英文:
Servlet error : HTTP Status 405 - HTTP method GET is not supported by this URL
问题
以下是翻译好的内容：
当我尝试在Eclipse中运行代码时，它不会传递参数。
请检查下面的图像网址。
但是当我单独运行HTML代码时，它会传递值。
我不知道为什么会这样。
我的Servlet代码是
package com.example;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class AddServelet extends HttpServlet
{
public void services(HttpServletRequest req, HttpServletResponse resp) {
int i = Integer.parseInt(req.getParameter("num1"));
int j = Integer.parseInt(req.getParameter("num2"));
int k = i+j;
System.out.println(k);
}
}
我的web.xml代码是
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd" id="WebApp_ID" version="4.0">
<servlet>
<servlet-name>abc</servlet-name>
<servlet-class>com.example.AddServelet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>abc</servlet-name>
<url-pattern>/add</url-pattern>
</servlet-mapping>
</web-app>
我的HTML代码是
<html>
<body>
<form action="add">
Number 1: <input type="number" name="num1"><br>
Number 2: <input type="number" name="num2"><br>
<input type="submit">
</form>
</body>
</html>
有人能帮助我解决这个问题吗？
提前谢谢您。
英文:
When i try to run the code in eclipse it won't pass the arguments.
please check the below images url.
But when i individually run the html code it passes the values.
I don't know why this work like this.
My Servlet code is
package com.example;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class AddServelet extends HttpServlet
{
public void services(HttpServletRequest req, HttpServletResponse resp) {
int i = Integer.parseInt(req.getParameter("num1"));
int j = Integer.parseInt(req.getParameter("num2"));
int k = i+j;
System.out.println(k);
}
}
and my web.xml code is
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd" id="WebApp_ID" version="4.0">
<servlet>
<servlet-name>abc</servlet-name>
<servlet-class>com.example.AddServelet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>abc</servlet-name>
<url-pattern>/add</url-pattern>
</servlet-mapping>
</web-app>
my html code
<html>
<body>
<form action="add">
Number 1: <input type="number" name="num1"><br>
Number 2: <input type="number" name="num2"><br>
<input type="submit">
</form>
</body>
</html>
Can anyone please help me with this.
Thank you in advance.
答案1
得分: 2
你需要重写doPost方法，因为你正在提交表单，而这是一个POST请求。
public class AddServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public AddServlet() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
int i = Integer.parseInt(request.getParameter("num1"));
int j = Integer.parseInt(request.getParameter("num2"));
int k = i + j;
System.out.println(k);
}
}
英文:
You need to override the doPost method because you are submitting the form and it's a post request.
public class AddServelet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public AddServelet() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
int i = Integer.parseInt(req.getParameter("num1"));
int j = Integer.parseInt(req.getParameter("num2"));
int k = i+j;
System.out.println(k);
}
}
答案2
得分: 1
覆盖 HttpServlet 的 protected void doGet(HttpServletRequest req, HttpServletResponse resp) 方法，如果你想处理 GET 请求。
英文:
override the HttpServlet protected void doGet(HttpServletRequest req, HttpServletResponse resp) method if you want to handle a GET request
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