Question

我正在撰写查询，以通过其产品ID查找订单的完全重复。

找到重复的条件是：

1）订单具有相同的产品数量。

2）所有产品ID都相同。

尝试过这样的事情，但它没有成功：

SELECT
  order.*,
  count(same_products.id),
  count(all_products.id)
FROM orders
  LEFT JOIN products AS all_products ON all_products.order_id = orders.id
  LEFT JOIN products AS same_products
    ON same_products.order_id = orders.id AND same_products.id IN (30868, 30862)
GROUP BY orders.id
HAVING count(same_products.id) = 4 AND count(all_products.id = 4)

Answer 1

如果你想要计算重复的行，你应该避免使用所有列（*）选择器，因为如果你的列中有增量值，这不会让你找到重复的行。

  SELECT
  order.id
  count(same_products.id),
  count(all_products.id)
FROM orders
  LEFT JOIN products AS all_products ON all_products.order_id = orders.id
  LEFT JOIN products AS same_products
    ON same_products.order_id = orders.id AND same_products.id IN (30868, 30862)
GROUP BY orders.id
HAVING count(same_products.id) >1 OR count(all_products.id )> 1

对于重复的行，您应该检查计数＆gt; 1（对于两个计数）并且要小心count（all_products.id = 4）如果你需要过滤这个值，你应该将它添加到相关表的on条件中，例如：

  SELECT
  order.id
  count(same_products.id),
  count(all_products.id)
FROM orders
  LEFT JOIN products AS all_products ON all_products.order_id = orders.id and all_products.id =4
  LEFT JOIN products AS same_products
    ON same_products.order_id = orders.id AND same_products.id IN (30868, 30862)
GROUP BY orders.id
HAVING count(same_products.id) >1

Answer 2

我不清楚你的副本是什么意思，我从你的描述中假设它是2个相同产品的订单。这似乎有点简单，例如给定

MariaDB [sandbox]> select * from orders;
+------+---------------------+-------------+
| id   | order_created       | customer_id |
+------+---------------------+-------------+
|    1 | 2016-01-01 00:00:00 |           1 |
|    2 | 2016-02-01 00:00:00 |           1 |
|    3 | 2016-03-01 00:00:00 |           1 |
|    4 | 2016-01-01 00:00:00 |           2 |
|    5 | 2016-02-01 00:00:00 |           2 |
|    6 | 2016-01-01 00:00:00 |           3 |
|   10 | 2016-12-01 00:00:00 |           4 |
+------+---------------------+-------------+
7 rows in set (0.00 sec)

MariaDB [sandbox]> select * from order_details;
+----+---------+-----------+------+
| id | orderid | productid | qty  |
+----+---------+-----------+------+
|  1 |       1 |      1213 |   10 |
|  2 |       1 |      9999 |   10 |
|  3 |       2 |      8888 |   10 |
|  4 |       3 |      1213 |   10 |
|  5 |       4 |      2222 |   10 |
|  6 |       5 |      9999 |   30 |
|  7 |       5 |      1213 |   30 |
|  8 |       6 |      9999 |   30 |
|  9 |       6 |      1213 |   30 |
+----+---------+-----------+------+
9 rows in set (0.00 sec)

select orders1.*,orders2.*,t.*
from
(select * from
(
select orderid o1orderid,group_concat(productid order by productid) o1grp,sum(qty) qty1
from order_details 
group by orderid
) o1
join
(select orderid o2orderid,group_concat(productid order by productid) o2grp, sum(qty) qty2
from order_details 
group by orderid
) o2
on o2grp = o1grp and qty2 = qty1 and o2orderid > o1orderid
) t
join orders orders1 on t.o1orderid = orders1.id
join orders orders2 on t.o2orderid = orders2.id

返回

+------+---------------------+-------------+------+---------------------+-------------+-----------+-----------+-------+-----------+-----------+-------+
| id   | order_created       | customer_id | id   | order_created       | customer_id | o1orderid | o1grp     | qty1  | o2orderid | o2grp     | qty2  |
+------+---------------------+-------------+------+---------------------+-------------+-----------+-----------+-------+-----------+-----------+-------+
|    5 | 2016-02-01 00:00:00 |           2 |    6 | 2016-01-01 00:00:00 |           3 |         5 | 1213,9999 | 30,30 |         6 | 1213,9999 | 30,30 |
+------+---------------------+-------------+------+---------------------+-------------+-----------+-----------+-------+-----------+-----------+-------+
1 row in set (0.03 sec)

但客户数量不同。

通过连接表字段找到完全相同的副本

2 个答案: