mysqli_fetch_assoc（）期望參數1是mysqli_result，null和mysqli_query（）中給出的參數1是mysqli，null給出

Question

我從這部分代碼中得到2個錯誤，請告訴我如何解決：

•mysqli_query（）預計參數1是mysqli的，在 空給出•mysqli_fetch_assoc（）預計參數1被mysqli_result，在

$sql = "SELECT * FROM user"; 
    $result = mysqli_query($conn, $sql); //connect to the database, then run the sql query 
    if (mysqli_num_rows($result) > 0) { 
     while($row = mysqli_fetch_assoc($result)) { 
      $id = $row['id']; 
      $sqlImg = "SELECT * FROM profileimg WHERE userid = '$id' "; 
      $resultImg = mysqli_query($sonn,$sqlImg); 
      while ($rowImg = mysqli_fetch_assoc($resultImg)) { 
       echo "<div class ='user-container'>"; 
        if ($rowImg['status'] == 0) { 
         echo "<img src='uploads/profile".$id.".jpg?".mt_rand()."'>"; 
        } else { 
         echo "<img src='uploads/profiledefault.jpg'>"; 
        } 
        echo "<p>".$row['username']."</p>"; 
       echo "</div>"; 
      } 
     } 
    } 

Answer 1

$result = mysqli_query($conn, $sql); //connect to the database, then run the sql query 

$康恩空給出沒有定義。