java无法使用servlet从索引访问不同的JSP文件

4 月，1 周 Questions & Answers 203

我正在尝试用TomCat建立一个简单的web应用程序，有一个servlet，它应该用指向两个不同JSP的两个按钮链接我的索引页面

它所做的事情是，当我点击这些按钮时，它不会更改页面，只会显示Servlet的名称“request”（抱歉，我不知道确切的术语）

另外，当我试图通过输入localhost:8080/reynumn直接访问时。jsp直接访问页面，它会停留在索引上，就好像它是那里唯一的页面一样，即使它看起来像是在尝试加载另一个页面

下面是Servlet和索引的代码

import javax.servlet.RequestDispatcher;
import javax.servlet.ServletConfig;
import javax.servlet.ServletContext;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;

@WebServlet("/Servlet")
public class Servlet extends HttpServlet {
    private ServletContext sc = this.getServletContext();

    public void init(ServletConfig config) throws ServletException{
        super.init(config);
        sc = this.getServletContext();
    }

    public Servlet(){
        super();
    }


    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        String op = request.getParameter("operation");

        if (op.equalsIgnoreCase("cv")){
            this.versPersonne(request, response);
        }
        if (op.equalsIgnoreCase("reunion")){
            this.versReunion(request, response);
        }

    }

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


    }

    private void versReunion(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        RequestDispatcher rd = sc.getRequestDispatcher("/reunion.jsp");
        rd.forward(request, response);
    }

    private void versPersonne(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        RequestDispatcher rd = sc.getRequestDispatcher("/cv.jsp");
        rd.forward(request, response);
    }


}

这是索引

<%@ page contentType="text/html;charset=UTF-8" %>
<html>
<head>
  <title>$Title$</title>
</head>
<body>

<form action="Servlet?operation=cv" method="post">
  <br><input type=submit value="Page personne"/>
</form>

<form action="Servlet?operation=reunion" method="post">
  <br><input type=submit value="Page réunion"/>
</form>

</body>
</html>

以防万一，这里是网络。XML文件：

<?xml version="1.0" encoding="UTF-8"?>
<web-app 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_4_0.xsd"
         version="4.0">

    <servlet>
        <servlet-name>Servlet</servlet-name>
        <jsp-file>/index.jsp</jsp-file>
    </servlet>
    <servlet-mapping>
        <servlet-name>Servlet</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>

</web-app>

我的JSP文件都在web文件夹中

我非常努力地寻找答案，尝试了许多不同的方法和配置，但似乎找不到答案

我确实认为问题在某种程度上在于servlet配置，但无法真正找到答案

更新：由于第一个答案，我删除了XML文件并更新了DoGet方法。此时，web应用确实尝试切换页面，但它向我抛出了一个500错误异常，告诉我它未能实例化servlet。下面是日志：

javax.servlet.ServletException: Erreur à l'instantiation de la classe servlet [com.example.Servlet]
org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:491)
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:92)
org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:668)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:343)
org.apache.coyote.http11.Http11Processor.service(Http11Processor.java:408)
org.apache.coyote.AbstractProcessorLight.process(AbstractProcessorLight.java:66)
org.apache.coyote.AbstractProtocol$ConnectionHandler.process(AbstractProtocol.java:764)
org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1388)
org.apache.tomcat.util.net.SocketProcessorBase.run(SocketProcessorBase.java:49)
java.base/java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1135)
java.base/java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:635)
org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
java.base/java.lang.Thread.run(Thread.java:844)

根本原因是：

java.lang.NullPointerException
javax.servlet.GenericServlet.getServletContext(GenericServlet.java:123)
com.example.Servlet.<init>(Servlet.java:15)
java.base/jdk.internal.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
java.base/jdk.internal.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62)
java.base/jdk.internal.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
java.base/java.lang.reflect.Constructor.newInstance(Constructor.java:488)
org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:491)
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:92)
org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:668)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:343)
org.apache.coyote.http11.Http11Processor.service(Http11Processor.java:408)
org.apache.coyote.AbstractProcessorLight.process(AbstractProcessorLight.java:66)
org.apache.coyote.AbstractProtocol$ConnectionHandler.process(AbstractProtocol.java:764)
org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1388)
org.apache.tomcat.util.net.SocketProcessorBase.run(SocketProcessorBase.java:49)
java.base/java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1135)
java.base/java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:635)
org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
java.base/java.lang.Thread.run(Thread.java:844)

这是项目结构：

https://imgur.com/a/YTpq4hD

提前感谢您的回答

Tags:

共 (2) 个答案

# 1 楼答案

你的web.xml文件就是罪魁祸首

您希望在WebServlet注释中将servlet用作/Servlet，但同时，xml文件将index.jsp映射到带有/*的任何URL

正如您已经用WebServlet注释声明了映射，并且index.jsp是默认页面，只需删除web.xml的映射，并将其拆分为：

<web-app 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_4_0.xsd"
         version="4.0">

</web-app>

或者更简单地移除网络。xml文件，因为它不包含任何更长的内容

servlet代码中还有一个错误：您试图在创建时初始化ServletContext，但不应该这样做。应该是：

@WebServlet("/Servlet")
public class Servlet extends HttpServlet {
    private ServletContext sc;    // delay initialization to init

    public void init(ServletConfig config) throws ServletException{
        super.init(config);
        sc = this.getServletContext();
    }

但实际上，您没有理由存储对ServletContext的引用。当您只需从Request中获取RequestDispatcher时，您仅使用它来获取RequestDispatcher：

private void versReunion(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    RequestDispatcher rd = request.getRequestDispatcher("/reunion.jsp");
    rd.forward(request, response);
}

就这些。在versPersonne中执行同样的操作，您可以删除任何引用sc的行

# 2 楼答案
调用方法doPost（请求、响应）；在如下所示的doGet方法中，这将解决您的问题
```
protected void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {

    doPost(request, response);

}
```
也在索引中。jsp您已经将表单操作指定为"Servlet?operation=reunion" 将其更改为"/Servlet?operation=reunion"

同样地，将"Servlet?operation=cv"更改为"/Servlet?operation=cv"

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java无法使用servlet从索引访问不同的JSP文件

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