To check the exit code of command in Bash the following logic is used:
if [ $? -ne 0 ]
then
echo "Error occurred"
return 1
fi
My problem is that adding this after every command makes the script very long with multiple copies of the same thing, and very hard to maintain.
The best thing here would be a function, that would be called from all the locations the the exit code needs to be checked. The problem is that the exit command can not be used, because it will kill current process (current Bash session will be killed), so only the return command can be used. But when using the return command, in a called function, the calling function must still check the exit code, and were back to the same problem.
Is there any thing like MACRO in Bash or any other way to the error checking more efficient?
Instead of this:
somecommand
if [ $? -ne 0 ]
then
echo "Error occurred"
return 1
fi
You don't need to write conditions on the $? variable,
you can use the command itself in if statements:
if ! somecommand
then
echo "Error occurred"
return 1
fi
Another alternative is to create a helper function to encapsulate the action on error, and use the || operator after the command to call the function and then return 1 after:
error() {
echo "Error occurred"
}
somecommand || { error; return 1; }
Finally, if you don't really need return 1 and you don't mind to exit the script in case of failure, then you can exit 1 inside the helper function and then the caller code can become more compact:
fatal() {
echo "Error occurred"
exit 1
}
somecommand || fatal
Related
I recently stumbled across some shell script code which is repeatedly doing the same thing over again and I was wondering if this could be a bit simplified.
Basically in a defined function, after each and every call the return code is checked. If it is not 0, some outputs are done and the function exits prematurely. In this cases we willfully use return to escape the function but do not want to stop the whole process by using exit.
e.g.:
function _myFunc {
typeset ret=0
_myFirstSubFunc "TEST"
ret=$?
if [[ $? -ne 0 ]]; then
echo "myFirstSubFunc produced an error: $ret"
return $ret
fi
_mySecondSubFunc "TEST"
ret=$?
if [[ $? -ne 0 ]]; then
echo "mySecondSubFunc produced an error: $ret"
return $ret
fi
return $ret
}
I'd like to reduce this repeated code of
checking the return code
printing out some information if the returncode is != 0
stopping the current function from continuing
to one call, so the code isn't that much cluttered with this checking. The execution of the function should continue if the previous call was successful.
Usually, when I'd like to abort (exit) the whole process, one function will do this trick. But when a value should be returned in the case of an error things get (at least for me) a bit tricky.
I tried to use traps or aliases, unfortunately without success. Additionally I'm limited to ksh88, so it seems I cannot use some conditionals following the call by using ||.
Any ideas, if there is something else so I can reduce this tedious error handling?
Thanks!
function orReport {
if "$#"; then
return
else
ret=$?
echo "$1 produced an error: $ret"
return "$ret"
fi
}
function _myFunc {
orReport _myFirstSubFunc "TEST" || return
orReport _mySecondSubFunc "TEST" || return
}
i have this code in my script :
check() {
if [ $? -eq 0 ]
then
DATA=`date +%Y%m%d_%H.%M.%S`
echo "O script $DATA Executou com sucesso" >>$log
else
DATA=`date +%Y%m%d_%H.%M.%S`
echo "O script $DATA Executou com erro"; >>$log
fi
}
but i want to put it in another script and when i execute in crontab once every 15 minutes. I want to check if the script executed successfully or not.
My question is in this first if section, how can I put the script ex3.sh for example
if script ex3.sh sucess
then
DATA=`date +%Y%m%d_%H.%M.%S`
echo "O script $DATA Executou com sucesso" >>$log
else
DATA=`date +%Y%m%d_%H.%M.%S`
echo "O script $DATA Executou com erro"; >>$log
fi
}
First to the terminology: What you show in your example (check) is not a script, but a function. It becomes a script, if you would add an actual invocation to check.
But to your proper question: You need to define somehow to the caller, whether or not you are successful. This can be done by havin your function return an exit code. This is done using the return statement. The convention is tor return 0 for success, and something between 1 and 127 for failure.
