如何在不重新启动 JVM 的情况下将外部 JAR 添加到 Spring 应用程序?

?我是否认为这是完全错误的，并且有更好的方法在类路径上放置不需要总是需要加载的 JAR(如果 JAR 被禁用"，则不应由 JVM 加载/编译，不应被 Spring 等捡起)?Is there a way to achieve this without too many awkard hacks? Or is recommended to utilize something like OSGi? Am I looking at this completely wrong and there is a better way to have JARs on the classpath that do not need always need loading (if the JAR is "disabled", it should not be loaded/compiled by the JVM, should not be picked up by Spring, etc.)?推荐答案如果在启动 Spring 应用程序之前复制 JAR 文件，这似乎是可能的.感觉很hackish，但它有效.使用风险自负！It looks like this is possible if the JAR files are copied before starting the Spring application. It feels hackish, but it works. Use at your own risk!您需要两个类，一个用于引导外部 JAR，然后将通过手动创建的 PropertiesLauncher 启动第二个类.引导类可以是普通的普通 Java 类(但也可以是 Spring Boot 应用程序)，只有第二个类需要是 SpringBootApplication.You need two classes, one for bootstrapping the external JARs, which will then start the second via a manually created PropertiesLauncher. The bootstrapping class can be a plain old regular Java class (but it can be a Spring Boot Application too) and only the second class needs to be a SpringBootApplication.// BootstrapApplication.javapublic class BootstrapApplication { public static void main(String[] args) { System.out.println("Please copy JAR files and press Enter ..."); System.in.read(); PropertiesLauncher.main(args); }}// Application.java@SpringBootApplicationpublic class Application { public static void main(String[] args) { SpringApplication.run(Application.class, args); }}在gradle文件中，我们可以通过删除bootJar任务清单配置并通过springBoot应用设置来切换回默认的JarLauncher> 配置块.mainClass 将在 MANIFEST.MF 文件中作为 Start-Class 结束.In the gradle file, we can switch back to the default JarLauncher, by removing the bootJar task manifest configuration and applying settings via the springBoot configuration block. mainClass will end up as Start-Class in the MANIFEST.MF file.// build.gradlespringBoot { mainClass = 'com.example.customlauncher.BootstrapApplication'}在加载器的属性文件中，需要设置一个新的属性，指向真正的应用程序类.此文件中的设置仅由 PropertiesLauncher 选取并被 JarLauncher 忽略.换句话说:JarLauncher 委托给来自清单文件的 Start-Class 和 PropertiesLauncher 委托给 loader.main 来自它的属性文件.In the properties file for the loader, a new property needs to be set, which points to the real application class. The settings in this file are only picked up by PropertiesLauncher and ignored by JarLauncher. In other words: JarLauncher delegates to Start-Class from the manifest file and PropertiesLauncher delegates to loader.main from its properties file.# .../resources/loader.propertiesloader.path=file:/path/to/dirloader.main=com.example.customlauncher.ApplicationSpring (Boot) 将首先调用 BootstrapApplication 的 main 方法，如 MANIFEST.MF 文件中指定的(通过 springBoot 配置控制)build.gradle 文件中的块).在这个 main 方法的实现中，一个新的 PropertiesLauncher 被创建，主类设置为real".应用程序(即Application).Spring (Boot) will first call the main method of BootstrapApplication, as specified in the MANIFEST.MF file (controlled via springBoot configuration block in the build.gradle file). In the implementation of this main method, a new PropertiesLauncher is created with the main class set to the "real" application (i.e. Application).执行应用程序仍然通过相同的调用完成:Executing the application is still done via the same invocation:java -jar build/libs/customlauncher-0.0.1-SNAPSHOT.jar在 JVM 启动之后添加到 /path/to/dir 的任何 JAR 文件，但之前在 BootstrapApplication 中调用 PropertiesLauncher#main 是然后在类路径和应用程序上下文中可用，如 Application 所示.Any JAR files added to /path/to/dir after the JVM has started, but before calling PropertiesLauncher#main in BootstrapApplication are then available in the classpath and application context as seen from Application. 这篇关于如何在不重新启动 JVM 的情况下将外部 JAR 添加到 Spring 应用程序?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！