Add (collect) exit codes in bash

Question

I need to depend on few separate executions in a script and don't want to bundle them all in an ugly 'if' statement. I would like to take the exit code '$?' of each execution and add it; at the end, if this value is over a threshold - I would like to execute a command.

Pseudo code:

ALLOWEDERROR=5

run_something
RESULT=$?
..other things..

run_something_else
RESULT=$RESULT + $?

if [ $RESULT -gt ALLOWEDERROR ] 
   then echo "Too many errors"
fi

Issue: Even though the Internet claims otherwise, bash refuses to treat the RESULT and $? as integer. What is the correct syntax?

Thanks.

Answer 1

You might want to take a look at the trap builtin to see if it would be helpful:

help trap

or

man bash

you can set a trap for errors like this:

#!/bin/bash

AllowedError=5

SomeErrorHandler () {
    (( errcount++ ))       # or (( errcount += $? ))
    if  (( errcount > $AllowedError ))
    then
        echo "Too many errors"
        exit $errcount
    fi
}

trap SomeErrorHandler ERR

for i in {1..6}
do
    false
    echo "Reached $i"     # "Reached 6" is never printed
done

echo "completed"          # this is never printed

If you count the errors (and only when they are errors) like this instead of using "$?", then you don't have to worry about return values that are other than zero or one. A single return value of 127, for example, would throw you over your threshold immediately. You can also register traps for other signals in addition to ERR.

Answer 2

Use the $(( ... )) construct.

$ cat st.sh
RESULT=0
true
RESULT=$(($RESULT + $?))
false
RESULT=$(($RESULT + $?))
false
RESULT=$(($RESULT + $?))
echo $RESULT
$ sh st.sh
2
$