Come posso scrivere SQL in Oracle nel mio caso?

Question

Così, qui sono il tavoli-

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); 

Qui, tavolo person_injuryPlace_map sarà solo mappare i due altri tavoli.

Come ho voluto mostrare i dati è -

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

Qui, il valore della cucina, toilette, sui tetti, colonna Garden è gli incidenti totali accaduto. Le colonne Pct mostreranno la percentuale del conteggio totale.

Come posso fare questo in Oracle SQL?

Answer 1

È necessario utilizzare la query standard PIVOT.

A seconda del Oracle database della versione, si potrebbe fare in due modi:

Utilizzando PIVOT per versione 11g e fino:

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 

Utilizzando MAX e DECODE per versione 10g e prima:

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 

Answer 2

Si dovrà prendere prima il conteggio e PCT in una sottoquery quindi utilizzare la funzione max + decodifica di riassumere entrambi nella richiesta della moda Verificare se la query di seguito aiuta:

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) 

Answer 3

Se si utilizza Oracle 11g o versione successiva, è possibile utilizzare la funzione di pivot per l'output richiesto.

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 
      )) 

Answer 4

ho fatto come il seguente -

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