servalet java中的tomcat错误

3 月 Questions & Answers 59

我是servlet的初学者。。。我想将一个简单的参数从html传递到servlet，但出现了一个错误“HTTP状态404-未找到”。。这是我的密码。。。网状物xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
    <servlet>
        <servlet-name>NewClass</servlet-name>
        <servlet-class>NewClass</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>NewClass</servlet-name>
        <url-pattern>/NewClass</url-pattern>
    </servlet-mapping>
    <welcome-file-list>
        <welcome-file>newhtml.html</welcome-file>
    </welcome-file-list>
</web-app>

newhtml。xml

<html>
    <head>
        <title>Simple Application</title>
        <meta charset="UTF-8">
        <meta name="viewport" content="width=device-width, initial-scale=1.0">
    </head>
    <body>
        <form action="/NewClass" >
            <input name="username" />
            <input type="submit"/>
                
        </form>
    </body>
</html><html>
    <head>
        <title>Simple Application</title>
        <meta charset="UTF-8">
        <meta name="viewport" content="width=device-width, initial-scale=1.0">
    </head>
    <body>
        <form action="/NewClass" >
            <input name="username" />
            <input type="submit"/>
                
        </form>
    </body>
</html><html>
    <head>
        <title>Simple Application</title>
        <meta charset="UTF-8">
        <meta name="viewport" content="width=device-width, initial-scale=1.0">
    </head>
    <body>
        <form action="/NewClass" >
            <input name="username" />
            <input type="submit"/>
                
        </form>
    </body>
</html>

新班级。爪哇$ public class NewClass extends HttpServlet{ protected void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException{ PrintWriter out = response.getWriter(); String req = request.getParameter("username"); out.println(req);}}

请任何人告诉我，为什么我会出现错误

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共 (1) 个答案

# 1 楼答案
你的<form action=""/>问题一定是
```
 <form action="NewClass" >
        <input name="username" />
        <input type="submit"/>
 </form>  
```
当你把/放在动作标记中时，它指向根，而不是指向上下文路径

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servalet java中的tomcat错误

共 (1) 个答案

# 1 楼答案