有人要求我制作一个简单的仪表板来显示个人资料视频。视频的位置保存在数据库中，所以我需要先检索它。
这是我在 index.php
<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，在视频屏幕中我收到消息
Video Format or MIME type is not supported
在浏览器控制台中，收到此错误消息:
HTTP load failed with status 404. Load of media resource
http://localhost/video_autoplay/undefined failed.
但是当我尝试在控制台中调用像 vid_list[1]; 这样的值时，该值存在，这意味着 readyState = 4 和 state = 200
有人可以帮我吗？
最佳答案
完成，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>
关于javascript - 自动播放从数据库检索的位置的视频，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/48249080/