使用Jquery和Ajax/Json上传视频

需要在此处添加代码才能添加上传视频的功能？使用Jquery和Ajax/Json上传视频

<div id="title">Upload Videos</div> 
<div class="vidUpload"> 
    <form id="vidUpload" enctype="multipart/form-data"> 
<input name="vidName" type="text" required="required" id="vidName" placeholder="Enter Video Name Here" title="Video Name"> 
<br> 
<textarea name="videoDescription" id="videoDescription" required class="videoDescription" placeholder="Enter Video Description Here" title="Enter Video Description Here"></textarea> 

<select name="select" required class="choosevidCat" id="choosevidCat"> 
<option value="">Choose the Catagory for your Video Here</option> 
      <?php 
    $sql = ("SELECT albumId, albumName, albumSelect FROM albums"); 
if (mysqli_connect_errno()) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$result = mysqli_query($mysqli, $sql); 

while($row = mysqli_fetch_array($result)) { 
$albumid = ($row['albumId']); 
$album_name = ($row['albumName']); 
$album_name1 = ($row['albumSelect']); 
echo "<option value=".$album_name1. ">$album_name</option>"; 
} 

?> 


<option id="createCat" value="createCatagory">Create New Catagory Here</option> 
</select> 

<input type="file" name="video" id="video"> 
<input type="button" name="videoToUpload" id="videoToUpload" value="Upload"> 

</form> 
<div id="loader"></div> 
<div id="viduploadResult"></div>

jQuery的

<script type="text/javascript"> 

$(document).ready(function() { 

    $("#videoToUpload").click(function() { 

var vidName = $("#vidName").val(); 
var videoDescription = $("#videoDescription").val(); 
var albumName1 = $("#choosevidCat").val(); 
var vidFile =$("#video").val(); 

    // Put an animated GIF image insight of content 
    $("#loader").empty().html('<img src="/images/loader.gif" class="vidloader1"/>'); 

     $.post("includes/vid_upload.inc.php",{vidName: vidName, videoDescription: videoDescription, albumName1: albumName1, vidFile: vidFile}, function(json) 
     { 

      if(json.result === "success") { 

     $("#viduploadResult").html("The Video "+vidName+" has been Uploaded!"); 


//  // First remove all the existing options 
//  $('#choosevidCat').empty(); 
// 
//  // Load the content: 
//  $('#choosevidCat').load(location.href + "#choosevidCat > *"); 

    }else{ 
     $("#viduploadResult").html(json.message); 
    } 

     }); 
    }); 
}) 
</script>

我花了几个小时在看API的喜欢blueimp等，我只是想上传一个视频文件，并把它放在我的服务器上，从我这里有形式。任何帮助，将不胜感激

来源

2015-04-26 Tiny

这个问题看起来像要求替代品（“建议我...”）的问题。这些在Stack Overflow上是不允许的，因为它们没有解决特定的编程问题，它们太宽泛，基本上取决于用户的选择而不是其他的东西。尝试提出问题，例如“我正在使用...并且我已经尝试过......但它不起作用。”而不是“我想这样做......告诉我一些替代方案”。在问之前，请阅读http://stackoverflow.com/help/on-topic。祝你好运！ –