Improve diagnostics when remote application does not start as expected

See gh-22909
pull/22996/head
Andy Wilkinson 4 years ago
parent 65ccb514d0
commit 2b1bb2f18f

@ -24,6 +24,7 @@ import java.util.List;
import java.util.function.BiFunction;
import org.awaitility.Awaitility;
import org.awaitility.core.ConditionTimeoutException;
import org.springframework.boot.devtools.RemoteSpringApplication;
import org.springframework.boot.devtools.tests.JvmLauncher.LaunchedJvm;
@ -77,7 +78,7 @@ abstract class RemoteApplicationLauncher extends AbstractApplicationLauncher {
createRemoteSpringApplicationClassPath(classesDirectory),
RemoteSpringApplication.class.getName(), "--spring.devtools.remote.secret=secret",
"http://localhost:" + port);
awaitRemoteSpringApplication(remoteSpringApplicationJvm.getStandardOut());
awaitRemoteSpringApplication(remoteSpringApplicationJvm);
return remoteSpringApplicationJvm.getProcess();
}
catch (Exception ex) {
@ -105,10 +106,22 @@ abstract class RemoteApplicationLauncher extends AbstractApplicationLauncher {
.getServerPort();
}
private void awaitRemoteSpringApplication(File standardOut) throws Exception {
FileContents contents = new FileContents(standardOut);
private void awaitRemoteSpringApplication(LaunchedJvm launchedJvm) throws Exception {
FileContents contents = new FileContents(launchedJvm.getStandardOut());
try {
Awaitility.waitAtMost(Duration.ofSeconds(30)).until(contents::get,
containsString("Started RemoteSpringApplication"));
}
catch (ConditionTimeoutException ex) {
if (!launchedJvm.getProcess().isAlive()) {
throw new IllegalStateException(
"Process exited with status " + launchedJvm.getProcess().exitValue()
+ " before producing expected standard output.\n\nStandard output:\n\n" + contents.get()
+ "\n\nStandard error:\n\n" + new FileContents(launchedJvm.getStandardError()).get(),
ex);
}
throw ex;
}
}
}

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