Spring Boot - How to specify an alternate start-class? (Multiple Entry Points)
Solution 1
I don't believe that property would apply in your case. There are 3 different "Launchers" (go back to the docs and see). If you are building a jar it uses the JarLauncher class. If you switch it to PropertiesLauncher then loader.main would be useful.
META-INF/MANIFEST.MF
Main-Class: org.springframework.boot.loader.PropertiesLauncher
Solution 2
I took a different approach and use a command line parameter to determine which class to use as my SpringApplication class. I only have a single main() method, but different Application classes with different configurations that are used based on a command line param.
I have a single class with a main() in it:
public static void main(String[] args) {
SpringApplication app;
if( ArrayUtils.contains(args, "--createdb")){
app = new SpringApplication(CreateDB.class);
args = (String[])ArrayUtils.add(args, "--spring.jpa.hibernate.ddl-auto=create");
} else {
app = new SpringApplication(Application.class);
}
app.setWebEnvironment(false);
app.setShowBanner(false);
app.addListeners(new ConfigurationLogger());
// launch the app
ConfigurableApplicationContext context = app.run(args);
// finished so close the context
context.close();
}
But I have 2 different SpringApplication classes: Application.class & CreateDB.class. Each class defines a different @ComponentScan
path as well as different @EnableAutoConfiguration
options and different @Configuration
options. Finally, based on my command line arguments, I can decide whether to programatically enable additional profiles/etc.
In my case, I want a different launcher to just create the DB schema and exit, so I've forced the command line parameter.
Solution 3
I would suggest having a single main
but using Spring profiles (or configuration properties) to select one or other "entry point" @Configuration
class.
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The Gilbert Arenas Dagger
Updated on September 15, 2022Comments
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The Gilbert Arenas Dagger over 1 year
I want to add an alternate entry point to my Spring-Boot application. I would prefer to keep this as a fat jar. Is this possible?
According to their documentation, the property
loader.main
specifies the name of the main class to launch.I tried
java -jar MyJar.jar --loader.main=com.mycompany.AlternateMain
but the start-class specified in my pom.xml was still run (and if I remove this from the pom.xml then I error during the packaging).Alternatively, I tried
java -cp MyJar.jar com.mycompany.AlternateMain
but I don't know of a good way to add all the nested jars to the classpath.Any suggestions?
Edit: Here is the solution that I used
As jst suggested, I changed my launcher to use the PropertiesLauncher. I did this by modifying the configuration of my spring-boot-maven-plugin.
<plugin> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-maven-plugin</artifactId> <configuration> <mainClass>${start-class}</mainClass> <layout>ZIP</layout> ...
The
<layout>ZIP</layout>
triggers Spring Boot to use thePropertiesLauncher
.I created my fat jar (mvn package) then called the alternate main like this:
java -jar -Dloader.main=com.mycompany.AlternateMain MyJar.jar
Thanks for the help!
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FGreg almost 9 yearsSpring Boot simply uses the JAR's Manifest to specify the main class and classpath. So I think the real question here is can there be more than one Main Class in an executable JAR? That question is also asked/answered here: stackoverflow.com/q/3976514/953327
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Makoto almost 9 yearsWhat are you intending to accomplish with this? Do you want to create multiple applications from within Spring Boot?
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The Gilbert Arenas Dagger almost 9 years@FGreg I can access the alternate main using the second command that I noted, which matches the answer in the SO thread you link to. In order to use this successfully, I would have to figure out how to add my nested jars to the classpath.
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The Gilbert Arenas Dagger almost 9 years@Makoto I do not want to create multiple applications with this. It's actually a proof of concept that I am doing now, but I have used multiple entry points in the past for a variety of reasons
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The Gilbert Arenas Dagger almost 9 yearsThis sounds very promising, but my results are the same. To use the PropertiesLauncher, I configured the spring-boot-maven-plugin, as described here stackoverflow.com/a/21328440/2860319. I could see in my manifest that it worked and the PropertiesLauncher was now being used, but the loader.main property specified via command line did not change the start-class... hmmmm
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The Gilbert Arenas Dagger almost 9 yearsI was adding the properties incorrectly, I'll update my post with what I did. Thanks for the great suggestion!
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ben3000 over 8 yearsThis is where I was heading too, presumably
CreateDB
andApplication
exist in separate packages, thus enabling you to use@ComponentScan
withbasePackages
set to that package? -
Eric B. over 8 yearsIndeed that is possible. In my case, they are actually in the same package, but my
@ComponentScan()
explicity sets different base packages to scan with different exclusion rules. It also allows you to have different@AutoConfiguration()
options, different Config classes, etc. And quite frankly - I find it much easier to do this way than to use the launcher and launcher parameters to specify long class names; this allows me to abstract it any way I want. I've basically packaged 2 separate SpringBoot apps (which share common classes) in the same jar. -
rmv almost 7 yearsCan you provide an example, please?