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java - Spring提交后Servlet无法访问POST方法

转载 作者:行者123 更新时间:2023-12-02 12:06:58 25 4
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我遇到一个问题,找不到答案。

通常情况下,我会使用 IDE 默认设置的所有内容,但根据要求,使用 Spring MVC,我必须删除 applicationContext.xmldispatcher-servlet.xml 只是为了满足标准。当我这样做时,它停止工作,因为 web.xml 文件,但我设法获取 Servlet 的 GET 方法,但是,当我使用 POST 方法时,它会向我抛出“请求的资源”不可用。”错误。

如何使用 POST 方法而不显示此错误?

这是我的index.jsp 文件

<%@taglib prefix="sql" uri="http://java.sun.com/jsp/jstl/sql"%>
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE html>

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Ejercicio usando FORM sin AJAX</title>
<link href="../css/bootstrap.min.css" rel="stylesheet">
<link rel="stylesheet" type="text/css" href="../css/main.css">
</head>

<body>
<nav class="navbar navbar-inverse navbar-fixed-top">
<div class="container">
<div class="navbar-header">
<button type="button" class="navbar-toggle collapsed" data-toggle="collapse" data-target="#navbar" aria-controls="navbar">
<span class="sr-only">Toggle navigation</span>
<span class="icon-bar"></span>
<span class="icon-bar"></span>
<span class="icon-bar"></span>
</button>
<a class="navbar-brand" href="./">Ejercicio</a>
</div>
<div id="navbar" class="collapse navbar-collapse">
<ul class="nav navbar-nav">
</ul>
</div>
</div>
</nav>

<div class="container">
<div class="col-md-12">
<h1>Ejercicio 1:- Usando Form y Action sin Ajax</h1>
</div>
<form action="../app/modulo/RecibirDatos.do" method="POST" id="form">
<div class="form-group col-md-6">
<label for="nombre">Nombre: </label>
<input type="text" name="nombre" id="nombre" class="form-control" placeholder="Inserte el nombre">
</div>
<div class="form-group col-md-6">
<label for="apellido">Apellido: </label>
<input type="text" name="apellido" id="apellido" class="form-control" placeholder="Inserte el apellido">
</div>
<input type="submit" value="Submit" class="btn btn-primary btn-block">
</form>
</div>

<script type="text/javascript" src="../js/jquery-3.2.1.js"></script>
<script type="text/javascript" src="../js/bootstrap.min.js"></script>
<script type="text/javascript" src="../js/bootbox.min.js"></script>
<script type="text/javascript" src="../js/main.js"></script>
</body>
</html>

这是我的 servlet 文件

package data.servlet;

import data.dataAccessObject.ManipulaPersona;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class RecibirDatos extends HttpServlet {

@Override
protected void service(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// Overriding service() usually isn't needed. - The default implementation mostly
// does the right thing&reg;
super.service(request, response);
}

/**
* Handles the HTTP <code>GET</code> method.
*
* @param request servlet request
* @param response servlet response
* @throws ServletException if a servlet-specific error occurs
* @throws IOException if an I/O error occurs
*/
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {

}

/**
* Handles the HTTP <code>POST</code> method.
*
* @param request servlet request
* @param response servlet response
* @throws ServletException if a servlet-specific error occurs
* @throws IOException if an I/O error occurs
*/
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
ManipulaPersona mp = new ManipulaPersona();
mp.crearPersona(request.getParameter("nombre"), request.getParameter("apellido"));
request.setAttribute("persona", mp.obtenerPersona());
request.getRequestDispatcher("../Views/resultado.jsp").forward(request, response);
}

/**
* Returns a short description of the servlet.
*
* @return a String containing servlet description
*/
@Override
public String getServletInfo() {
return "Short description";
}

}

这是我的 web.xml 文件:

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
<context-param>
<param-name></param-name>
<param-value></param-value>
</context-param>
<!-- <listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>2</load-on-startup>
</servlet> -->
<servlet>
<servlet-name>Inicio</servlet-name>
<servlet-class>data.servlet.Inicio</servlet-class>
</servlet>
<servlet>
<servlet-name>RecibirDatos</servlet-name>
<servlet-class>data.servlet.RecibirDatos</servlet-class>
</servlet> <!--
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.lol</url-pattern>
</servlet-mapping> -->
<servlet-mapping>
<servlet-name>Inicio</servlet-name>
<url-pattern>/Inicio</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>RecibirDatos</servlet-name>
<url-pattern>/app/modulo/RecibirDatos.do</url-pattern>

</servlet-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
<welcome-file-list>
<welcome-file>Inicio</welcome-file>
</welcome-file-list>
</web-app>

最佳答案

如果GET请求使用相同的代码,那么POST也应该可以工作。

您可能需要查看下面的代码行,这可能会导致失败

request.getRequestDispatcher("../Views/resultado.jsp").forward(request, response);

您需要检查 /app/Views/resultado.jsp 是否可以访问?

关于java - Spring提交后Servlet无法访问POST方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46835521/

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