Hence, if you detect an error, do a
return 1
in your script. If everything goes well, exit it by
return 0
If you keep to this convention, you can indeed make use of the if statement you had in mind:
if check ARGUMENTS
then
.... # check function successful
else
.... # check function not successful
fi
The same principle does not only apply to functions, but also to scripts, with the main difference that with scripts, you signal the exit code not with return, but with an exit statement, i.e.
exit 0 # Success
exit 1 # Failure
I have several functions in my .bashrc and I want to terminate the whole "run" on error. For example:
function assert_env()
{
var_name=$1
if [ -z "${!var_name}" ]; then
printf "Missing environment variable: $var_name\n"
exit 1
fi
}
function my_test()
{
assert_env "abc"
print $abc
}
If i type my_test in terminal I want to terminate the execution, but this script closes the terminal (as expected - I run exit 1).
How may I terminate the execution without closing the current terminal?
EDIT: To be more specific - I don't want to return from assert_env, I want my following commands from my_test to not be executed if the condition from assert_env is not met, something like C++ assert
Note that exit returns the exit code specified to the shell itself, so using exit 1 would obviously exit your current shell. You need to use return here.
As far as you requirement is considered, just use the exit code from the function assert_env to decide if you want to run further statements. Here using return will cause the current function to go out of scope and what ever code you are returning back to the callee can be used to check if its successful/failure
function assert_env() {
local ret_val=0; var_name=$1
if [ -z "${!var_name}" ]; then
printf "Missing environment variable: $var_name\n"
ret_val=1
fi
return ${ret_val}
}
Now using it
function my_test() {
if ! assert_env "abc"; then
return
fi
# Or could be just written as assert_env "abc" || return
}
This way, if the assert_env returns 0, the if-clause in my_test asserts failure and rest of the code gets executed. And on returning 1 the condition becomes true and the function call is returned without exiting the main scipt.
Note that function keyword is non-standard and not POSIX compliant and may not work across shells. Just drop the keyword if you want to make it portable.
I've found a workaround - execute the function content in a subshell
Example:
function assert_env()
{
var_name=$1
if [ -z "${!var_name}" ]; then
printf "Missing environment variable: $var_name\n"
exit 1
fi
}
function my_test()
{
(
printf "$1\n"
assert_env "abc"
printf $abc
)
}
This question already has answers here:
Automatic exit from Bash shell script on error [duplicate]
(8 answers)
Closed 6 years ago.
#Example Script
wget http://file1.com
cd /dir
wget http://file2.com
wget http://file3.com
I want to execute the bash script line by line and test the exit code ($?) of each execution and determine whether to proceed or not:
It basically means I need to add the following script below every line in the original script:
if test $? -eq 0
then
echo "No error"
else
echo "ERROR"
exit
fi
and the original script becomes:
#Example Script
wget http://file1.com
if test $? -eq 0
then
echo "No error"
else
echo "ERROR"
exit
fi
cd /dir
if test $? -eq 0
then
echo "No error"
else
echo "ERROR"
exit
fi
wget http://file2.com
if test $? -eq 0
then
echo "No error"
else
echo "ERROR"
exit
fi
wget http://file3.com
if test $? -eq 0
then
echo "No error"
else
echo "ERROR"
exit
fi
But the script becomes bloated.
Is there a better method?
One can use set -e but it's not without it's own pitfalls. Alternative one can bail out on errors:
command || exit 1
And an your if-statement can be written less verbose:
if command; then
The above is the same as:
command
if test "$?" -eq 0; then
set -e makes the script fail on non-zero exit status of any command. set +e removes the setting.
There are many ways to do that.
For example can use set in order to automatically stop on "bad" rc; simply by putting
set -e
on top of your script. Alternatively, you could write a "check_rc" function; see here for some starting points.
Or, you start with this:
check_error () {
if [ $RET == 0 ]; then
echo "DONE"
echo ""
else
echo "ERROR"
exit 1
fi
}
To be used with:
echo "some example command"
RET=$? ; check_error
As said; many ways to do this.
