我有一张交易表:
Transactions
------------
id | account | type | date_time | amount
----------------------------------------------------
1 | 001 | 'R' | '2012-01-01 10:01:00' | 1000
2 | 003 | 'R' | '2012-01-02 12:53:10' | 1500
3 | 003 | 'A' | '2012-01-03 13:10:01' | -1500
4 | 002 | 'R' | '2012-01-03 17:56:00' | 2000
5 | 001 | 'R' | '2012-01-04 12:30:01' | 1000
6 | 002 | 'A' | '2012-01-04 13:23:01' | -2000
7 | 003 | 'R' | '2012-01-04 15:13:10' | 3000
8 | 003 | 'R' | '2012-01-05 12:12:00' | 1250
9 | 003 | 'A' | '2012-01-06 17:24:01' | -1250
并且我希望选择所有特定类型('R')的商品，但不要立即(按照date_time字段的顺序)针对同一帐户提交另一种其他类型('A')交易的商品...
因此，在给出前面的示例的情况下，查询应引发以下行:
id | account |type | date | amount
----------------------------------------------------
1 | 001 | 'R' | '2012-01-01 10:01:00' | 1000
5 | 001 | 'R' | '2012-01-04 12:30:01' | 1000
7 | 003 | 'R' | '2012-01-04 15:13:10' | 3000
(如您所见，第2行未显示，因为第3行“取消了” ...第4行也被“第6行“取消了”；第7行出现了(即使帐户003属于已取消的第2行) ，这一次在第7行中不会被任何“A”行取消；并且不会出现第8行(对于003帐户来说也是如此，因为现在该帐户被9取消了，也不会取消7，只是先前的一个一:8 ...
我已经尝试过Joins，Where子句中的子查询，但是我真的不确定如何进行查询...
我尝试过的是:
尝试加入:
SELECT trans.type as type,
trans.amount as amount,
trans.date_time as dt,
trans.account as acct,
FROM Transactions trans
INNER JOIN ( SELECT t.type AS type, t.acct AS acct, t.date_time AS date_time
FROM Transactions t
WHERE t.date_time > trans.date_time
ORDER BY t.date_time DESC
) AS nextTrans
ON nextTrans.acct = trans.acct
WHERE trans.type IN ('R')
AND nextTrans.type NOT IN ('A')
ORDER BY DATE(trans.date_time) ASC
这引发了错误，因为我无法在MySQL中向JOIN引入外部值。
在以下位置尝试子查询:
SELECT trans.type as type,
trans.amount as amount,
trans.date_time as dt,
trans.account as acct,
FROM Transactions trans
WHERE trans.type IN ('R')
AND trans.datetime <
( SELECT t.date_time AS date_time
FROM Transactions t
WHERE t.account = trans.account
ORDER BY t.date_time DESC
) AS nextTrans
ON nextTrans.acct = trans.acct
ORDER BY DATE(trans.date_time) ASC
这是错误的，我可以在MySQL的WHERE中引入外部值，但是我无法设法找到正确过滤所需方法的方法...
重要编辑:
我设法实现了一个解决方案，但现在需要进行认真的优化。这里是:
SELECT *
FROM (SELECT t1.*, tFlagged.id AS cancId, tFlagged.type AS cancFlag
FROM transactions t1
LEFT JOIN (SELECT t2.*
FROM transactions t2
ORDER BY t2.date_time ASC ) tFlagged
ON (t1.account=tFlagged.account
AND
t1.date_time < tFlagged.date_time)
WHERE t1.type = 'R'
GROUP BY t1.id) tCanc
WHERE tCanc.cancFlag IS NULL
OR tCanc.cancFlag <> 'A'
我自己加入了表格，只是考虑到相同的帐户和不错的date_time。 Join按date_time排序。按ID分组，我设法仅获得加入的第一个结果，该结果恰好是同一帐户的下一个交易。
然后在外部选择中，我将那些具有“A”的内容过滤掉，因为这意味着下一个交易实际上是对它的取消。换句话说，如果同一个帐户没有下一个交易，或者下一个交易是“R”，则该交易不会被取消，并且必须在结果中显示...
