使用Jquery和Ajax/Json上傳視頻

需要在此處添加代碼才能添加上傳視頻的功能？使用Jquery和Ajax/Json上傳視頻

<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>

我花了幾個小時在看API的喜歡blueimp等，我只是想上傳一個視頻文件，並把它放在我的服務器上，從我這裏有形式。任何幫助，將不勝感激

來源

2015-04-26 Tiny

這個問題看起來像要求替代品（「建議我...」）的問題。這些在Stack Overflow上是不允許的，因爲它們沒有解決特定的編程問題，它們太寬泛，基本上取決於用戶的選擇而不是其他的東西。嘗試提出問題，例如「我正在使用...並且我已經嘗試過......但它不起作用。」而不是「我想這樣做......告訴我一些替代方案」。在問之前，請閱讀http://stackoverflow.com/help/on-topic。祝你好運！ –