Question

我的mysql数据库中有一个名为“events”的表：

+-----+-----------+------------------------------+------------+
| ID  | CATEGORY  | NAME                         | TYPE       |
+-----+-----------+------------------------------+------------+
| 1   | 1         | Concert                      | music      |
| 2   | 2         | Basketball match             | indoors    |
| 3   | 1         | Theather play                | outdoors   |
| 4   | 1         | Concert                      | outdoors   |
+-----+-----------+------------------------------+------------+

我需要一个查询来计算类别1的事件，哪种类型是音乐，还有室外意味着从上面的表中计数应该只有1：有三个事件与类别1 但只有“音乐会”有户外和音乐类型（ID 1和ID 4）。

那个查询是什么？可以这样做吗？

Answer 1

试试这个：

SELECT count(DISTINCT e1.name)
FROM `events` AS e1
JOIN `events` AS e2 ON e1.name = e2.name
WHERE e1.category = 1 
    AND e2.category = 1 
    AND e1.type = 'music' 
    AND e2.type = 'outdoor'

或者更难理解的方式，但比前一种方式更快：

SELECT count(*) FROM (
    SELECT `name`
    FROM `events`
    WHERE `category` = 1
    GROUP BY `name`
    HAVING SUM( `type` = 'music') * SUM( `type` = 'outdoor' ) >= 1
) AS notNeeded

Answer 2

对于2个标准，我会使用Alin的答案。您可以使用更多数字的方法。

SELECT COUNT(*)
FROM   (SELECT `name`
        FROM   `events`
        WHERE  `category` = 1
               AND `type` IN ( 'outdoors', 'music' )
        GROUP  BY `name`
        HAVING COUNT(DISTINCT `type`) = 2) t

Answer 3

尝试此查询

Select count(*), group_concat(TYPE SEPARATOR ',') as types 
from events where category = 1 
HAVING LOCATE('music', types) and  LOCATE('outdoors', types)

Answer 4

尝试：

SELECT * FROM `events` AS e1
LEFT JOIN `events` AS e2 USING (`name`)
WHERE e1.`category` = 1 AND e2.`category` = 1 AND e1.`type` = 'music' AND e2.`type` = 'outdoors'

Answer 5

SELECT COUNT(*)
    FROM table
    WHERE category=1
    AND type='music' AND type IN (SELECT type
                                  FROM table
                                  WHERE type = 'outdoor')

一行保持重置我的连接。用？我会尝试发表评论

Answer 6

从Category ='1'并输入（'music'，'outdoor'）

的事件中选择计数（不同ID）作为'eventcount'

一个mysql查询：在同一列中匹配的几个条件

6 个答案: