2015-12-19 63 views
-2

我正在編寫一個腳本,將mp3上傳到我的服務器中的文件夾,並將標題,網址和藝術家添加到「音頻」表中。PHP MySQL多個mp3上傳數據庫

,但它不工作

數據庫

<? 
$objConnect = mysql_connect("localhost","root","root") or die("Error Connect to Database"); 
$objDB = mysql_select_db("mydatabase"); 

for($i=0;$i<count($_FILES["filUpload"]["name"]);$i++) 
$target_dir = "/mounted-storage/home150/sub007/sc80538-VHHY/website.com/audio/files/"; 
$trackName = $_FILES['foto']['name']; 
$titel  = htmlspecialchars($_POST['titel']); 
$artist  = $_POST['artiest']; 
{ 
    if($_FILES["filUpload"]["name"][$i] != "") 
    { 
     if(move_uploaded_file($_FILES["filUpload"]["tmp_name"][$i],"myfile/".$_FILES["filUpload"]["name"][$i])) 
     { 
      //*** Insert Record ***// 
      mysql_query("INSERT INTO `audio` (titel, url, categorie) values ('".$titel."', 'http://website.com/audio/files/".$trackName."','".$artist."')"); 
     } 
    } 
} 

echo "Copy/Upload Complete<br>"; ?> 

和HTML表單

PHP MySQL的多個MP3上傳:

\t <form name="form1" method="post" action="upload.php" enctype="multipart/form-data"> 
 
Track1 : <input name="titel" type="text" id="Titel" size="63" /><p> 
 
\t <input type="file" name="filUpload[]"><br> 
 
    <br><br> 
 
    Track2 : <input name="titel" type="text" id="Titel" size="63" /><p> 
 
\t <input type="file" name="filUpload[]"><br> 
 
    <br><br> 
 
    Track3 : <input name="titel" type="text" id="Titel" size="63" /><p> 
 
\t <input type="file" name="filUpload[]"><br> 
 
    <br><br> 
 
    Track4 : <input name="titel" type="text" id="Titel" size="63" /><p> 
 
\t <input type="file" name="filUpload[]"><br> 
 
    <br><br> 
 
    Track5 : <input name="titel" type="text" id="Titel" size="63" /><p> 
 
\t <input type="file" name="filUpload[]"><br> 
 
    <br><br> 
 
    Track6 : <input name="titel" type="text" id="Titel" size="63" /><p> 
 
\t <input type="file" name="filUpload[]"><br> 
 
    <br><br> 
 
    <select name="artiest"> 
 
       <?php 
 
$query = mysql_query("SELECT * FROM artiesten ORDER BY naam ASC"); 
 
while ($array = mysql_fetch_assoc($query)){ 
 
\t echo "<option value=\"". $array['naam'] ."\">". $array['naam']. "</option>"; 
 
} 
 
?> 
 
      </select> 
 
    
 
\t <input name="btnSubmit" type="submit" value="Submit"> 
 
\t </form>

+0

那你希望我們猜測甚至是錯誤的信息? – Shadow

+0

仔細看看這個'$ trackName = $ _FILES ['foto'] ['name'];''和'$ _POST ['artiest']'等等。你需要學習如何調試代碼。 –

+0

http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index –

回答

0

這只是一個示例代碼你必須適合你的網站D收藏MySQL連接潔具我留下評論或只包括外部文件

PHP代碼

if (isset($_POST["upload"])){ 


    $userdir = "mp3/songs/"; 
    for($i=0; $i<count($_FILES['mp3']['name']); $i++) { 

    $target_file = $userdir . basename($_FILES["mp3"]["name"][$i]); 
    $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION); 




      if (!file_exists($userdir)){ 
      mkdir($userdir, 0777, true); 
      } 
     if (move_uploaded_file($_FILES["mp3"]["tmp_name"][$i], $target_file)) { 

      // your mysql connect code or you have to be included external file with $connect variable or just rename it 
      $song = $_FILES["mp3"]["name"][0]; 
      $sql = sprintf(
          "INSERT INTO music (mp3) VALUES ('%s')", 
          mysqli_escape_string($connect, $song) 
          ); 

          $res = mysqli_query($connect,$sql); 

      echo "The file has been uploaded.\n"; 
     } else { 

      echo "Sorry, there was an error uploading your file.\n"; 
     } 

    } 
} 

的html代碼:

<form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post" enctype="multipart/form-data"> 
<input type="file" name="mp3[]" multiple="true" > 
<input type="submit" name="upload" value="upload"> 
</form> 
+0

嘿謝謝你我的朋友爲你的答案,但我還需要添加數據庫中每個軌道的標題名稱 – Fouad

+0

第一我認爲你應該命名每個輸入字段像title1 title2和ect只是爲了識別女巫是女巫,後來只是輸入它們同樣的方式,我用一個mp3示例演示你的'$ sql = sprintf( 「INSERT INTO music(title,mp3)VALUES('%s','%s')」, mysqli_escape_string($ connect,$ title), mysqli_escape_string($ connect,$ song) );'標題可能等於$ _POST [「title1」];做到這一點將所有的領域,或使用'或'來做這個' – yahoo5000