mysqli_fetch_array（）取数组错误，期望参数1是资源

-2

这个问题似乎重复，但我跟着我搜索的所有问题，但没有帮助，我得到mysql取数组错误在result2，虽然$email变量我在查询中使用是作为一个会话变量初始化之前，当我输入的所有电子邮件，是不是在member表，我获得所需的结果，但我得到的错误查询advert_account表，下面是我的代码mysqli_fetch_array（）取数组错误，期望参数1是资源

$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"; 
} 
}

来源

2017-04-19 Abdullahi Yusuf

在'SELECT * advert_account'中丢失'FROM' – Saty

是@Saty这是错误，已修复它，它的工作，发布它作为答案，所以我可以标记 –

检查下面的答案！ – Saty

您选择的sql语句写入正确。尝试使用以下代码：

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

而不是检查！

来源

2017-04-19 10:19:20 mageDev0688

mysqli_fetch_array（）取数组错误，期望参数1是资源

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