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自动播放从数据库中检索到的位置的视频

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我曾经有人要求制作一个显示 Profiles 视频的简单仪表板 . 视频的位置保存在数据库中，因此我需要先检索它 . 这是我在 index.php 的代码

<body onload="onload();">
    <video style="border: 5px solid black" controls="" id="video" width="720" height="480" onended="onVideoEnded();">
    </video>
    <script>
        var vid_list = [];
        var list_index = 0;
        var video_player = null;
        function onload(){
            console.log("body loaded");
            var xmlhttp = new XMLHttpRequest();
            xmlhttp.open("GET", "get_location.php", true);
            xmlhttp.send();
            xmlhttp.onreadystatechange = function(){
                if (this.readyState == 4 && this.status == 200) {
                    var data = JSON.parse(this.responseText);
                    vid_list = [];
                    for(var i = 0; i < data.length; i++){
                        vid_list.push(data[i].lokasi);
                        console.log(vid_list[i]); //running
                    }
                }
            }
            for(var x = 0; x < vid_list.length; x++){
                console.log(vid_list[x]);
            }
            video_player = document.getElementById("video");
            video_player.setAttribute("src", vid_list[list_index]);
            video_player.play();
        }

        function onVideoEnded(){
            //console.log("video ended");
            if(list_index < vid_list.length - 1){
                list_index++;
            }
            else{
                list_index = 0;
            }
            video_player.setAttribute("src", vid_list[list_index]);
            video_player.play();
        }
</script>
</body>

这是 get_location.php

<?php
    $con = mysqli_connect("localhost","root","","video");

    $sql = "SELECT * FROM video";
    $result = mysqli_query($con,$sql);

    $data = array();
    while($row = mysqli_fetch_assoc($result)) {
        $data[] = $row;
    }
    echo json_encode($data);
?>

问题是视频无法播放，在视频屏幕上我收到消息

不支持视频格式或MIME类型

并在浏览器控制台中，收到此错误消息：

HTTP加载失败，状态为404.媒体资源加载http：// localhost / video_autoplay / undefined失败 .

但是当我尝试在控制台中调用类似 vid_list[1]; 的值时，该值存在，这意味着 readyState = 4 和 state = 200

有人能帮助我吗？

javascript php autoplay

1 回答

  • 1

    完成后，我将代码更改为：

    <body>
    <video style="border: 5px solid black" controls="" id="video" width="720" height="480" onended="onVideoEnded();">
    </video>
</body>
    <script>
        var vid_list = [];
        var xmlhttp = new XMLHttpRequest();
        xmlhttp.open("GET", "get_location.php", true);
        xmlhttp.send();
        xmlhttp.onreadystatechange = function(){
            if (this.readyState == 4 && this.status == 200) {
                var data = JSON.parse(this.responseText);
                vid_list = [];
                for(var i = 0; i < data.length; i++){
                    vid_list.push(data[i].lokasi);
                    console.log(vid_list[i]);//running
                }
                testing(vid_list);
            }
        }
        //console.log(vid_list);
        var list_index = 0;
        var video_player = null;
        function testing(param){
            console.log(param);
            // console.log("body loaded");
            video_player = document.getElementById("video");
            video_player.setAttribute("src", param[list_index]);
            console.log(list_index);
            console.log(param);
            video_player.play();
            video_player.volume = 0.30;
        }

        function onVideoEnded(){
            //console.log("video ended");
            if(list_index < vid_list.length - 1){
                list_index++;
            }
            else{
                list_index = 0;
            }
            video_player.setAttribute("src", vid_list[list_index]);
            video_player.play();
        }
        //document.getElementById("txtHint").innerHTML = vid_list[2];
</script>
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