Getting exit code from curl in bash script

Question

I want to take the output of simple.sh, a script from the internet and check its exit code. I on OSX

#!/bin/bash
$(curl -s http://127.0.0.1:8000/simple.sh)
if [ -z "$?" ]; then
    echo "Good"
    exit 0
else
    echo "Bad"
    exit 1
fi

simple.sh:

#!/bin/bash
exit 0

The problem I am getting is:

./test.sh: line 2: #!/bin/bash: No such file or directory
Bad

Answer 1

Can't say it's elegant, but this is the way I would do it:

#!/bin/bash
curl -s http://127.0.0.1:8000/simple.sh | /bin/bash -s >/dev/null 2>&1
rc=$?
if [ -z "$rc" ]
then
    echo "Good"
    exit 0
else
    echo "Bad"
    exit 1
fi

Seems to me, the way you are doing it is similar to executing a here-file inside the $( ... ) construct. Never tried that, not sure bash works that way.

Letting curl echo the contents of the file and piping it to bash accounts for the text output of the curl command and allows bash to execute it.

I'll bet that, if you try this, you will get the same results:

$( cat /[path]/simple.sh ); echo $?

Answer 2

Your idea is right, but you seem to have defined a wrong conditional for checking the return code with [ -z "$?" ] which checks if the return code string is empty or not. Irrespective of the result of the curl output, your $? will carry a value which means, you'll never assert the if condition of your script. You probably need to check the return code that curl returns directly in your script

$(curl -s http://127.0.0.1:8000/simple.sh)
if [ "$?" -eq 0 ]; then

or even more tersely written without the test operator to allow the return code to be directly used in the if condition

if curl -s http://127.0.0.1:8000/simple.sh 2>&1 >/dev/null; then