Run Commands if Previous Successful (or not) in Bash

Question

Having read the docs and following this answer, I wanted to test it with a simple example:

test_1.sh

#!/usr/bin/bash

let x=1/0
echo $x

test_2.sh

#!/usr/bin/bash

printf "\nThis is 2\n"

But both of these implementations of control flow fail: run test_1, if unsuccessful, run test_2

# Just errors on line 1 of course
./test_1.sh 
if [ $? -ne 0 ]; then
    ./test_2.sh
fi

Also fails:

./test_1.sh || ./test_2.sh

Any tips for controlling flow between multiple scripts?

Answer 1

Based on your comment:

I just have a process like test_1.sh that could result in success (return code 0), or failure (anything other than 0). I want something like test_2.sh to run only in the former case.

you want the opposite of the snippets in your question.

./test_1.sh
if [ $? -ne 0 ]; then
    ./test_2.sh
fi

runs test_2.sh if test_1.sh indicates failure, as does

./test_1.sh || ./test_2.sh 

To run test_2.sh if test_1.sh indicates success, any of these will work:

./test_1.sh
if [ $? -eq 0 ]; then
    ./test_2.sh
fi

if ./test_1.sh; then
    ./test_2.sh
fi

./test_1.sh && ./test_2.sh

See What are the shell's control and redirection operators? for details.

Your current test_1.sh always indicates success, in spite of the division by zero. See Ignore or catch division by zero for details.