如何从 PHP 中的此 MySQL 查询响应中获取值

Question

以下是我的查询。我评论说我在代码中尝试过

$query = "SELECT SUM(paymentAmount) From salepayment WHERE adminUserId ='$adminUserId' AND invoiceId = '$salesInvoiceId' ";
                    $this->res = $this->db->prepare($query);
                    $this->result = $this->res->execute();
                    $countAmount = $this->res->rowCount();
                    if ($countAmount > 0) {
                        $sumOfPaymentAmount = $this->res->fetch(PDO::FETCH_OBJ);
                        echo json_encode($sumOfPaymentAmount);
                        // $totalPaid = $sumOfPaymentAmount->"SUM(paymentAmount)";
                        // echo json_encode($totalPaid);
                    }else{
                        $paid = 0;
                    }

这是我的回复

{
    "SUM(paymentAmount)": "350"
}

我想将此值放入 $paid 变量中。

感谢您的帮助。我是 PHP 和服务器端编码的初学者，请原谅我的错误。

Answer 1

这应该可以通过在查询中的 SUM(paymentAmount) 之后添加 as paid 来完成，如下所示：

SELECT SUM(paymentAmount) as paid From...

如何从 PHP 中的此 MySQL 查询响应中获取值

How to fetch value from this MySQL query response in PHP

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