php - mysqli_query 返回不满足条件

Question

这个问题已经有答案了:

Object can't be converted to a string in MySQLi PHP [duplicate] (2 个回答)

已关闭 4 年前。

如果total_money，我应该完成以下条件在表total_balance中大于10对于确切的用户消息应该显示但不起作用

有任何语法或逻辑错误吗？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";
      }
      ?> 

表: total_balance

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

Answer 1

改变

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

至

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