0

给定以下表格,我如何构建一个 SQL 查询,其中包括“items”表中所有项目的列表,以及“colors”表中每种颜色的列,对于列出的每个项目,指示什么颜色item 有关系。

如果这根本不清楚,请让我知道哪些附加信息将有助于澄清。表信息和所需的 SQL 结果如下:

物品表:

id | item_name
1  | 'item 1'
2  | 'item 2'
3  | 'item 3'

颜色表:

id | color_name
1  | 'red'
2  | 'blue'
3  | 'green'

item_color 表:

item_id | color_id
1       | 1
1       | 3
2       | 2
2       | 3
3       | 2

所需的 SQL 查询结果:

item_name | red | blue | green
'item 1'  |  1  | null |   1
'item 2'  | null|   1  |   1
'item 3'  | null|   1  | null

谢谢,科林

4

4 回答 4

5

利用:

SELECT item_name,
       MAX(red) 'red',
       MAX(blue) 'blue',
       MAX(green) 'green'
  FROM (SELECT t.item_name,
         CASE
           WHEN c.color_name = 'red' THEN
             1
           ELSE
             NULL
         END 'red',
         CASE
           WHEN c.color_name = 'blue' THEN
             1
           ELSE
             NULL
         END 'blue',
         CASE
           WHEN c.color_name = 'green' THEN
             1
           ELSE
             NULL
         END 'green'       
    FROM ITEMS t
    JOIN ITEM_COLOR ic ON ic.item_id = t.item_id
    JOIN COLORS c ON c.id = ic.color_id)
GROUP BY item_name 

如果您想要与某个项目关联的红色/蓝色/绿色的总数,请将 MAX 更改为 COUNT。

使用子查询因子替代:

WITH icolors AS (
   SELECT t.item_name,
          CASE
           WHEN c.color_name = 'red' THEN
             1
           ELSE
             NULL
         END 'red',
     CASE
       WHEN c.color_name = 'blue' THEN
         1
       ELSE
         NULL
     END 'blue',
     CASE
       WHEN c.color_name = 'green' THEN
         1
       ELSE
             NULL
     END 'green'       
    FROM ITEMS t
    JOIN ITEM_COLOR ic ON ic.item_id = t.item_id
    JOIN COLORS c ON c.id = ic.color_id)
  SELECT t.item_name,
         MAX(t.red) 'red',
         MAX(t.blue) 'blue',
         MAX(t.green) 'green'
    FROM icolors t
GROUP BY t.item_name
于 2009-09-01T17:07:55.550 回答
3

你在oracle 11g上吗?

这似乎是11g中新枢轴功能的理想用途

于 2009-09-01T17:05:42.840 回答
2

如果你事先知道所有可能的颜色,你可以做的很乱但很有效。如果您事先不知道所有可能的颜色,那就更难了——您必须运行一些查询来找出结果表中将出现哪些列,然后制作 SQL 来创建这些列(动态 SQL) .

因此,假设您知道结果表中的列:

SELECT i.item_name, r.red, b.blue, g.green
  FROM items i
       LEFT JOIN
       (SELECT item_name, COUNT(*) AS red
          FROM item_color
         WHERE color_id = 1
         GROUP BY item_name) AS r
       ON i.item_name = r.item_name
       LEFT JOIN
       (SELECT item_name, COUNT(*) AS green
          FROM item_color
         WHERE color_id = 3
         GROUP BY item_name) AS g
       ON i.item_name = g.item_name
       LEFT JOIN
       (SELECT item_name, COUNT(*) AS blue
          FROM item_color
         WHERE color_id = 2
         GROUP BY item_name) AS b
       ON i.item_name = b.item_name

请注意,在这个公式中,我在构建查询时使用了颜色表中的数据。并且替代形式将构建子查询作为(内部)连接到颜色表,使用颜色名称而不是 WHERE 子句中的代码。

于 2009-09-01T17:11:31.283 回答
1
create table item (id number not null, item_name varchar2(200) not null);
create table color (id number not null, color_name varchar2(200) not null);
create table item_color (item_id number not null, color_id number not null);

insert into item values (1, 'item 1');
insert into item values (2, 'item 2');
insert into item values (3, 'item 3');


insert into color values (1, 'red');
insert into color values (2, 'blue');
insert into color values (3, 'green');

insert into item_color values (1, 1);
insert into item_color values (1, 3);
insert into item_color values (2, 2);
insert into item_color values (2, 3);
insert into item_color values (3, 2);

commit;

然后选择:

select * from
(
select
  i.item_name
  , c.color_name
from
  item i
  , color c
  , item_color ic
where
  ic.item_id = i.id
  and ic.color_id = c.id
) pivot (
  count(color_name) cnt
  for color_name in ('red', 'blue', 'green')
);

给出:

item 1    1    0    1
item 2    0    1    1
item 3    0    1    0

如果您事先不知道颜色列表,您可以先选择颜色表,然后动态构建枢轴选择(for color_name in (select color_name from color)不可能进行子选择),或者您可以使用pivot xml并后处理结果:

select * from
(
select
  i.item_name
  , c.color_name
from
  item i
  , color c
  , item_color ic
where
  ic.item_id = i.id
  and ic.color_id = c.id
) pivot xml (
  count(color_name) cnt
  for color_name in (any)
)

给出:

item 1  <PivotSet><item><column name = "COLOR_NAME">green</column><column name = "CNT">1</column></item><item><column name = "COLOR_NAME">red</column><column name = "CNT">1</column></item></PivotSet>
item 2  <PivotSet><item><column name = "COLOR_NAME">blue</column><column name = "CNT">1</column></item><item><column name = "COLOR_NAME">green</column><column name = "CNT">1</column></item></PivotSet>
item 3  <PivotSet><item><column name = "COLOR_NAME">blue</column><column name = "CNT">1</column></item></PivotSet>
于 2011-11-10T13:28:29.710 回答