我懂了:
+----+---------+------+---------------------+--------+--------+----------+
| id | account | type | date_time | amount | cancId | cancFlag |
+----+---------+------+---------------------+--------+--------+----------+
| 1 | 001 | R | 2012-01-01 10:01:00 | 1000 | 5 | R |
| 5 | 001 | R | 2012-01-04 12:30:01 | 1000 | NULL | NULL |
| 7 | 003 | R | 2012-01-04 15:13:10 | 3000 | 8 | R |
+----+---------+------+---------------------+--------+--------+----------+
它将每个交易与下一个交易及时关联到同一帐户，然后筛选出已取消的交易...成功!
正如我所说，现在的问题是优化。我的真实数据有很多行(因为期望通过时间来保存事务的表具有)，并且对于现在约10,000行的表，此查询在1分44秒内得到了肯定的结果。我想连接就是问题……(对于那些在这里了解协议(protocol)的人，我应该怎么做？在这里发起一个新问题，并将其发布为该问题的解决方案？或者只是在这里等待更多答案？)
最佳答案
这是基于嵌套子查询的解决方案。首先，我添加了几行来处理更多案例。例如，事务10不应被事务12取消，因为事务11介于两者之间。
> select * from transactions order by date_time;
+----+---------+------+---------------------+--------+
| id | account | type | date_time | amount |
+----+---------+------+---------------------+--------+
| 1 | 1 | R | 2012-01-01 10:01:00 | 1000 |
| 2 | 3 | R | 2012-01-02 12:53:10 | 1500 |
| 3 | 3 | A | 2012-01-03 13:10:01 | -1500 |
| 4 | 2 | R | 2012-01-03 17:56:00 | 2000 |
| 5 | 1 | R | 2012-01-04 12:30:01 | 1000 |
| 6 | 2 | A | 2012-01-04 13:23:01 | -2000 |
| 7 | 3 | R | 2012-01-04 15:13:10 | 3000 |
| 8 | 3 | R | 2012-01-05 12:12:00 | 1250 |
| 9 | 3 | A | 2012-01-06 17:24:01 | -1250 |
| 10 | 3 | R | 2012-01-07 00:00:00 | 1250 |
| 11 | 3 | R | 2012-01-07 05:00:00 | 4000 |
| 12 | 3 | A | 2012-01-08 00:00:00 | -1250 |
| 14 | 2 | R | 2012-01-09 00:00:00 | 2000 |
| 13 | 3 | A | 2012-01-10 00:00:00 | -1500 |
| 15 | 2 | A | 2012-01-11 04:00:00 | -2000 |
| 16 | 2 | R | 2012-01-12 00:00:00 | 5000 |
+----+---------+------+---------------------+--------+
16 rows in set (0.00 sec)
首先，创建一个查询以针对每笔交易获取“同一帐户中最近一笔交易的日期”:
SELECT t2.*,
MAX(t1.date_time) AS prev_date
FROM transactions t1
JOIN transactions t2
ON (t1.account = t2.account
AND t2.date_time > t1.date_time)
GROUP BY t2.account,t2.date_time
ORDER BY t2.date_time;
+----+---------+------+---------------------+--------+---------------------+
| id | account | type | date_time | amount | prev_date |
+----+---------+------+---------------------+--------+---------------------+
| 3 | 3 | A | 2012-01-03 13:10:01 | -1500 | 2012-01-02 12:53:10 |
| 5 | 1 | R | 2012-01-04 12:30:01 | 1000 | 2012-01-01 10:01:00 |
| 6 | 2 | A | 2012-01-04 13:23:01 | -2000 | 2012-01-03 17:56:00 |
| 7 | 3 | R | 2012-01-04 15:13:10 | 3000 | 2012-01-03 13:10:01 |
| 8 | 3 | R | 2012-01-05 12:12:00 | 1250 | 2012-01-04 15:13:10 |
| 9 | 3 | A | 2012-01-06 17:24:01 | -1250 | 2012-01-05 12:12:00 |
| 10 | 3 | R | 2012-01-07 00:00:00 | 1250 | 2012-01-06 17:24:01 |
| 11 | 3 | R | 2012-01-07 05:00:00 | 4000 | 2012-01-07 00:00:00 |
| 12 | 3 | A | 2012-01-08 00:00:00 | -1250 | 2012-01-07 05:00:00 |
| 14 | 2 | R | 2012-01-09 00:00:00 | 2000 | 2012-01-04 13:23:01 |
| 13 | 3 | A | 2012-01-10 00:00:00 | -1500 | 2012-01-08 00:00:00 |
