我该如何在Oracle中编写SQL？

小编典典

我该如何在Oracle中编写SQL？

sql

所以，这是表格-

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);

在这里，表格person_injuryPlace_map只会映射其他两个表格。

我想要显示数据的方式是-

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

在这里，“厨房”，“洗手间”，“屋顶”，“花园”列的值是发生的事件总数。“百分比”列将显示总计数的百分比。

如何在Oracle SQL中做到这一点？

阅读 165

2021-05-05

共1个答案

小编典典

您需要使用标准的 PIVOT 查询。

根据您的 Oracle数据库版本 ，您可以通过以下两种方式进行操作：

将 PIVOT 用于 11g 及更高版本：

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

在版本10g* 及之前的 版本中 使用 MAX 和 DECODE ： *

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

2021-05-05