我有一个更复杂的查询，发现我必须在外部查询中使用GROUP_NCAT才能使其工作：

原始查询：

SELECT DISTINCT userID 
FROM event GROUP BY userID 
HAVING count(distinct(cohort))=2);

隐含的：

SELECT GROUP_CONCAT(sub.userID SEPARATOR ', ') 
FROM (SELECT DISTINCT userID FROM event 
GROUP BY userID HAVING count(distinct(cohort))=2) as sub;

希望这能帮助到某人。

本例中另一个有趣的例子-

下面是people_hobbies示例表的结构-

DESCRIBE people_hobbies;
+---------+--------------+------+-----+---------+----------------+
| Field   | Type         | Null | Key | Default | Extra          |
+---------+--------------+------+-----+---------+----------------+
| id      | int unsigned | NO   | PRI | NULL    | auto_increment |
| ppl_id  | int unsigned | YES  | MUL | NULL    |                |
| name    | varchar(200) | YES  |     | NULL    |                |
| hby_id  | int unsigned | YES  | MUL | NULL    |                |
| hobbies | varchar(50)  | YES  |     | NULL    |                |
+---------+--------------+------+-----+---------+----------------+

该表的填充方式如下-

SELECT * FROM people_hobbies;
+----+--------+-----------------+--------+-----------+
| id | ppl_id | name            | hby_id | hobbies   |
+----+--------+-----------------+--------+-----------+
|  1 |      1 | Shriya Jain     |      1 | reading   |
|  2 |      4 | Shirley Setia   |      4 | coding    |
|  3 |      2 | Varsha Tripathi |      7 | gardening |
|  4 |      3 | Diya Ghosh      |      2 | fishing   |
|  5 |      4 | Shirley Setia   |      3 | gaming    |
|  6 |      1 | Shriya Jain     |      6 | cycling   |
|  7 |      2 | Varsha Tripathi |      1 | reading   |
|  8 |      3 | Diya Ghosh      |      5 | shopping  |
|  9 |      3 | Diya Ghosh      |      4 | coding    |
| 10 |      4 | Shirley Setia   |      1 | reading   |
| 11 |      1 | Shriya Jain     |      4 | coding    |
| 12 |      1 | Shriya Jain     |      3 | gaming    |
| 13 |      4 | Shirley Setia   |      2 | fishing   |
| 14 |      4 | Shirley Setia   |      7 | gardening |
| 15 |      2 | Varsha Tripathi |      3 | gaming    |
| 16 |      2 | Varsha Tripathi |      2 | fishing   |
| 17 |      1 | Shriya Jain     |      5 | shopping  |
| 18 |      1 | Shriya Jain     |      7 | gardening |
| 19 |      3 | Diya Ghosh      |      1 | reading   |
| 20 |      4 | Shirley Setia   |      5 | shopping  |
+----+--------+-----------------+--------+-----------+

现在，生成了一个表hobby_list，其中包含所有人的列表和每个人的爱好列表，每个爱好在一行中-

CREATE TABLE hobby_list AS
    -> SELECT ppl_id, name,
    -> GROUP_CONCAT(hobbies ORDER BY hby_id SEPARATOR "\n")
    -> AS hobbies
    -> FROM people_hobbies
    -> GROUP BY ppl_id
    -> ORDER BY ppl_id;

SELECT * FROM hobby_list;

这里，我的意图是在不使用group_concat（）函数的情况下应用字符串连接：

Set @concatHobbies = '';
SELECT TRIM(LEADING ', ' FROM T.hobbies ) FROM 
(
   select 
   Id, @concatHobbies := concat_ws(', ',@concatHobbies,hobbies) as hobbies
   from peoples_hobbies
)T
Order by Id DESC
LIMIT 1

Here

   select 
   Id, @concatHobbies := concat_ws(', ',@concatHobbies,hobbies) as hobbies
   from peoples_hobbies

将返回

  Id    hobbies
  1     , shopping
  2     , shopping, fishing
  3     , shopping, fishing, coding

现在我们的预期结果是第三。所以我用

  Order by Id DESC 
  LIMIT 1
  

然后我也将第一个“，”从字符串中删除

 TRIM(LEADING ', ' FROM T.hobbies )

有一个GROUP聚合函数GROUP_CONCAT。

连接多个单独行的替代语法

警告：这篇文章会让你饿。

鉴于：

我发现自己想要选择多个单独的行，而不是一个组，并在某个字段上进行连接。

假设您有一个产品ID及其名称和价格表：

+------------+--------------------+-------+
| product_id | name               | price |
+------------+--------------------+-------+
|         13 | Double Double      |     5 |
|         14 | Neapolitan Shake   |     2 |
|         15 | Animal Style Fries |     3 |
|         16 | Root Beer          |     2 |
|         17 | Lame T-Shirt       |    15 |
+------------+--------------------+-------+

然后，你有一些花哨的聊天ajax，将这些小狗列为复选框。

饥饿的河马用户选择13、15、16。她今天没有甜点。。。

查找：

一种用纯mysql在一行中总结用户订单的方法。

解决方案：

将GROUP_NCAT和IN子句一起使用：

mysql> SELECT GROUP_CONCAT(name SEPARATOR ' + ') AS order_summary FROM product WHERE product_id IN (13, 15, 16);

哪些输出：

+------------------------------------------------+
| order_summary                                  |
+------------------------------------------------+
| Double Double + Animal Style Fries + Root Beer |
+------------------------------------------------+

奖金解决方案：

如果您也想要总价，请输入SUM（）：

mysql> SELECT GROUP_CONCAT(name SEPARATOR ' + ') AS order_summary, SUM(price) AS total FROM product WHERE product_id IN (13, 15, 16);
+------------------------------------------------+-------+
| order_summary                                  | total |
+------------------------------------------------+-------+
| Double Double + Animal Style Fries + Root Beer |    10 |
+------------------------------------------------+-------+

通过设置GROUP_CONCAT_max_len参数，可以更改GROUP_CONCAT值的最大长度。

请参阅MySQL文档中的详细信息。