我创建Restfull Api以流式传输视频.mp4格式它按我的意愿运行。之后我添加下拉列表在dir中有视频列表。当用户选择视频时它会流式传输我用jquery做了但我有问题当我想制作精选视频时它给我发现了404
html
<!DOCTYPE html>
<html>
<head>
<title> </title>
<script src="Scripts/jquery-2.1.1.js"></script>
<script src="Scripts/Script.js"></script>
</head>
<body>
<select id="ddlcas"></select>
<br />
<video id="mainPlayer" width="1280" height="720"
autoplay="autoplay" controls="controls" onloadeddata="onLoad()">
<source src="" />
</video>
</body>
</html>
的javascript
$(function () {
//variables
var selectedFile = "";
////load data into table to view all car
$("#ddlcas").load(GetFiles());
function GetFiles() {
jQuery.support.cors = true;
$.ajax({
url: 'api/media/GetFiles',
type: 'GET',
dataType: 'json',
success: function (data) {
var appenddata;
$.each(data, function (key, value) {
appenddata += "<option value = '" + value + " '>" + value + " </option>";
});
$('#ddlcas').html(appenddata);
},
error: function (x, y, z) {
alert(x + '\n' + y + '\n' + z);
}
});
}
$("#ddlcas").change(function () {
selectedFile = $('#ddlcas').val().replace(/^.*[\\\/]/, '');
$("source").load(Paly(selectedFile));
});
function Paly(src)
{
jQuery.support.cors = true;
$.ajax({
url: 'api/media/play?f=' + selectedFile,
type: 'GET',
dataType: 'json',
success: function (data) {
$('source').attr('src', data);;
},
error: function (x, y, z) {
alert(x + '\n' + y + '\n' + z);
}
});
}
});
答案 0 :(得分:0)
试试这个
$("#ddlcas").load(GetFiles());
function GetFiles() {
jQuery.support.cors = true;
$.ajax({
url: 'api/media/GetFiles',
type: 'GET',
dataType: 'json',
success: function (data) {
var appenddata;
$.each(data, function (key, value) {
appenddata += "<option value = '" + value + " '>" + value + " </option>";
});
$('#ddlcas').html(appenddata);
},
error: function (x, y, z) {
alert(x + '\n' + y + '\n' + z);
}
});
}
$("#ddlcas").change(function () {
var selectedFile = $('#ddlcas').val().replace(/^.*[\\\/]/, '');
$('#mainPlayer source').attr('src', 'api/media/play?f=' + selectedFile);;
$('#mainPlayer')[0].load();
});