I have the following problem in MySQL 5.5. Here's my table.
Suppose i have a table With 'Names' with columns Rank,NAME and data in table
When i run the query it will give result as
select *from name
Rank | NAME
-------------
1 | A
1 | B
1 | C
2 | D
2 | E
2 | F
3 | G
3 | H
3 | I 现在,这当然是一种非常不方便的组织数据的方式,但这就是数据发生的方式(并将继续进入)。
我需要能够扔在它的名字列表对应于他们各自的排名如下所示
Rank | Name | Name | Name
-----------------------------------
1 | A | B | C
2 | D | E | F
3 | G | H | I我有类似这样的查询
select
case when rank=1 then name else null end as 1,
case when rank=2 then name else null end as 2,
case when rank=3 then name else null end as 3
from name 具有相同等级的名字需要被带出并显示在同一行中。我不能估计学生将获得的最后一个等级,因此我不能使用' in‘operator.Since手动通过等级值。等级值不可预测,我需要根据它们的等级动态地将它们放在交叉表视图中。
我尝试过所有类型的动态交叉表生成查询(是的,我都看过了),但没有任何成功。请帮帮我。谢谢!
发布于 2014-08-12 20:37:59
DROP TABLE IF EXISTS my_table;
CREATE TABLE my_table
(rank INT NOT NULL
,name CHAR(1) NOT NULL
,PRIMARY KEY(rank,name)
);
INSERT INTO my_table VALUES
(1 ,'A'),
(1 ,'B'),
(1 ,'C'),
(2 ,'D'),
(2 ,'E'),
(2 ,'F'),
(3 ,'G'),
(3 ,'H'),
(3 ,'I');
SELECT * FROM my_table;
+------+------+
| rank | name |
+------+------+
| 1 | A |
| 1 | B |
| 1 | C |
| 2 | D |
| 2 | E |
| 2 | F |
| 3 | G |
| 3 | H |
| 3 | I |
+------+------+
SELECT x.*,COUNT(*) subrank FROM my_table x JOIN my_table y ON y.rank = x.rank AND y.name <= x.name GROUP BY x.rank,x.name;
+------+------+---------+
| rank | name | subrank |
+------+------+---------+
| 1 | A | 1 |
| 1 | B | 2 |
| 1 | C | 3 |
| 2 | D | 1 |
| 2 | E | 2 |
| 2 | F | 3 |
| 3 | G | 1 |
| 3 | H | 2 |
| 3 | I | 3 |
+------+------+---------+
SELECT rank
, MAX(CASE WHEN subrank = 1 THEN name END) name1
, MAX(CASE WHEN subrank = 2 THEN name END) name2
, MAX(CASE WHEN subrank = 3 THEN name END) name3
FROM
( SELECT x.*
, COUNT(*) subrank
FROM my_table x
JOIN my_table y
ON y.rank = x.rank
AND y.name <= x.name
GROUP
BY x.rank
, x.name
) a
GROUP
BY rank;
+------+-------+-------+-------+
| rank | name1 | name2 | name3 |
+------+-------+-------+-------+
| 1 | A | B | C |
| 2 | D | E | F |
| 3 | G | H | I |
+------+-------+-------+-------+https://stackoverflow.com/questions/25264290
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