SQLI查詢

Question

我建立一個音樂播放器，試圖打印僅我的數據庫的第一行使用它作爲播放列表，這將是當前播放的第一首歌曲只需要第一個值和其他人都將被存儲在一個JQuery ListView。我收到此錯誤代碼：

注意：類mysqli_result的對象不能轉換在C INT：\用戶\上線39 Xaimer \桌面\互動\ USBWebserver V8.6 \ ROOT \ playlist.php 39

行是if語句

<?php 
echo'<p>Playlist</p>'; 
$sql="SELECT * FROM co1706assigment.tracks INNER JOIN co1706assigment.playlist ON tracks.track_id=playlist.track_id "; 
$playlistcounter=mysqli_query($conn,"SELECT count(*) FROM co1706assigment.playlist")or die(mysqli_error($conn)); 
$result = mysqli_query($conn, $sql) or die(mysqli_error($conn)); 
while (($row = mysqli_fetch_array($result))&&($row1=mysqli_fetch_array($playlistcounter))) // it takes all the results from sql query 
{ 
    if($playlistcounter<2) 
    { 
     echo'<ul data-role="listview" data-filter="true"> 
          <li><img id="song" src="'.$row["image"].'"/><p class="name">Track Name: </p>'.$row["name"].' 
          <p class="name">Album Name: </p>'.$row["album"].' 
          <p class="name">Sample</p> 
          <audio controls> 
          <source src="'.$row["sample"].'" type="audio/mpeg"> 
          </audio> 
          </li> 
          </ul>'; 
    } 
} 
echo'<p> My Playlist</p>'; 
while ($row = mysqli_fetch_array($result)) // it takes all the results from sql query 
{ 
    echo'<ul data-role="listview"> 
    <li><a href="TrackDescription.php" data-ajax="false">'.$row["name"].'</a></li> 
    </ul>'; 
} 
?> 

Answer 1

使用mysqli_fetch_assoc看看
http://php.net/manual/en/mysqli-result.fetch-assoc.php

OR

使用這樣while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)){}