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Don't ignore test.mjs child process exit codes in the Gulpfile

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In the Gulpfile only the exit codes of `test.mjs` child processes
erroneously aren't checked. This causes failures in `test.mjs` to be
logged but not propagated to the master process, which in turn causes
test runners such as GitHub Actions to succeed because they only
monitor the master process. This is easy to reproduce by throwing an
error at the top of `test.mjs` and running `gulp makeref` or `gulp
unittest`: the error is logged, but the task that spawned the child
process succeeds and the master process exits with exit code 0. This is
problematic because it can easily cause errors to go by unnoticed.

This commit fixes the issue by making sure that the `test.mjs`
invocations are handled in the same way as the other child processes
in the file, i.e., if the child process exits with a non-zero exit code
then the master process also exits with a non-zero exit code. After this
patch the error is still logged, but the task now also fails and the
master process exits with exit code 1 to properly signal failure.
This commit is contained in:
Tim van der Meij 2024-01-21 18:49:55 +01:00
parent fce822cde0
commit 6dd77b3c14
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1 changed files with 7 additions and 0 deletions
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gulpfile.mjs
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@ -691,6 +691,9 @@ function createTestSource(testsName, { bot = false, xfaOnly = false } = {}) {
const testProcess = startNode(args, { cwd: TEST_DIR, stdio: "inherit" });
testProcess.on("close", function (code) {
if (code !== 0) {
throw new Error(`Running ${testsName} tests failed.`);
}
source.push(null);
});
return undefined;
@ -722,6 +725,10 @@ function makeRef(done, bot) {
const testProcess = startNode(args, { cwd: TEST_DIR, stdio: "inherit" });
testProcess.on("close", function (code) {
if (code !== 0) {
done(new Error("Creating reference images failed."));
return;
}
done();
});
}