Best bet is to use set -e to terminate the script as soon as any non-zero return code is observed. Alternatively you can write a function to deal with error traps and call it after every command, this will reduce the if...else part and you can print any message before exiting.
trap errorsRead ERR;
function errorsRead() {
echo "Some none-zero return code observed..";
exit 1;
}
somecommand #command of your need
errorsRead # calling trap handling function
You can do this contraption:
wget http://file1.com || exit 1
This will terminate the script with error code 1 if a command returns a non-zero (failed) result.
I'd like to return an exit code from a BASH script that is called within another script, but could also be called directly. It roughly looks like this:
#!/bin/bash
dq2-get $1
if [ $? -ne 0 ]; then
echo "ERROR: ..."
# EXIT HERE
fi
# extract, do some stuff
# ...
Now in the line EXIT HERE the script should exit and return exit code 1. The problem is that
I cannot use return, because when I forget to source the script instead of calling it, return will not exit, and the rest of the script will be executed and mess things up.
I cannot use exit, because this closes the shell.
I cannot use the nice trick kill -SIGINT $$, because this doesn't allow to return an exit code.
Is there any viable alternative that I have overlooked?
The answer to the question title (not in the body as other answers have addressed) is:
Return an exit code without closing shell
(exit 33)
If you need to have -e active and still avoid exiting the shell with a non-zero exit code, then do:
(exit 33) && true
The true command is never executed but is used to build a compound command that is not exited by the -e shell flag.
That sets the exit code without exiting the shell (nor a sourced script).
For the more complex question of exiting (with an specific exit code) either if executed or sourced:
#!/bin/bash
[ "$BASH_SOURCE" == "$0" ] &&
echo "This file is meant to be sourced, not executed" &&
exit 30
return 88
Will set an exit code of 30 (with an error message) if executed.
And an exit code of 88 if sourced.
Will exit both the execution or the sourcing without affecting the calling shell.
Use this instead of exit or return:
[ $PS1 ] && return || exit;
Works whether sourced or not.
You can use x"${BASH_SOURCE[0]}" == x"$0" to test if the script was sourced or called (false if sourced, true if called) and return or exit accordingly.
Another option is to use a function and put the return values in that and then simply either source the script (source processStatus.sh) or call the script (./processStatus.sh) . For example consider the processStatus.sh script that needs to return a value to the stopProcess.sh script but also needs to be called separately from say the command line without using source (only relevant parts included)
Eg:
check_process ()
{
if [ $1 -eq "50" ]
then
return 1
else
return 0
fi
}
and
source processStatus.sh $1
RET_VALUE=$?
if [ $RET_VALUE -ne "0" ]
then
exit 0
fi
You can use return if you use set -e in the beginning of the script.
If you just want to check if the function returned no errors, I'd rather suggest rewriting your code like this:
#!/bin/bash
set -e # exit program if encountered errors
dq2-get ()
{
# define the function here
# ...
if [ $1 -eq 0 ]
then
return 0
else
return 255
# Note that nothing will execute from this point on,
# because `return` terminates the function.
}
# ...
# lots of code ...
# ...
# Now, the test:
# This won't exit the program.
if $(dq2-get $1); then
echo "No errors, everything's fine"
else
echo "ERROR: ..."
fi
# These commands execute anyway, no matter what
# `dq2-get $1` returns (i.e. {0..255}).
# extract, do some stuff
# ...
Now, the code above won't leave the program if the function dq2-get $1 returns errors. But, implementing the function all by itself will exit the program because of the set -e. The code below describes this situation:
# The function below will stop the program and exit
# if it returns anything other than `0`
# since `set -e` means stop if encountered any errors.
$(dq2-get $1)
# These commands execute ONLY if `dq2-get $1` returns `0`
# extract, do some stuff
# ...
Thanks for the question, my case was to source a file for some setup, but end the script and skip the setup actions if certain conditions were not met.
I had hit the issue of an attempt to use exit() actually causing the closing of my terminal, and found myself here :D
After reviewing the options for the specific solution i just went with something like the below, I also think Deepaks answer is worth reviewing if this approach works in your case.
if [ -z "$REQUIRED_VAR" ]; then
echo "please check/set \$REQUIRED_VAR ..."
echo "skipping logic"
else
echo "starting logic"
doStuff()
echo "completed logic"
fi