我建立在Java中,可以接收Java源文件的服务器,它应该使用JavaCompiler进行动态编译它,然后加载类。 但问题是,如果服务器收到名称相同,但不同内容的文件,它仍然会加载前一类,并给出了相同的输出。 我注意到一些答案提示创建类,我试图加载并使用不同的类加载器的超类,但它仍的情况下,如果Java源文件是动态发送到服务器?
这里是我的编译和FileServer.java负载的方法:
public final static int FILE_SIZE = 1022386;
public static void compile(String fileName)
{
// Save source in .java file.
File sourceFile = new File(fileName);
// Compile source file.
JavaCompiler compiler = ToolProvider.getSystemJavaCompiler();
DiagnosticCollector <JavaFileObject> diagnostics =
new DiagnosticCollector<JavaFileObject>();
StandardJavaFileManager fileManager =
compiler.getStandardFileManager(diagnostics, null, null);
File [] files = new File [] {sourceFile};
Iterable<? extends JavaFileObject> compilationUnits =
fileManager.getJavaFileObjectsFromFiles(Arrays.asList(files));
String [] compileOptions = new String[] {"-classpath", "runtime.jar"};
Iterable<String> compilationOptions = Arrays.asList(compileOptions);
JavaCompiler.CompilationTask task =
compiler.getTask(null, fileManager, diagnostics, compilationOptions,
null, compilationUnits);
task.call();
}
public static void compileLoad (String fileName)
{
compile(fileName);
String className = "";
int i = 0;
while(fileName.charAt(i) != '.') {
className += fileName.charAt(i);
i++;
}
ClassLoader classLoader = FileServer.class.getClassLoader();
// Dynamically load class and invoke its main method.
try {
//Class<?> cls = Class.forName(className);
Class<?> cls = classLoader.loadClass(className);
Method meth = cls.getMethod("main", String[].class);
String[] params = null;
meth.invoke(null, (Object) params);
} catch (Exception e) {
e.printStackTrace();
}
}