这是我有两个表标签和客户的情况，结构如下
Tags Table
ID Name
1 Tag1
2 Tag2
Customers Table
ID Tag_ID Name
1 1 C1
2 2 C2
3 1 C3
我想要一条 SQL 语句来获取每个标签的前 10 个客户（按字母顺序）？是否可以在一个查询中完成。
P.S表格中的数据是样本数据，不是实际数据
【问题讨论】：
标签： mysql
考虑以下几点：
DROP TABLE IF EXISTS tags;
CREATE TABLE tags
(tag_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,name VARCHAR(12) NOT NULL
);
INSERT INTO tags VALUES
(1,'One'),
(2,'Two'),
(3,'Three'),
(4,'Four'),
(5,'Five'),
(6,'Six');
DROP TABLE IF EXISTS customers;
CREATE TABLE customers
(customer_id INT NOT NULL
,customer VARCHAR(12)
);
INSERT INTO customers VALUES
(1,'Dave'),
(2,'Ben'),
(3,'Charlie'),
(4,'Michael'),
(5,'Steve'),
(6,'Clive'),
(7,'Alice'),
(8,'Ken'),
(9,'Petra');
DROP TABLE IF EXISTS customer_tag;
CREATE TABLE customer_tag
(customer_id INT NOT NULL
,tag_ID INT NOT NULL
,PRIMARY KEY(customer_id,tag_id)
);
INSERT INTO customer_tag VALUES
(1,1),
(1,2),
(1,4),
(2,3),
(2,2),
(3,1),
(4,4),
(4,2),
(5,2),
(5,5),
(5,6),
(6,6);
以下查询返回与每个标签关联的所有客户，以及按字母顺序排序时他们各自的“排名”...
SELECT t.*, c1.*, COUNT(ct2.tag_id) rank
FROM tags t
JOIN customer_tag ct1
ON ct1.tag_id = t.tag_id
JOIN customers c1
ON c1.customer_id = ct1.customer_id
JOIN customer_tag ct2
ON ct2.tag_id = ct1.tag_id
JOIN customers c2
ON c2.customer_id = ct2.customer_id
AND c2.customer <= c1.customer
GROUP
BY t.tag_id, c1.customer_id
ORDER
BY t.tag_id,rank;
+--------+-------+-------------+----------+------+
| tag_id | name | customer_id | customer | rank |
+--------+-------+-------------+----------+------+
| 1 | One | 3 | Charlie | 1 |
| 1 | One | 1 | Dave | 2 |
| 2 | Two | 2 | Ben | 1 |
| 2 | Two | 1 | Dave | 2 |
| 2 | Two | 4 | Michael | 3 |
| 2 | Two | 5 | Steve | 4 |
| 3 | Three | 2 | Ben | 1 |
| 4 | Four | 1 | Dave | 1 |
| 4 | Four | 4 | Michael | 2 |
| 5 | Five | 5 | Steve | 1 |
| 6 | Six | 6 | Clive | 1 |
| 6 | Six | 5 | Steve | 2 |
+--------+-------+-------------+----------+------+
如果我们只想要前 2 个，比如说，对于每个标签，我们可以重写如下...
SELECT t.*
, c1.*
FROM tags t
JOIN customer_tag ct1
ON ct1.tag_id = t.tag_id
JOIN customers c1
ON c1.customer_id = ct1.customer_id
JOIN customer_tag ct2
ON ct2.tag_id = ct1.tag_id
JOIN customers c2
ON c2.customer_id = ct2.customer_id
AND c2.customer <= c1.customer
GROUP
BY t.tag_id, c1.customer_id
HAVING COUNT(ct2.tag_id) <=2
ORDER
BY t.tag_id, c1.customer;
+--------+-------+-------------+----------+
| tag_id | name | customer_id | customer |
+--------+-------+-------------+----------+
| 1 | One | 3 | Charlie |
| 1 | One | 1 | Dave |
| 2 | Two | 2 | Ben |
| 2 | Two | 1 | Dave |
| 3 | Three | 2 | Ben |
| 4 | Four | 1 | Dave |
| 4 | Four | 4 | Michael |
| 5 | Five | 5 | Steve |
| 6 | Six | 6 | Clive |
| 6 | Six | 5 | Steve |
+--------+-------+-------------+----------+
这很好，但是在性能存在问题的情况下，类似以下的解决方案会更快 - 尽管您可能需要在构建表之前运行 SET NAMES utf8; （我必须这样做）才能使其正常工作：
SELECT tag_id, name, customer_id,customer
FROM
(
SELECT t.*
, c.*
, CASE WHEN @prev=t.tag_id THEN @i:=@i+1 ELSE @i:=1 END rank
, @prev := t.tag_id
FROM tags t
JOIN customer_tag ct
ON ct.tag_id = t.tag_id
JOIN customers c
ON c.customer_id = ct.customer_id
JOIN ( SELECT @i:=1, @prev:=0) vars
ORDER
BY t.tag_id
, c.customer
) x
WHERE rank <=2
ORDER
BY tag_id,customer;
+--------+-------+-------------+----------+
| tag_id | name | customer_id | customer |
+--------+-------+-------------+----------+
| 1 | One | 3 | Charlie |
| 1 | One | 1 | Dave |
| 2 | Two | 2 | Ben |
| 2 | Two | 1 | Dave |
| 3 | Three | 2 | Ben |
| 4 | Four | 1 | Dave |
| 4 | Four | 4 | Michael |
| 5 | Five | 5 | Steve |
| 6 | Six | 6 | Clive |
| 6 | Six | 5 | Steve |
+--------+-------+-------------+----------+
【讨论】：
为此，我们必须使用两个会话变量，一个用于行号，另一个用于存储旧客户 ID 以将其与当前 ID 进行比较，如下查询：
select c.name, @row_number:=CASE
WHEN @cid = c.id THEN @row_number + 1
ELSE 1
END AS rows,
@id:=c.id as CustomerId from tags t, customers c where t.id=c.id group by c.name where Rows<=10
我们在查询中使用了 CASE 语句。如果客户编号不变，我们增加row_number变量
【讨论】：
你的问题让我想起了this one（尤其是投票最多的答案），所以我想出了这个问题：
SELECT Tags.ID,
Tags.Name,
SUBSTRING_INDEX(GROUP_CONCAT(Customers.Name
ORDER BY Customers.Name),
',', 10) AS Customers
FROM Customers
INNER JOIN Tags
ON Tags.ID = Customers.Tag_ID
GROUP BY Tags.ID
ORDER BY Tags.Id;
It works，但这显然是一种 hacky 方式，因为 MySQL 还没有提供更自然的工具。
【讨论】：