php - 从 mysqli_query 返回不满足条件

Question

total_money如果表中的total_balance 大于10应显示的确切用户消息但不起作用，我应该完成以下条件

任何语法或逻辑错误？？ 怎么解决？

 <?php
     $db = mysqli_connect('localhost', '********', '**********', '*********');

      $user_check_query =  "SELECT total_money FROM total_balance WHERE username = '" . $_SESSION['username'] . "' ";
    $result = mysqli_query($db, $user_check_query);
      $row = mysqli_fetch_array($result);

      if($result >10){
          echo "Enough";
        } else {
            echo "Not Enough Money";
      }
      ?> 

桌子： total_balance

 ID   username        total_money
+----+--------------+---------------+
|1   | John         |  100          |
+----+--------------+---------------+
|2   | Alex         |  10           |
+----+--------------+---------------+
|3   | Pani         |  5            |
+----+--------------+---------------+

标签： phpmysqlmysqli

Answer 1

改变

if($result >10){
          echo "Enough";
        } else {
            echo "Not Enough Money";
      }

到

if($row['total_money']  >10){
          echo "Enough";
        } else {
            echo "Not Enough Money";
      }