mysqli_fetch_array（）は、配列エラーを取得します。パラメータ1がリソースになることを期待しています。

Question

この質問は繰り返されているようですが、私は検索しても助けにはならず、$emailでもmysqlフェッチ配列エラーがresult2私はセッション変数として前に初期化されたクエリで使用される、と私はmemberテーブルにない任意のメールアドレスを入力したときに、私は必要な結果を得るが、私はadvert_accountテーブルを照会し、エラーを取得しています、以下の私のコードです

$email = $_SESSION['email']; 
if(count($_POST)>0){ 
$email = $_SESSION['email']; 
$transfered = $_REQUEST['transfer_amount']; 
$user_to = $_REQUEST["user_to"]; 
$result1=mysqli_query($db_handle,"SELECT * FROM members WHERE email='".$user_to."'"); 
$row1=mysqli_fetch_array($result1); 
if(is_array($row1)){ 
     $result2 = mysqli_query($db_handle,"SELECT * advert_account WHERE email='".$email."'"); 
     $row2=mysqli_fetch_array($result2); 
     if(is_array($row2)){ 
      $user_balance = $row2['balance']; 
      if($user_balance > 0 && $_REQUEST["transfer_amount"] <= $user_balance){ 
      $user_balance = $user_balance - $_REQUEST["transfer_amount"]; 
      mysqli_query($db_handle,"UPDATE advert_account SET balance='$user_balance' WHERE email='".$email."'"); 
      $result3 = mysqli_query($db_handle,"SELECT * advert_account WHERE email='".$user_to."'"); 
      $row3=mysqli_fetch_array($result3); 
      if(is_array($row3)){ 
       $to_balance = $row3['balance']; 
       $to_balance = $to_balance + $_REQUEST["transfer_amount"]; 
       mysqli_query($db_handle,"UPDATE advert_account SET balance='$to_balance' WHERE email='".$user_to."'"); 
       $transfer = "Your transfer of " .$transfered. "to" .$user_to. " was successful"; 

      } 
      else{ 

       mysqli_query($db_handle,"INSERT INTO advert_account(email,balance)VALUES('".$user_to."','".$transfered."')"); 
       $transfer = "Your transfer of " .$transfered. "to" .$user_to. " was successful"; 
      } 


     } 
     else{ 
      $transfer = "Insufficient fund to perform transaction, please top up your account and try again"; 
     } 
     } 

     else{ 
      $transfer = "Opps, it seems like you don't have advert account yet, please top up your account to start enjoying this service"; 
     } 

} 
else{ 
    $transfer = "Invalid destination account, please verify and try again"; 
} 
} 

Answer 1

選択したsql文が正しいと書かれています。以下のコードを試してください：

$result3 = mysqli_query($db_handle,"SELECT * FROM advert_account WHERE email='".$user_to."'"); 

チェックしてください！