无法获取 servlet 转发的 jsp 中的属性

Question

我有一个简单的 jsp 应用程序。那使用Maven.

Pom.xml:

<project xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://maven.apache.org/POM/4.0.0"
     xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd">
    <modelVersion>4.0.0</modelVersion>
    <groupId>Lab3.web_application</groupId>
    <artifactId>web_application</artifactId>
    <packaging>war</packaging>
    <version>1.0-SNAPSHOT</version>
    <name>web_application Maven Webapp</name>
    <url>http://maven.apache.org</url>

    <dependencies>
      <dependency>
        <groupId>junit</groupId>
        <artifactId>junit</artifactId>
        <version>3.8.1</version>
        <scope>test</scope>
      </dependency>

      <dependency>
        <groupId>javax</groupId>
        <artifactId>javaee-web-api</artifactId>
        <version>6.0</version>
        <scope>provided</scope>
      </dependency>
    </dependencies>

    <build>
    <finalName>web_application</finalName>
    <plugins>
        <plugin>
            <groupId>org.apache.maven.plugins</groupId>
            <artifactId>maven-compiler-plugin</artifactId>
            <version>2.3.2</version>
            <configuration>
                <source>1.6</source>
                <target>1.6</target>
            </configuration>
        </plugin>
        <plugin>
            <groupId>com.oracle.weblogic</groupId>
            <artifactId>wls-maven-plugin</artifactId>
            <version>12.1.3.0</version>
            <configuration>
                <middlewareHome>/home/bluebird/Oracle/Middleware/Oracle_Home</middlewareHome>
                <adminurl>t3://localhost:7001</adminurl>
                <domainHome>/home/bluebird/Oracle/Middleware/Oracle_Home/user_projects/domains/base_domain
                </domainHome>
                <user>weblogic</user>
                <password>yamahar6</password>
                <upload>true</upload>
                <remote>false</remote>
                <verbose>true</verbose>
                <source>./target/web_application.war</source>
                <name>${project.build.finalName}</name>
            </configuration>
            <executions>
                <execution>
                    <id>startserver</id>
                    <phase>initialize</phase>
                    <goals>
                        <goal>start-server</goal>
                    </goals>
                </execution>
                <execution>
                    <phase>pre-integration-test</phase>
                    <goals>
                        <goal>deploy</goal>
                    </goals>
                  </execution>
                </executions>
            </plugin>
         </plugins>
     </build>
  <distributionManagement>
    <snapshotRepository>
        <id>localSnapshot</id>
        <name>local snapshot repository</name>
        <url>file:///home/bluebird/.m2/repository</url>
    </snapshotRepository>
  </distributionManagement>
</project>

索引jsp：

<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<html>
  <head><title>Hello World ${requestScope.name}</title></head>
    <body>
        Hello World!<br/>
    <%
      out.println("Your IP address is " + request.getRemoteAddr());
      out.println("Your name is " +     request.getAttribute("name"));
     %>
    </body>
</html>

web.xml:

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >

<web-app>
  <display-name>Archetype Created Web Application</display-name>

  <servlet>
    <servlet-name>MainServlet</servlet-name>
    <servlet-class>MainServlet</servlet-class>
  </servlet>

  <servlet-mapping>
    <servlet-name>MainServlet</servlet-name>
    <url-pattern>/processForm</url-pattern>
  </servlet-mapping>

</web-app>

MainServlet.java:

@WebServlet("/processForm")
public class MainServlet extends HttpServlet {

    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp)
        throws ServletException, IOException {


        req.setAttribute("name", "Devcolibri");
        req.getRequestDispatcher("index.jsp").forward(req, resp);


    }

    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp)
        throws ServletException, IOException {

        super.doPost(req, resp);
    }

}

服务器成功运行并且应用程序正常运行。但是，我无法获取 servlet 设置的属性。在网络浏览器中我收到：

世界，您好！你的 IP 地址是 127.0.0.1 你的名字是 null

我读过这个但这对我没有帮助。

感谢所有阅读到最后的人。希望对你有所帮助。 P.S。对不起我的英语不好。我来自乌克兰。

Answer 1

我认为您阅读的上的答案解释了这种情况，但没有告诉您确切的 url 用途。 /processForm 是您必须请求才能调用 servlet 的映射路径。

因此，如果您的 Web 应用程序的上下文路径是 /web_application，您需要浏览到 /web_application/processForm。

如果您单独浏览 /web_application，您的容器将搜索 welcome files，并且由于 index.jsp 默认存在，您将请求 jsp无需调用 servlet。因此，servlet 中设置的请求属性将为空。

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