Question

java字符串如何解析成能运行的java代码?

Answer 1

java字符串如何解析成运行的java代码
有些情况下，不得不动态运行Java代码，以便提供更加灵活的方式，以下代码可参考（在JDK 1.5+平台上运行通过）：

public static void main(String[] args) {
int i = 10;
String code = "System.out.println(\"Hello World!\"+(13+2*5/3));";
code += "for(int i=0;i<" + i + ";i++){";
code += " System.out.println(Math.pow(i,2));";
code += "}";
try {
run(code);
} catch (Exception e) {
e.printStackTrace();
}
}

private synchronized static File compile(String code) throws Exception {
File file = File.createTempFile("JavaRuntime", ".java", new File(System.getProperty("user.dir")));
file.deleteOnExit();
// 获得类名
String classname = getBaseFileName(file);
// 将代码输出到文件
PrintWriter out = new PrintWriter(new FileOutputStream(file));
out.println(getClassCode(code, classname));
out.close();
// 编译生成的java文件
String[] cpargs = new String[] { "-d",
System.getProperty("user.dir") + "\\WebRoot\\WEB-INF\\classes",
file.getName() };
int status = Main.compile(cpargs);

if (status != 0) {
throw new Exception("语法错误！");
}
return file;

}

private static synchronized void run(String code) throws Exception {
String classname = getBaseFileName(compile(code));
new File(System.getProperty("user.dir")
+ "\\WebRoot\\WEB-INF\\classes\\" + classname + ".class")
.deleteOnExit();
try {
Class cls = Class.forName(classname);
Method main = cls.getMethod("method", null);
main.invoke(cls, null);
} catch (Exception se) {
se.printStackTrace();
}
}

private static String getClassCode(String code, String className) {
StringBuffer text = new StringBuffer();
text.append("public class " + className + "{\n");
text.append(" public static void method(){\n");
text.append(" " + code + "\n");
text.append(" }\n");
text.append("}");
return text.toString();
}

private static String getBaseFileName(File file) {
String fileName = file.getName();
int index = fileName.indexOf(".");
String result = "";
if (index != -1) {
result = fileName.substring(0, index);
} else {
result = fileName;
}
return result;
}

Answer 2

这个还是比较难实现的。建议你换个思路。字符串可以放在js中用eval函数执行。