Question

我在生命中第一次尝试在我的数据库表中使用PIVOT。这是表格：

Answers (id_user, id_question, topic, value) VALUES
(1, 1, 'a', 3),
(1, 1, 'b', 3),
(1, 1, 'c', 2),

(1, 2, 'a', 1),
(1, 2, 'b', 2),
(1, 2, 'c', 3),

[...]

(1, 14, 'a', 1),
(1, 14, 'b', 2),
(1, 14, 'c', 1);

但我正试图解决这个问题：

Answers (id_user, id_question, topic_A, topic_B, topic_C) VALUES
(1, 1, 3, 3, 2),
(1, 2, 1, 2, 3),
[...]
(1, 14, 1, 2, 1);

使用此查询，我在PIVOT附近获得了一般语法错误

SELECT *
FROM
(     SELECT id_user, id_question, topic, value
      FROM Answers
      WHERE id_user=98
) as risp
PIVOT 
(     FOR id_question IN ([a],[b],[c])
)     AS pvt

Answer 1

在MySQL中使用聚合的一种方式：

select id_user, id_question,
       max(case when topic = 'A' then value end) as topic_A,
       max(case when topic = 'B' then value end) as topic_B,
       max(case when topic = 'C' then value end) as topic_C
from Answers a
group by id_user, id_question;

顺便说一句，把它放回同一张桌子是不合理的。查询应该足够了。您也可以将其放入视图或另一个表中。

编辑：

另一种方法是使用连接。查询有点复杂：

select uq.id_user, uq.id_question, aa.value as topic_A,
       ab.value as topic_B, ac.value as topic_C
from (select distinct id_user, id_question from answers) uq left outer join
     Answers aa
     on aa.id_user = uq.id_user and aa.id_question = uq.id_question and
        aa.topic = 'A' left outer join
     Answers ab
     on ab.id_user = uq.id_user and ab.id_question = uq.id_question and
        ab.topic = 'B' left outer join
     Answers aa
     on ac.id_user = uq.id_user and ac.id_question = uq.id_question and
        ac.topic = 'C';

数据透视查询不起作用

1 个答案: