如何从MySQL中的另一个表中只选择第一个匹配值？

以下是我的数据库架构：

付款表：

+------------+--------+--------+---------------------+
| payment_id | tab_id | amount | created             |
+------------+--------+--------+---------------------+
| 1          | 1      |  5     | 2017-05-22 12:14:27 |
| 2          | 2      |  10    | 2017-05-22 12:15:21 |
| 3          | 2      |  1     | 2017-05-22 13:11:14 |
+------------+--------+--------+---------------------+

选项卡表：

+------------+----------------+
| tab_id     | service_charge |
+------------+----------------+
| 1          | 1              |
| 2          | 3              |
+------------+----------------+

我需要计算每次付款的总金额+服务费，但服务费应仅包含在第一次付款匹配选项卡id中

我当前的查询：

SELECT
  payment.payment_id,
  (payment.amount + tab.service_charge) as total_amount,
  payment.created
FROM payment
  INNER JOIN tab
    ON payment.tab_id = tab.tab_id;

实际结果： 正如您在下面看到的，从选项卡2收取的服务费包括两次付款和两次付款

预期结果： 总金额不应包括付款中的服务费，如下图所示

+------------+-----------------+---------------------+
| payment_id | total_amount    | created             |
+------------+-----------------+---------------------+
| 1          | 6               | 2017-05-22 12:14:27 |
| 2          | 13              | 2017-05-22 12:15:21 |
| 3          | 1               | 2017-05-22 13:11:14 |
+------------+-----------------+---------------------+

---

您应该确定与tab_id匹配的第一笔付款，然后根据该信息决定是否使用服务费：

SELECT
  payment.payment_id,
  payment.amount + if (payment.created=m.mintime, tab.service_charge, 0) as total_amount,
  payment.created
FROM payment
  INNER JOIN tab
    ON payment.tab_id = tab.tab_id
  JOIN (
    SELECT tab_id, min(created) as 'mintime'
    FROM payment
    GROUP BY tab_id
  ) AS m on m.tab_id = payment.tab_id;

---

请看：@草莓我看了一下，但我仍然不清楚这个话题如何帮助我解决我遇到的问题。请阅读已被提升106次的公认答案。按照简单的说明操作。看看会发生什么，谢谢。朴素典雅。

SELECT
  payment.payment_id,
  payment.amount + if (payment.created=m.mintime, tab.service_charge, 0) as total_amount,
  payment.created
FROM payment
  INNER JOIN tab
    ON payment.tab_id = tab.tab_id
  JOIN (
    SELECT tab_id, min(created) as 'mintime'
    FROM payment
    GROUP BY tab_id
  ) AS m on m.tab_id = payment.tab_id;