways that $? can be set:
When a called function terminates and returns to its caller then $? in the caller will be equal to the final value of $? in the terminated function.
When a parent shell implicitly or explicitly waits on a single sub shell and is released by termination of that sub shell, then $? in the parent shell will be equal to the final value of $? in the terminated sub shell.
Some built-in functions can modify $? depending upon their result. But some don't.
Built-in functions "return" and "exit", when followed by a numerical argument both set $? with their argument, and terminate execution.
It is worth noting that $? can be assigned a value by calling exit in a sub shell, like this:
# (exit 259)
# echo $?
3
In simple words (mainly for newbie in coding), we can say,
`return`: exits the function,
`exit()`: exits the program (called as process while running)
Also if you observed, this is very basic, but...,
`return`: is the keyword
`exit()`: is the function
If you convert a Bash script into a function, you typically replace exit N with return N. The code that calls the function will treat the return value the same as it would an exit code from a subprocess.
Using exit inside the function will force the entire script to end.
Adding an actionable aspect to a few of the other answers:
Both can give exit codes - default or defined by the function, and the only 'default' is zero for success for both exit and return. Any status can have a custom number 0-255, including for success.
Return is used often for interactive scripts that run in the current shell, called with . script.sh for example, and just returns you to your calling shell. The return code is then accessible to the calling shell - $? gives you the defined return status.
Exit in this case also closes your shell (including SSH connections, if that's how you're working).
Exit is necessary if the script is executable and called from another script or shell and runs in a subshell. The exit codes then are accessible to the calling shell - return would give an error in this case.
First of all, return is a keyword and exit is a function.
That said, here's a simplest of explanations.
return
It returns a value from a function.
exit
It exits out of or abandons the current shell.