Question

我正在使用jsp和amp;创建一个Web项目。 jsp页面中的servlet我在网页上播放视频时声明了视频标签然后我使用javaScript获得了播放位置但播放了未在servlet中传递的位置值。当我运行web项目时，只运行Index.jsp但值不在servlet中传递。

请帮助我们解决此问题

这是我的代码：

的index.jsp

    <%@page contentType="text/html" pageEncoding="UTF-8"%>


    <html>
    <body>

        <video id="myVideo" width="320" height="176" controls>
      <source src="mov_bbb.mp4" type="video/mp4">
      <source src="mov_bbb.ogg" type="video/ogg">
      Your browser does not support HTML5 video.
    </video>
    <p>Playback position: <span id="demo"></span></p>

    <script>
    // Get the video element with id="myVideo"
    var vid = document.getElementById("myVideo");


    vid.ontimeupdate = function() {myFunction()};

    function myFunction() {
    // Display the current position of the video in a <p> element with id="demo"
        var time=document.getElementById("demo").innerHTML = vid.currentTime;
        $.ajax({
           url:'http://localhost:8084/demotime/NewServlet',
           data:{
               myPost:time
           },
           type:'POST'
       });
    }

    </script>

    <p>Video courtesy of <a href="https://www.bigbuckbunny.org/" target="_blank">Big Buck Bunny</a>.</p>

    </body>
    </html>

这是Servlet文件

  package servlet;

    import java.io.IOException;
    import java.io.PrintWriter;
    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 javax.servlet.http.HttpSession;

    /**
     *
     * @author Administrator
     */
    @WebServlet(name ="NewServlet")
    public class NewServlet extends HttpServlet {

        protected void doGet(HttpServletRequest request, HttpServletResponse response)
                throws ServletException, IOException {
            PrintWriter out = response.getWriter();
            try {
                HttpSession session = request.getSession();
                String time="";
                if(time!=null){
                    time=request.getParameter("myPost");
                    out.print(time);
                }else{
                    System.out.println("time is null");
                }
            } catch (Exception e) {
            }
    {

           }
        }


    }

这是web.xml文件

    <?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" 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_3_1.xsd">
    <servlet>
        <servlet-name>NewServlet</servlet-name>
        <servlet-class>servlet.NewServlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>NewServlet</servlet-name>
        <url-pattern>/NewServlet</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
    <welcome-file-list>  
        <welcome-file>index.jsp</welcome-file>  
        </welcome-file-list>  
</web-app>

Answer 1

如果是单参数，您可以尝试传递如下面的查询参数

<强> JS：

var time=document.getElementById("demo").innerHTML = vid.currentTime;
$.ajax({
   url:'http://localhost:8084/demotime/NewServlet? myPost'+time,
   data:{
       myPost:time
   },
   type:'POST'
});

<强>的Servlet

String time=request.getParameter("myPost");

祝你好运

如何在servlet中播放视频时获取javaScript变量值

1 个答案: