Question

下面是web.xml，servlet和jsp代码

import java.io.IOException;
import java.io.PrintWriter;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

/**
 * Servlet implementation class MyServlet
 */

public class MyServlet extends HttpServlet {
 private static final long serialVersionUID = 1L;

   /**
  * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
  */
 protected void doGost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
  // TODO Auto-generated method stub
  String userName = request.getParameter("userName");
  PrintWriter out = response.getWriter();
  out.println("hello "+userName+" how are you ?");
 }

}

JSPCode：

<!DOCTYPE html>
<html>
<head>
<meta charset="ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
 This is First Servlet
 <form action="firstTest">
  <table>
   <tr>
    <td>Username:</td>
    <td><input type="text" name="userName"/></td>
   </tr>
   <tr>
    <td> Password:</td>
    <td><input type="password" name="password" /></td>
   </tr>
   <tr>
    <td colspan="2"><input name = "sumbit "type="submit" /></td>
   </tr>
  </table>
 </form>
</body>
</html>

Web.xml代码：

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
  <display-name>Trial</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
  </welcome-file-list>

  <servlet>
   <servlet-name>FirstServlet</servlet-name>
   <servlet-class>MyServlet</servlet-class>
  </servlet>

  <servlet-mapping>
   <servlet-name>FirstServlet</servlet-name>
   <url-pattern>/firstTest</url-pattern>
  </servlet-mapping>
</web-app>

当我提供URL http：localhost：8080 /试用时，Index.jsp即将到来，但是当我提供用户名和密码时，URL正在转为http：localhost：8080 / firstTest而不是http：localhost：8080 / Trial / firstTest我得到405错误＆＃34; Tomcat错误HTTP状态405 - 此方法不支持HTTP方法GET＆＃34;在我的代码中出了什么问题

Answer 1

...错字

protected void doGost

更改为doPost或doGet ...

Answer 2

在你的jsp代码表单标签中你没有指定方法名，默认情况下它使用GET方法，所以只需替换这段代码，我希望它可以工作：

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;


public class MyServlet extends HttpServlet {
private static final long serialVersionUID = 1L;

protected void doGet(HttpServletRequest request,
        HttpServletResponse response) throws ServletException, IOException {
    String userName = request.getParameter("userName");
    PrintWriter out = response.getWriter();
    out.println("hello " + userName + " how are you ?");
}

protected void doPost(HttpServletRequest request,
        HttpServletResponse response) throws ServletException, IOException {
    doGet(request, response);
}

}

Answer 3

在我们部署到服务器之前解决此类拼写错误

我们使用@override注释来测试方法是否是重载方法。什么时候你编译java它会抛出异常。实际上编译器会检查语法错误等。但是在这种情况下编译器处理的doGost（）也是一种方法，通过在编译级别使用上面的注释，只有开发人员才能理解问题发生的位置。

因此。我的建议是@override用于重写方法。

表单提交不在servlet应用程序中工作

3 个答案: