[英]JSP failed to direct to the Post method in the Servlet class in JavaEE and tomcat 9
我在jsp中有以下表单,其中包含一个文本字段和一个按钮
<form action="/login.do" method="Post">
Enter
<br/><br/>
Your Name <input type="text" name="name"/>
<br/><br/>
Password <input type="text" name="password"/>
<br/><br/>
<select name="Gender">
<option value="Male">Male</option>
<option value="Female">Female</option>
</select>
<br/><br/>
<input type="submit" value="Login"/>
我的Servlet:
@WebServlet(urlPatterns = "/login.do")
public class LoginServlet extends HttpServlet{
protected void doGet(HttpServletRequest request, javax.servlet.http.HttpServletResponse response)
throws ServletException, IOException{
request.getRequestDispatcher("/WEB-INF/view/login.jsp").forward(request, response);
}
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String name =request.getParameter("name");
String password = request.getParameter("password");
String gender = ((String)request.getParameter("Gender")=="Female")? "Ms.":"Mr.";
if(new ValidateUser().validate(name, password)){
request.setAttribute("name", name);
request.setAttribute("password", password);
request.setAttribute("gender", gender);
request.getRequestDispatcher("/WEB-INF/view/welcome.jsp").forward(request, response);;
} else {
request.setAttribute("errorMes", "Login Failed");
request.getRequestDispatcher("/WEB-INF/view/login.jsp").forward(request, response);;
}
}
根据一个教程,当我单击jsp页面中的按钮时,它将触发Servlet类中的doPost()方法。
但是我一直收到HTTP状态404-/login.do错误
请帮忙! ...
我的web.xml
<?xml version="1.0" encoding="UTF-8"?>
<!-- webapp/WEB-INF/web.xml -->
<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" version="3.0">
<display-name>To do List</display-name>
<welcome-file-list>
<welcome-file>login.do</welcome-file>
</welcome-file-list>
运行项目时,我正在连接到以下URL: http:// localhost:8081 / in28Minutes-first-webapp /
当我单击按钮时,我已连接到以下URL: http:// localhost:8081 / login.do ,它显示“ HTTP状态404-/login.do”错误
我已经尝试过“ $ {pageContext.request.contextPath} /login.do”,它仍然给出错误
我有这个pom.xml
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.in28minutes</groupId>
<artifactId>in28Minutes-first-webapp</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>war</packaging>
<dependencies>
<dependency>
<groupId>javax</groupId>
<artifactId>javaee-web-api</artifactId>
<version>6.0</version>
<scope>provided</scope>
</dependency>
</dependencies>
<build>
<pluginManagement>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.2</version>
<configuration>
<verbose>true</verbose>
<source>1.7</source>
<target>1.7</target>
<showWarnings>true</showWarnings>
</configuration>
</plugin>
<plugin>
<groupId>org.apache.tomcat.maven</groupId>
<artifactId>tomcat7-maven-plugin</artifactId>
<version>2.2</version>
<configuration>
<path>/</path>
<contextReloadable>true</contextReloadable>
</configuration>
</plugin>
</plugins>
</pluginManagement>
</build>
我的建议是:
您应该在表单标签集位置的操作参数中使用页面localhost:8081 / in28Minutes-first-webapp / login.do而不是localhost:8081 / login.do。 尝试设置以下设置之一: action="login.do"
, action="localhost:8081/in28Minutes-first-webapp/login.do"
或action="/in28Minutes-first-webapp/login.do"
。
我已经尝试过“ $ {pageContext.request.contextPath} /login.do”,它仍然给出错误
您给定的错误如何? ${pageContext.request.contextPath}
的值如何?
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.