需要某种形式的;“条件分组”;在MySQL中

需要某种形式的;“条件分组”;在MySQL中,sql,mysql,group-by,Sql,Mysql,Group By,我有一个文章表: id | type | date ----------------------- 1 | A | 2010-01-01 2 | A | 2010-01-01 3 | B | 2010-01-01 字段类型可以是A、B或C 我需要运行一个报告,该报告将返回每天每种类型的文章数量,如下所示: date | count(type="A") | count(type="B") | count(type="C") ------------------

我有一个文章表:

id | type | date
-----------------------
 1 | A    | 2010-01-01
 2 | A    | 2010-01-01
 3 | B    | 2010-01-01
字段
类型
可以是A、B或C

我需要运行一个报告,该报告将返回每天每种类型的文章数量,如下所示:

date       | count(type="A") | count(type="B") | count(type="C")
-----------------------------------------------------
2010-01-01 | 2               | 1               | 0
2010-01-02 | 5               | 6               | 7
目前,我正在为每种类型运行3个查询,然后手动合并结果

select date, count(id) from article where type="A" group by date
是否可以在一个查询中执行此操作?(在纯sql中,没有存储过程或类似的东西)


谢谢

总和和大小写相结合就行了

select date
    , sum(case when type ='A' then 1 else 0 end) as count_type_a 
    , sum(case when type ='B' then 1 else 0 end) as count_type_b
    , sum(case when type ='C' then 1 else 0 end) as count_type_c 
from article group by date

总和和大小写的结合应该可以

select date
    , sum(case when type ='A' then 1 else 0 end) as count_type_a 
    , sum(case when type ='B' then 1 else 0 end) as count_type_b
    , sum(case when type ='C' then 1 else 0 end) as count_type_c 
from article group by date
编辑:使用了比本答案中的方法更好的方法。我之所以把它留在这里,是因为它也以另一种方式满足了这个问题:


您应该能够使用子查询,如下所示:

SELECT    DATE(a.date) as date,
          (SELECT COUNT(a1.id) FROM articles a1 WHERE a1.type = 'A' AND a1.date = a.date) count_a,
          (SELECT COUNT(a2.id) FROM articles a2 WHERE a2.type = 'B' AND a2.date = a.date) count_b,
          (SELECT COUNT(a3.id) FROM articles a3 WHERE a3.type = 'C' AND a3.date = a.date) count_c
FROM      articles a
GROUP BY  a.date;
测试用例:

CREATE TABLE articles (id int, type char(1), date datetime);

INSERT INTO articles VALUES (1, 'A', '2010-01-01');
INSERT INTO articles VALUES (2, 'A', '2010-01-01');
INSERT INTO articles VALUES (3, 'B', '2010-01-01');
INSERT INTO articles VALUES (4, 'B', '2010-01-02');
INSERT INTO articles VALUES (5, 'B', '2010-01-02');
INSERT INTO articles VALUES (6, 'B', '2010-01-03');
INSERT INTO articles VALUES (7, 'B', '2010-01-01');
INSERT INTO articles VALUES (8, 'C', '2010-01-05');
结果:

+------------+---------+---------+---------+
| date       | count_a | count_b | count_c |
+------------+---------+---------+---------+
| 2010-01-01 |       2 |       2 |       0 |
| 2010-01-02 |       0 |       2 |       0 |
| 2010-01-03 |       0 |       1 |       0 |
| 2010-01-05 |       0 |       0 |       1 |
+------------+---------+---------+---------+
4 rows in set (0.00 sec)
编辑:使用了比本答案中的方法更好的方法。我之所以把它留在这里,是因为它也以另一种方式满足了这个问题:


您应该能够使用子查询,如下所示:

SELECT    DATE(a.date) as date,
          (SELECT COUNT(a1.id) FROM articles a1 WHERE a1.type = 'A' AND a1.date = a.date) count_a,
          (SELECT COUNT(a2.id) FROM articles a2 WHERE a2.type = 'B' AND a2.date = a.date) count_b,
          (SELECT COUNT(a3.id) FROM articles a3 WHERE a3.type = 'C' AND a3.date = a.date) count_c
FROM      articles a
GROUP BY  a.date;
测试用例:

CREATE TABLE articles (id int, type char(1), date datetime);

INSERT INTO articles VALUES (1, 'A', '2010-01-01');
INSERT INTO articles VALUES (2, 'A', '2010-01-01');
INSERT INTO articles VALUES (3, 'B', '2010-01-01');
INSERT INTO articles VALUES (4, 'B', '2010-01-02');
INSERT INTO articles VALUES (5, 'B', '2010-01-02');
INSERT INTO articles VALUES (6, 'B', '2010-01-03');
INSERT INTO articles VALUES (7, 'B', '2010-01-01');
INSERT INTO articles VALUES (8, 'C', '2010-01-05');
结果:

+------------+---------+---------+---------+
| date       | count_a | count_b | count_c |
+------------+---------+---------+---------+
| 2010-01-01 |       2 |       2 |       0 |
| 2010-01-02 |       0 |       2 |       0 |
| 2010-01-03 |       0 |       1 |       0 |
| 2010-01-05 |       0 |       0 |       1 |
+------------+---------+---------+---------+
4 rows in set (0.00 sec)

子查询与日期不相关;它们报告整个表中每种类型的总数,而不仅仅是与
a.date
@Bill:Fixed匹配的行。但我看到亚历克斯的答案显然更好。你认为我应该把我的答案留在这里,还是这是一个非常负面的例子?我会把它留在这里。查看替代解决方案的外观通常很有用;它们报告整个表中每种类型的总数,而不仅仅是与
a.date
@Bill:Fixed匹配的行。但我看到亚历克斯的答案显然更好。你认为我应该把我的答案留在这里,还是这是一个非常负面的例子?我会把它留在这里。了解替代解决方案的外观通常很有用。