为什么在Java中这么难做到这一点?如果您希望拥有任何类型的模块系统,您需要能够动态加载JAR文件。有人告诉我,有一种方法可以通过编写自己的ClassLoader来实现,但这需要做很多工作,因为(至少在我看来)应该像调用以JAR文件作为参数的方法一样简单。
对实现这一功能的简单代码有什么建议吗?
为什么在Java中这么难做到这一点?如果您希望拥有任何类型的模块系统,您需要能够动态加载JAR文件。有人告诉我,有一种方法可以通过编写自己的ClassLoader来实现,但这需要做很多工作,因为(至少在我看来)应该像调用以JAR文件作为参数的方法一样简单。
对实现这一功能的简单代码有什么建议吗?
当前回答
下面是一个未弃用的版本。我修改了原始版本,删除了不推荐使用的功能。
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/*
* Created on Oct 6, 2004
*/
package tools;
import java.io.File;
import java.io.IOException;
import java.lang.reflect.Constructor;
import java.lang.reflect.InvocationTargetException;
import java.lang.reflect.Method;
import java.net.URL;
import java.net.URLClassLoader;
/**
* Useful class for dynamically changing the classpath, adding classes during runtime.
*/
public class ClasspathHacker {
/**
* Parameters of the method to add an URL to the System classes.
*/
private static final Class<?>[] parameters = new Class[]{URL.class};
/**
* Adds a file to the classpath.
* @param s a String pointing to the file
* @throws IOException
*/
public static void addFile(String s) throws IOException {
File f = new File(s);
addFile(f);
}
/**
* Adds a file to the classpath
* @param f the file to be added
* @throws IOException
*/
public static void addFile(File f) throws IOException {
addURL(f.toURI().toURL());
}
/**
* Adds the content pointed by the URL to the classpath.
* @param u the URL pointing to the content to be added
* @throws IOException
*/
public static void addURL(URL u) throws IOException {
URLClassLoader sysloader = (URLClassLoader)ClassLoader.getSystemClassLoader();
Class<?> sysclass = URLClassLoader.class;
try {
Method method = sysclass.getDeclaredMethod("addURL",parameters);
method.setAccessible(true);
method.invoke(sysloader,new Object[]{ u });
} catch (Throwable t) {
t.printStackTrace();
throw new IOException("Error, could not add URL to system classloader");
}
}
public static void main(String args[]) throws IOException, SecurityException, ClassNotFoundException, IllegalArgumentException, InstantiationException, IllegalAccessException, InvocationTargetException, NoSuchMethodException{
addFile("C:\\dynamicloading.jar");
Constructor<?> cs = ClassLoader.getSystemClassLoader().loadClass("test.DymamicLoadingTest").getConstructor(String.class);
DymamicLoadingTest instance = (DymamicLoadingTest)cs.newInstance();
instance.test();
}
}
其他回答
这可能是一个延迟响应,我可以这样做(fastutil-8.2.2.jar的一个简单示例)使用jhplot。来自DataMelt的Web类(http://jwork.org/dmelt)
import jhplot.Web;
Web.load("http://central.maven.org/maven2/it/unimi/dsi/fastutil/8.2.2/fastutil-8.2.2.jar"); // now you can start using this library
根据文档,这个文件将在“lib/user”中下载,然后动态加载,因此您可以立即在同一个程序中使用这个jar文件中的类。
下面的解决方案有点笨拙,因为它使用反射来绕过封装,但它工作得完美无缺:
File file = ...
URL url = file.toURI().toURL();
URLClassLoader classLoader = (URLClassLoader)ClassLoader.getSystemClassLoader();
Method method = URLClassLoader.class.getDeclaredMethod("addURL", URL.class);
method.setAccessible(true);
method.invoke(classLoader, url);
你应该看一看OSGi,比如在Eclipse平台中实现的。它就是这么做的。您可以安装、卸载、启动和停止所谓的包,这些包实际上是JAR文件。但是它做的更多,例如,它提供了可以在运行时在JAR文件中动态发现的服务。
或参阅Java模块系统的规范。
困难的原因在于安全。类加载器是不可变的;您不应该在运行时随意地向它添加类。我很惊讶它能与系统类加载器一起工作。下面是如何创建你自己的子类加载器:
URLClassLoader child = new URLClassLoader(
new URL[] {myJar.toURI().toURL()},
this.getClass().getClassLoader()
);
Class classToLoad = Class.forName("com.MyClass", true, child);
Method method = classToLoad.getDeclaredMethod("myMethod");
Object instance = classToLoad.newInstance();
Object result = method.invoke(instance);
很痛苦,但事实就是这样。
请看看我开始的这个项目:proxy-object lib
这个库将从文件系统或任何其他位置加载jar。它将为jar专用一个类装入器,以确保没有库冲突。用户将能够从加载的jar中创建任何对象,并在其上调用任何方法。 这个库被设计用来从支持Java 7的代码库中加载用Java 8编译的jar。
创建一个对象:
File libDir = new File("path/to/jar");
ProxyCallerInterface caller = ObjectBuilder.builder()
.setClassName("net.proxy.lib.test.LibClass")
.setArtifact(DirArtifact.builder()
.withClazz(ObjectBuilderTest.class)
.withVersionInfo(newVersionInfo(libDir))
.build())
.build();
String version = caller.call("getLibVersion").asString();
ObjectBuilder支持工厂方法、调用静态函数和回调接口实现。 我将在自述页面上发布更多的例子。