java - 这有什么问题？我找不到符号

Question

这是一个类的代码方法的前四行：

这是在“ CrudController ”类中

public String create(Object entity, Class clazz, String req_id, HttpServletResponse response) throws NullPointerException {
    if (entity == null ) throw NullPointerException("ENTITY");
    else if(clazz == null) throw NullPointerException("CLAZZ");
            else if(response == null) throw NullPointerException("RESPONSE");

我收到这三行的这些错误：

[ERROR] /path/to/CrudController.java:[42,29] error: cannot find symbol
[ERROR]  class CrudController
[ERROR] /path/to/CrudController.java:[43,31] error: cannot find symbol
[ERROR]  class CrudController
[ERROR]/pathto/CrudController.java:[44,48] error: cannot find symbol
[ERROR]  class CrudController

位置编号（29、31 和 48 在 '==' 的开头是正确的

Answer 1

全部更改throw NullPointerException(STRING);为throw new NullPointerException(STRING);

在这里，您正在抛出异常。异常是相应异常类型类的实例，因此您需要抛出一个需要为什么new关键字的实例。

public String create(Object entity, Class clazz, String req_id, HttpServletResponse     response) throws NullPointerException {
if (entity == null ) throw new NullPointerException("ENTITY");
else if(clazz == null) throw new NullPointerException("CLAZZ");
        else if(response == null) throw new NullPointerException("RESPONSE");

Answer 2

throw NullPointerException("CLAZZ");不是法律声明。您必须使用new运算符来创建NullPointerException

例如

throw new NullPointerException(STRING);