php - php中如何从对象中获取对象的值？

Question

我正在从成员表中选择一列。现在我想从该列中获取值。但是当我尝试获取该值时，我收到一个异常:

 <br />
<b>Notice</b>:  Trying to get property of non-object in
<b>C:\Program Files (x86)\Ampps\www\MLMapi\PurchaseProduct.php</b> on line
<b>54</b>
<br />

代码:

    $stmt = $dbConnection->prepare("select Member.sales_hold_fund from Member where member_id=?");
    $stmt->execute(array($member_id));
    $member = $stmt->fetch(PDO::FETCH_ASSOC);

        $fund = $member-> sales_hold_fund;// line54

        $sales_hold_fund = $fund + $amount;

我想从数据库获取 sales_hold_fund 值并将金额值添加到 sales_hold_fund。

成员(member)值(value):

 "Member": {
    "sales_hold_fund": "1000"
  }

有人可以帮忙吗？谢谢..

Answer 1

尝试

 $fund = $member['sales_hold_fund'];