java - 未找到操作，为什么？

Question

我使用如下方法创建了一个操作:

public class NomenclatureAction extends ActionSupport {

    // ...

    @Actions({
      @Action(value = "ajaxDoStuff",
              results = {
                    @Result(name = ActionSupport.SUCCESS, location = "success.jsp"),
                    @Result(name = ActionSupport.INPUT, location = "fail.jsp") 
              }),
      @Action(value = "index.action",
              results = {
                    @Result(name = ActionSupport.SUCCESS, location = "success.jsp"),
                    @Result(name = ActionSupport.INPUT, location = "fail.jsp") 
              })
    })
    public final String doStuff() {
        // ...
        return ActionSupport.SUCCESS;
    }

}

我想使用以下 URL 之一调用相同的方法 doStuff:

• http://my-server.com/public/namespace/ajaxDoStuff
• http://my-server.com/public/namespace/ajaxDoStuff.action
• http://my-server.com/public/namespace/index
• http://my-server.com/public/namespace/index.action

到目前为止，它适用于前两个 URL，但不适用于后两个 URL。
我错过了什么？

Answer 1

value 属性不应包含扩展名。

@Action(value = "index<del>.action</del>",
  results = {
        @Result(name = ActionSupport.SUCCESS, location = "success.jsp"),
        @Result(name = ActionSupport.INPUT, location = "fail.jsp") 
  })

您也可以尝试 http://my-server.com/public/namespace ，它应该由默认启用的约定 uknown 处理程序处理。