SQL。销售排名加入

Question

我有桌子：订单和产品。 如何在MySQL 5.6中生成排名？

如果产品价值相同，则排名也必须相同。 下面我需要按计数排名

SELECT
  count(productpk), productpk,
  @prev := @curr,
  @curr := count(productpk),
  @rank := IF(@prev = @curr, @rank, @rank+1) AS rank
FROM orders AS om
JOIN products AS p ON om.PK=p.p_order,
  (SELECT @curr := null, @prev := null, @rank := 0) sel1
GROUP BY productpk ORDER BY count(productpk);

有效结果是（count - ＆gt; rank）：

• 2 - ＆gt; 3
• 2 - ＆gt; 3
• 4 - ＆gt; 2
• 4 - ＆gt; 2
• 6 - ＆gt; 1

Answer 1

此查询应该可以解决问题。

SELECT
sq.productpk,
sq.cp,
@rank := IF(@prev = sq.cp, @rank, @rank + 1) AS rank,
@prev := sq.cp
FROM
(
    SELECT
    productpk,
    COUNT(productpk) AS cp
    FROM orders o
    JOIN products p ON o.PK = p.p_order
    GROUP BY productpk 
) sq
, (SELECT @prev := NULL, @rank := 0) var_init_subquery
ORDER BY sq.cp DESC

SELECT子句中的顺序很重要。当你第一次做这样的事情时

@prev := @curr,

然后是这样的

@rank := IF(@prev = @curr,...

没有意义，因为@prev总是等于@curr。顺便说一下，@curr在这种情况下也毫无意义。

您必须将@prev与IF()函数中的当前行进行比较。之后，将当前行分配给@prev。读取下一行时，@prev仍保留上一行的值。

最后，您必须将分组查询放在子查询中。人们不会认为这是必要的，因为查询在逻辑上像

一样处理

1. FROM
2. WHERE
3. GROUP BY
4. HAVING
5. ORDER BY
6. 选择

但MySQL并不是这样做的，至少在涉及用户定义的变量时不会这样做。将此简单测试视为证据：

root@localhost:playground > select a, @r:=@r+1 as r from bar, (select @r := 0) sq;
+------+------+
| a    | r    |
+------+------+
|    1 |    1 |
|    1 |    2 |
|    1 |    3 |
|    1 |    4 |
|    1 |    5 |
|    1 |    6 |
|    1 |    7 |
|    1 |    8 |
|    1 |    9 |
|    1 |   10 |
|    2 |   11 |
|    2 |   12 |
|    2 |   13 |
|    2 |   14 |
|    2 |   15 |
+------+------+
15 rows in set (0.00 sec)

root@localhost:playground > select a, @r:=@r+1 as r from bar, (select @r := 0) sq group by a;
+------+------+
| a    | r    |
+------+------+
|    1 |    1 |
|    2 |   11 |
+------+------+
2 rows in set (0.00 sec)

Answer 2

您可以使用具有相同结果集的内部联接

来执行此操作

 var query = from t in root.Descendants("Action") where (string)t.Attribute("Attribute") == "urn:oasis:names:tc:xacml:1.0:action:action-id" select t;