| 15 | 2 | A | 2012-01-11 04:00:00 | -2000 | 2012-01-09 00:00:00 |
| 16 | 2 | R | 2012-01-12 00:00:00 | 5000 | 2012-01-11 04:00:00 |
+----+---------+------+---------------------+--------+---------------------+
13 rows in set (0.00 sec)
使用它作为子查询，以使每个事务及其前身在同一行上。使用一些过滤条件提取出我们感兴趣的交易-即，其前身是被完全取消的“R”交易的“A”交易-
SELECT
t3.*,transactions.*
FROM
transactions
JOIN
(SELECT t2.*,
MAX(t1.date_time) AS prev_date
FROM transactions t1
JOIN transactions t2
ON (t1.account = t2.account
AND t2.date_time > t1.date_time)
GROUP BY t2.account,t2.date_time) t3
ON t3.account = transactions.account
AND t3.prev_date = transactions.date_time
AND t3.type='A'
AND transactions.type='R'
AND t3.amount + transactions.amount = 0
ORDER BY t3.date_time;
+----+---------+------+---------------------+--------+---------------------+----+---------+------+---------------------+--------+
| id | account | type | date_time | amount | prev_date | id | account | type | date_time | amount |
+----+---------+------+---------------------+--------+---------------------+----+---------+------+---------------------+--------+
| 3 | 3 | A | 2012-01-03 13:10:01 | -1500 | 2012-01-02 12:53:10 | 2 | 3 | R | 2012-01-02 12:53:10 | 1500 |
| 6 | 2 | A | 2012-01-04 13:23:01 | -2000 | 2012-01-03 17:56:00 | 4 | 2 | R | 2012-01-03 17:56:00 | 2000 |
| 9 | 3 | A | 2012-01-06 17:24:01 | -1250 | 2012-01-05 12:12:00 | 8 | 3 | R | 2012-01-05 12:12:00 | 1250 |
| 15 | 2 | A | 2012-01-11 04:00:00 | -2000 | 2012-01-09 00:00:00 | 14 | 2 | R | 2012-01-09 00:00:00 | 2000 |
+----+---------+------+---------------------+--------+---------------------+----+---------+------+---------------------+--------+
4 rows in set (0.00 sec)
从上面的结果可以明显看出，我们已经快到了-我们已经确定了不需要的交易。使用
LEFT JOIN我们可以将它们从整个交易集中过滤掉:SELECT
transactions.*
FROM
transactions
LEFT JOIN
(SELECT
transactions.id
FROM
transactions
JOIN
(SELECT t2.*,
MAX(t1.date_time) AS prev_date
FROM transactions t1
JOIN transactions t2
ON (t1.account = t2.account
AND t2.date_time > t1.date_time)
GROUP BY t2.account,t2.date_time) t3
ON t3.account = transactions.account
AND t3.prev_date = transactions.date_time
AND t3.type='A'
AND transactions.type='R'
AND t3.amount + transactions.amount = 0) t4
USING(id)
WHERE t4.id IS NULL
AND transactions.type = 'R'
ORDER BY transactions.date_time;
+----+---------+------+---------------------+--------+
| id | account | type | date_time | amount |
+----+---------+------+---------------------+--------+
| 1 | 1 | R | 2012-01-01 10:01:00 | 1000 |
| 5 | 1 | R | 2012-01-04 12:30:01 | 1000 |
| 7 | 3 | R | 2012-01-04 15:13:10 | 3000 |
| 10 | 3 | R | 2012-01-07 00:00:00 | 1250 |
| 11 | 3 | R | 2012-01-07 05:00:00 | 4000 |
| 16 | 2 | R | 2012-01-12 00:00:00 | 5000 |
+----+---------+------+---------------------+--------+
关于mysql - SQL:选择事务不在同一表内的条件的事务，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/9475177/