在我的情况下，如何在 Oracle 中编写 SQL?

2022-01-22 00:00:00 pivot sql oracle

所以，这里是表格-

create table person (
id number,
name varchar2(50)
);

create table injury_place (
id number,
place varchar2(50)
);

create table person_injuryPlace_map (
person_id number,
injury_id number
);

insert into person values (1, 'Adam');
insert into person values (2, 'Lohan');
insert into person values (3, 'Mary');
insert into person values (4, 'John');
insert into person values (5, 'Sam');


insert into injury_place values (1, 'kitchen');
insert into injury_place values (2, 'Washroom');
insert into injury_place values (3, 'Rooftop');
insert into injury_place values (4, 'Garden');


insert into person_injuryPlace_map values (1, 2);
insert into person_injuryPlace_map values (2, 3);
insert into person_injuryPlace_map values (1, 4);
insert into person_injuryPlace_map values (3, 2);
insert into person_injuryPlace_map values (4, 4);
insert into person_injuryPlace_map values (5, 2);
insert into person_injuryPlace_map values (1, 1);

这里，person_injuryPlace_map 表只会映射其他两个表.

Here, table person_injuryPlace_map will just map the both other tables.

我想要显示数据的方式是 -

How I wanted to show data is -

Kitchen   Pct      Washroom   Pct     Rooftop   Pct     Garden   Pct
-----------------------------------------------------------------------
1         14.29%   3          42.86%   1        14.29%   2        28.57%

这里，厨房、洗手间、屋顶、花园柱的值是发生的事件总数.Pct 列将显示占总数的百分比.

Here, the value of Kitchen, Washroom, Rooftop, Garden column is the total incidents happened. Pct columns will show the percentage of the total count.

如何在 Oracle SQL 中执行此操作?

How can I do this in Oracle SQL?

推荐答案

您需要使用标准的 PIVOT 查询.

You need to use the standard PIVOT query.

根据您的 Oracle 数据库版本，您可以通过两种方式进行:

Depending on your Oracle database version, you could do it in two ways:

将 PIVOT 用于 11g 版 及更高版本:

Using PIVOT for version 11g and up:

SQL> SELECT *
  2  FROM
  3    (SELECT c.place place,
  4      row_number() OVER(PARTITION BY c.place ORDER BY NULL) cnt,
  5      (row_number() OVER(PARTITION BY c.place ORDER BY NULL)/
  6      COUNT(place) OVER(ORDER BY NULL))*100 pct
  7    FROM person_injuryPlace_map A
  8    JOIN person b
  9    ON(A.person_id = b.ID)
 10    JOIN injury_place c
 11    ON(A.injury_id = c.ID)
 12    ORDER BY c.place
 13    ) PIVOT (MAX(cnt),
 14             MAX(pct) pct
 15             FOR (place) IN ('kitchen' AS kitchen,
 16                             'Washroom' AS Washroom,
 17                             'Rooftop' AS Rooftop,
 18                             'Garden' AS Garden));

   KITCHEN KITCHEN_PCT   WASHROOM WASHROOM_PCT    ROOFTOP ROOFTOP_PCT     GARDEN GARDEN_PCT
---------- ----------- ---------- ------------ ---------- ----------- ---------- ----------
         1  14.2857143          3   42.8571429          1  14.2857143          2 28.5714286

MAX 和 DECODE 用于 10g 版 及之前的版本:

Using MAX and DECODE for version 10g and before:

SQL> SELECT MAX(DECODE(t.place,'kitchen',cnt)) Kitchen ,
  2    MAX(DECODE(t.place,'kitchen',pct)) Pct ,
  3    MAX(DECODE(t.place,'Washroom',cnt)) Washroom ,
  4    MAX(DECODE(t.place,'Washroom',pct)) Pct ,
  5    MAX(DECODE(t.place,'Rooftop',cnt)) Rooftop ,
  6    MAX(DECODE(t.place,'Rooftop',pct)) Pct ,
  7    MAX(DECODE(t.place,'Garden',cnt)) Garden ,
  8    MAX(DECODE(t.place,'Garden',pct)) Pct
  9  FROM
 10    (SELECT b.ID bid,
 11      b.NAME NAME,
 12      c.ID cid,
 13      c.place place,
 14      row_number() OVER(PARTITION BY c.place ORDER BY NULL) cnt,
 15      ROUND((row_number() OVER(PARTITION BY c.place ORDER BY NULL)/
 16      COUNT(place) OVER(ORDER BY NULL))*100, 2) pct
 17    FROM person_injuryPlace_map A
 18    JOIN person b
 19    ON(A.person_id = b.ID)
 20    JOIN injury_place c
 21    ON(A.injury_id = c.ID)
 22    ORDER BY c.place
 23    ) t;

   KITCHEN        PCT   WASHROOM        PCT    ROOFTOP        PCT     GARDEN        PCT
---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------
         1      14.29          3      42.86          1      14.29          2      28.57

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