1391

使用MySQL,我可以执行以下操作:

SELECT hobbies FROM peoples_hobbies WHERE person_id = 5;

我的输出:

shopping
fishing
coding

但我只想要 1 行,1 列:

预期输出:

shopping, fishing, coding

原因是我从多个表中选择多个值,并且在所有连接之后,我得到的行数比我想要的多得多。

我在MySQL Doc上寻找了一个函数,它看起来不像CONCATorCONCAT_WS函数接受结果集。

那么这里有人知道怎么做吗?

4

15 回答 15

1975

您可以使用GROUP_CONCAT

SELECT person_id,
   GROUP_CONCAT(hobbies SEPARATOR ', ')
FROM peoples_hobbies
GROUP BY person_id;

正如路德维希在他的评论中所说,您可以添加DISTINCT运算符以避免重复:

SELECT person_id,
   GROUP_CONCAT(DISTINCT hobbies SEPARATOR ', ')
FROM peoples_hobbies
GROUP BY person_id;

正如 Jan 在他们的评论中所说,您还可以在使用以下方法对值进行排序之前对其进行排序ORDER BY

SELECT person_id, 
       GROUP_CONCAT(hobbies ORDER BY hobbies ASC SEPARATOR ', ')
FROM peoples_hobbies
GROUP BY person_id;

正如 Dag 在他的评论中所说,结果有 1024 字节的限制。要解决此问题,请在查询之前运行此查询:

SET group_concat_max_len = 2048;

当然,您可以2048根据自己的需要进行更改。要计算和分配值:

SET group_concat_max_len = CAST(
                     (SELECT SUM(LENGTH(hobbies)) + COUNT(*) * LENGTH(', ')
                           FROM peoples_hobbies
                           GROUP BY person_id) AS UNSIGNED);
于 2008-11-10T02:48:11.677 回答
124

看看GROUP_CONCAT你的 MySQL 版本 (4.1) 是否支持它。有关更多详细信息,请参阅文档

它看起来像:

  SELECT GROUP_CONCAT(hobbies SEPARATOR ', ') 
  FROM peoples_hobbies 
  WHERE person_id = 5 
  GROUP BY 'all';
于 2008-11-10T02:48:54.560 回答
84

连接多个单独行的替代语法

警告:这篇文章会让你感到饥饿。

鉴于:

我发现自己想要选择多个单独的行——而不是一个组——并在某个字段上连接。

假设您有一张包含产品 ID 及其名称和价格的表格:

+------------+--------------------+-------+
| 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 ajax 将这些小狗列为复选框。

您的饥饿河马用户选择13, 15, 16. 今天没有给她甜点...

寻找:

一种使用纯 mysql 在一行中总结用户订单的方法。

解决方案:

GROUP_CONCAT子句一起使用IN

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

哪个输出:

+------------------------------------------------+
| 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 |
+------------------------------------------------+-------+
于 2014-09-23T20:26:46.890 回答
41

GROUP_CONCAT您可以通过设置group_concat_max_len参数来更改值的最大长度。

请参阅MySQL 文档中的详细信息。

于 2010-04-08T11:32:55.043 回答
31

有一个 GROUP 聚合函数GROUP_CONCAT

于 2008-11-10T02:48:38.227 回答
28

就我而言,我有一排 Id,必须将其转换为 char,否则,结果将被编码为二进制格式:

SELECT CAST(GROUP_CONCAT(field SEPARATOR ',') AS CHAR) FROM table
于 2014-06-10T09:24:31.350 回答
18

使用 MySQL(5.6.13) 会话变量和赋值运算符,如下所示

SELECT @logmsg := CONCAT_ws(',',@logmsg,items) FROM temp_SplitFields a;

那么你可以得到

test1,test11
于 2013-09-27T18:47:25.257 回答
16

我有一个更复杂的查询,发现我必须GROUP_CONCAT在外部查询中使用才能让它工作:

原始查询:

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

内爆:

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

希望这可能对某人有所帮助。

于 2015-04-29T16:35:34.430 回答
12

对于在这里查看如何使用GROUP_CONCAT子查询的人-发布此示例

SELECT i.*,
(SELECT GROUP_CONCAT(userid) FROM favourites f WHERE f.itemid = i.id) AS idlist
FROM items i
WHERE i.id = $someid

所以GROUP_CONCAT必须在子查询中使用,而不是包装它。

于 2019-04-26T09:26:01.907 回答
10

试试这个:

DECLARE @Hobbies NVARCHAR(200) = ' '

SELECT @Hobbies = @Hobbies + hobbies + ',' FROM peoples_hobbies WHERE person_id = 5;

TL;博士;

set @sql='';
set @result='';
set @separator=' union \r\n';
SELECT 
@sql:=concat('select ''',INFORMATION_SCHEMA.COLUMNS.COLUMN_NAME  ,''' as col_name,',
INFORMATION_SCHEMA.COLUMNS.CHARACTER_MAXIMUM_LENGTH ,' as def_len ,' ,
'MAX(CHAR_LENGTH(',INFORMATION_SCHEMA.COLUMNS.COLUMN_NAME , '))as  max_char_len',
' FROM ',
INFORMATION_SCHEMA.COLUMNS.TABLE_NAME
) as sql_piece, if(@result:=if(@result='',@sql,concat(@result,@separator,@sql)),'','') as dummy
FROM INFORMATION_SCHEMA.COLUMNS 
WHERE 
INFORMATION_SCHEMA.COLUMNS.DATA_TYPE like '%char%'
and INFORMATION_SCHEMA.COLUMNS.TABLE_SCHEMA='xxx' 
and INFORMATION_SCHEMA.COLUMNS.TABLE_NAME='yyy';
select @result;
于 2015-02-26T15:25:38.097 回答
2

我们有两种方法来连接 MySql 中的列

select concat(hobbies) as `Hobbies` from people_hobbies where 1

或者

select group_concat(hobbies) as `Hobbies` from people_hobbies where 1
于 2018-11-13T13:37:54.953 回答
0

已经很晚了,但对于那些正在搜索“使用数据透视表将多个 MySQL 行连接到一个字段”的人会有所帮助:)

询问:

SELECT pm.id, pm.name, GROUP_CONCAT(c.name) as channel_names
FROM payment_methods pm
LEFT JOIN payment_methods_channels_pivot pmcp ON pmcp.payment_method_id = pm.id
LEFT JOIN channels c ON c.id = pmcp.channel_id
GROUP BY pm.id

payment_methods 
  id  | name
  1   | PayPal

channels
  id  | name
  1   | Google
  2   | Faceook

payment_methods_channels_pivot
   payment_method_id | channel_id
   1                 |  1
   1                 |  2

输出:

在此处输入图像描述

于 2021-07-13T06:21:25.790 回答
0

在 sql server 中使用 string_agg 将行字段值转换为列:

select string_agg(field1, ', ') a FROM mytable 

or

select string_agg(field1, ', ') within group (order by field1 dsc) a FROM mytable group by field2
于 2021-08-13T17:48:18.787 回答
0

在这里,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

这里

   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排在第三位。所以我使用最后一行

  Order by Id DESC 
  LIMIT 1
  

然后我还要从我的字符串中删除第一个','

 TRIM(LEADING ', ' FROM T.hobbies )
于 2021-11-08T19:04:38.767 回答
-1

在这种情况下另一个有趣的例子 -

以下是示例表的结构people_hobbies-

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    |                |
+---------+--------------+------+-----+---------+----------------+

该表填充如下 -

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

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;

CONCAT_GROUP()

于 2021-10-04T09:29:34.020 回答