在我的情况下,如何在 Oracle 中编写 SQL?
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【中文标题】在我的情况下,如何在 Oracle 中编写 SQL?【英文标题】:How can I write SQL in Oracle in my case? 【发布时间】:2015-11-23 06:07:36 【问题描述】: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,表 person_injuryPlace_map
只会映射其他两个表。
我想显示数据的方式是 -
Kitchen Pct Washroom Pct Rooftop Pct Garden Pct
-----------------------------------------------------------------------
1 14.29% 3 42.86% 1 14.29% 2 28.57%
这里,Kitchen, Washroom, Rooftop, Garden column 的值是发生的事件总数。 Pct 列将显示占总数的百分比。
如何在 Oracle SQL 中执行此操作?
【问题讨论】:
很高兴看到有人提供创建和插入语句。 【参考方案1】:您需要使用标准的PIVOT查询。
根据您的 Oracle 数据库版本,您可以通过两种方式进行:
将 PIVOT 用于 11g 版 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
MAX 和 DECODE 用于 10g 版 及之前的版本:
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
【讨论】:
【参考方案2】:如果您使用 Oracle 11g 或更高版本,您可以使用数据透视函数来获得所需的输出。
SELECT *
FROM (
SELECT id
,place
,round((
cnt / sum(cnt) OVER (
ORDER BY NULL
)
) * 100, 2) AS pct
FROM (
SELECT a.id
,a.place
,count(a.id) AS cnt
FROM injury_place a
JOIN person_injuryPlace_map b ON a.id = b.injury_id
GROUP BY a.id
,a.place
)
)
pivot(max(id) , max(pct) pct FOR place IN (
'kitchen' AS kitchen
,'Washroom' Washroom
,'Rooftop' Rooftop
,'Garden' Garden
))
【讨论】:
【参考方案3】:您必须首先在子查询中获取 count 和 pct,然后使用 max+decode 函数以所需的方式汇总它们 检查以下查询是否有帮助:
SELECT Max(Decode(i.place,'Kitchen',cnt)) AS "Kitchecn"
, Max(Decode(i.place,'Kitchen',pct)) AS "Pct"
, Max(Decode(i.place,'Washroom',cnt)) AS "Washroom"
, Max(Decode(i.place,'Washroom',pct)) AS "Pct"
, Max(Decode(i.place,'Rooftop',cnt)) AS "Rooftop"
, Max(Decode(i.place,'Rooftop',pct)) AS "Pct"
, Max(Decode(i.place,'Garden',cnt)) AS "Garden"
, Max(Decode(i.place,'Garden',pct)) AS "Pct"
FROM (SELECT i.place
, Count(pim.injury_id) AS cnt
, (Count(pim.injury_id)*100/(SELECT Count(*) FROM person_injuryPlace_map)) AS pct
FROM person_injuryPlace_map pim
INNER JOIN injury_place i ON i.id = pim.injury_id
GROUP BY i.place)
【讨论】:
使用列别名时出现语法错误。使用单引号使其成为字符串,而不是别名。 谢谢两位。更正了语法。【参考方案4】:我是这样做的 -
select a."kitchen"
, round((100/"Total")*a."kitchen") "Pct"
, a."Garden"
, round((100/"Total")*a."Garden") "Pct"
, a."Washroom"
, round((100/"Total")*a."Washroom") "Pct"
, a."Rooftop"
, round((100/"Total")*a."Rooftop") "Pct"
from
(
select
sum(decode(ip.place, 'kitchen', 1, 0)) "kitchen"
, sum(decode(ip.place, 'Garden', 1, 0)) "Garden"
, sum(decode(ip.place, 'Washroom', 1, 0)) "Washroom"
, sum(decode(ip.place, 'Rooftop', 1, 0)) "Rooftop"
, sum(decode(ip.place, 'kitchen', 1, 0))
+ sum(decode(ip.place, 'Garden', 1, 0))
+ sum(decode(ip.place, 'Washroom', 1, 0))
+ sum(decode(ip.place, 'Rooftop', 1, 0)) "Total"
from
person p
join person_injuryPlace_map pim on pim.person_id = p.id
join injury_place ip on ip.id = pim.injury_id
) a
【讨论】:
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