Question

考虑以下查询：

SELECT DISTINCT ON (ser.id) *
FROM server ser
LEFT JOIN subscription sub ON ser.id = sub.server_id
WHERE (
    COUNT(SELECT err.id FROM error err WHERE ser.id = err.id) > 0
    OR SUM(SELECT pay.amount FROM payment pay WHERE ser.id = pay.id) > 0
);

此处，将返回正在订阅并且有错误或付款的唯一服务器列表。

但是，我想要返回服务器ID，错误数量和付款总额，而不是返回所有服务器列（*）。例如，初始选择应如下所示：

SELECT DISTINCT ON (ser.id) ser.id, countErrors, sumPayments

选择ser.id是直截了当的，但如何从聚合函数中选择countErrors和sumPayments＆＃34; count＆＃34;和＆＃34;总和＆＃34; （考虑到它们是WHERE子句中的条件）？

我想象着＆＃34;其中＆＃34;条件看起来像这样：

    COUNT(SELECT err.id FROM error err WHERE ser.id = err.id) AS countErrors  > 0
    OR SUM(SELECT pay.amount FROM payment pay WHERE ser.id = pay.id) AS sumPayments > 0

有可能这样做吗？如果是这样，怎么能实现呢？

测试数据如下所示：

server
+----+
| id |
+----+
| 1  |
+----+
| 2  |
+----+
| 3  |
+----+
| 4  |
+----+

subscription
+----+-----------+
| id | server_id |
+----+-----------+
| 1  | 1         |
+----+-----------+
| 2  | 2         |
+----+-----------+
| 3  | 2         |
+----+-----------+
| 4  | 3         |
+----+-----------+
| 5  | 3         |
+----+-----------+

error
+----+-----------+
| id | server_id |
+----+-----------+
| 1  | 1         |
+----+-----------+
| 3  | 4         |
+----+-----------+

payment
+----+-----------+--------+
| id | server_id | amount |
+----+-----------+--------+
| 1  | 1         | 200    |
+----+-----------+--------+
| 2  | 2         | 200    |
+----+-----------+--------+
| 3  | 2         | 100    |
+----+-----------+--------+

测试数据的结果：

+-----------+-------------+-------------+
| server_id | countErrors | sumPayments |
+-----------+-------------+-------------+
| 1         | 1           | 200         |
+-----------+-------------+-------------+
| 2         | 0           | 300         |
+-----------+-------------+-------------+

服务器＃4没有订阅，因此应该省略。
服务器＃3有订阅，但没有错误或付款，因此应该省略。
服务器＃1和服务器＃2都有订阅和付款和/或错误。

Answer 1

除非我遗漏了什么，否则我会按如下方式编写您的查询。在单独的真实子查询中执行错误和付款的聚合，并加入它们。此外，还有subscription表的连接，但这仅用于过滤掉没有订阅的服务器。最后，WHERE子句删除任何没有错误或付款的服务器。

SELECT
    s.id AS server_id,
    COALESCE(e.countErrors, 0) AS countErrors,
    COALESCE(p.sumPayments, 0) AS sumPayments
FROM server s
INNER JOIN
(
    SELECT DISTINCT server_id
    FROM subscription
) su
    ON s.id = su.server_id
LEFT JOIN
(
    SELECT server_id, COUNT(*) AS countErrors
    FROM error
    GROUP BY server_id
) e
    ON s.id = e.server_id
LEFT JOIN
(
    SELECT server_id, SUM(amount) AS sumPayments
    FROM payment
    GROUP BY server_id
) p
    ON s.id = p.server_id
WHERE
    p.sumPayments > 0 OR
    e.countErrors > 0
ORDER BY
    s.id;

Demo

Answer 2

这里的错误是将COUNT放在SELECT之外，它需要进入内部：

(SELECT COUNT(err.id) FROM error err WHERE ser.id = err.id) > 0
    OR (SELECT SUM(pay.amount) FROM payment pay WHERE ser.id = pay.id) > 0

是否可以选择子查询

2 个答案:

Demo