SQL将行转置为列

Question

我有一个有趣的难题,我认为可以用纯粹的SQL解决.我有类似于以下的表:

responses:

user_id | question_id | body
----------------------------
1       | 1           | Yes
2       | 1           | Yes
1       | 2           | Yes
2       | 2           | No
1       | 3           | No
2       | 3           | No


questions:

id | body
-------------------------
1 | Do you like apples?
2 | Do you like oranges?
3 | Do you like carrots?

我想得到以下输出

user_id | Do you like apples? | Do you like oranges? | Do you like carrots?
---------------------------------------------------------------------------
1       | Yes                 | Yes                  | No
2       | Yes                 | No                   | No

我不知道会有多少问题,而且它们是动态的,所以我不能只为每个问题编码.我正在使用PostgreSQL,我相信这称为转置,但我似乎无法找到任何说明在SQL中执行此操作的标准方法.我记得在大学期间在我的数据库课上做这个,但它是在MySQL中,老实说我不记得我们是怎么做到的.

我假设它将是连接和GROUP BY声明的组合,但我甚至无法弄清楚如何开始.

有人知道怎么做吗？非常感谢!

编辑1:我发现了一些关于使用交叉表的信息,这似乎是我想要的,但我无法理解它.更好的文章的链接将不胜感激!

Answer 1

使用:

  SELECT r.user_id,
         MAX(CASE WHEN r.question_id = 1 THEN r.body ELSE NULL END) AS "Do you like apples?",
         MAX(CASE WHEN r.question_id = 2 THEN r.body ELSE NULL END) AS "Do you like oranges?",
         MAX(CASE WHEN r.question_id = 3 THEN r.body ELSE NULL END) AS "Do you like carrots?"
    FROM RESPONSES r
    JOIN QUESTIONS q ON q.id = r.question_id
GROUP BY r.user_id

这是一个标准的数据透视查询,因为您将数据从行"旋转"到列数据.

Answer 2

我实现了一个真正的动态函数来处理这个问题,而无需硬编码任何特定类的答案或使用外部模块/扩展.它还可以完全控制列排序,并支持多个键和类/属性列.

你可以在这里找到它:https://github.com/jumpstarter-io/colpivot

解决此特定问题的示例:

begin;

create temporary table responses (
    user_id integer,
    question_id integer,
    body text
) on commit drop;

create temporary table questions (
    id integer,
    body text
) on commit drop;

insert into responses values (1,1,'Yes'), (2,1,'Yes'), (1,2,'Yes'), (2,2,'No'), (1,3,'No'), (2,3,'No');
insert into questions values (1, 'Do you like apples?'), (2, 'Do you like oranges?'), (3, 'Do you like carrots?');

select colpivot('_output', $$
    select r.user_id, q.body q, r.body a from responses r
        join questions q on q.id = r.question_id
$$, array['user_id'], array['q'], '#.a', null);

select * from _output;

rollback;

这输出:

 user_id | 'Do you like apples?' | 'Do you like carrots?' | 'Do you like oranges?' 
---------+-----------------------+------------------------+------------------------
       1 | Yes                   | No                     | Yes
       2 | Yes                   | No                     | No

Answer 3

您可以通过这种方式使用交叉表功能解决此示例

drop table if exists responses;
create table responses (
user_id integer,
question_id integer,
body text
);

drop table if exists questions;
create table questions (
id integer,
body text
);

insert into responses values (1,1,'Yes'), (2,1,'Yes'), (1,2,'Yes'), (2,2,'No'), (1,3,'No'), (2,3,'No');
insert into questions values (1, 'Do you like apples?'), (2, 'Do you like oranges?'), (3, 'Do you like carrots?');

select * from crosstab('select responses.user_id, questions.body, responses.body from responses, questions where questions.id = responses.question_id order by user_id') as ct(userid integer, "Do you like apples?" text, "Do you like oranges?" text, "Do you like carrots?" text);

首先,您必须安装tablefunc扩展.从9.1版本开始,您可以使用create extension来实现:

CREATE EXTENSION tablefunc;