java如何传递HttpServletRequest参数？

Question

@RequestMapping("/returnformv2")
public String serveletTest(HttpServletRequest request){        
    String firstname = request.getParameter("fname");
    String lastname = request.getParameter("lname");
    String fullname = lastname.concat(firstname);
    request.setAttribute("fullname", fullname);     
    return "helloworld";
}

现在，如果我从JSP检索，${fullname}是lastname。康卡特（名字）。 让我展示一些基本的java代码，比如我调用serveletTest（“hello”）

public String serveletTest(String myString){        
    System.out.println(myString);
    myString = myString.concat(world);
    return "bye world";
}

myString应该仍然是“hello”对吗

对我来说，HttpServletRequest中的getParameter（）有意义，但不是请求。setAttribute（）。有人能解释一下这是怎么回事吗

Answer 1

我找到了我一直在寻找的答案。X( 请求调度员正在做所有的魔术

据javadoc说,

void javax.servlet.ServletRequest.setAttribute(String name, Object o)

Stores an attribute in this request. Attributes are reset between requests. This method is most often used in conjunction with RequestDispatcher.

网络。xml>；配置Spring MVC Dispatcher Servlet并设置URL映射

<servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring-mvc-demo-servlet.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>dispatcher</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>