第1步:「如何炸燬數據庫」
來源:
SQL Fiddle
的Oracle 11g R2架構設置:
查詢1:
select * from films11
Results:
| YEAR | DIRECTOR | MOVIETITLE | ACTORNAME |
|------|----------|------------|----------------|
| 2000 | dir1 | title1 | act1,act2 |
| 2001 | dir2 | title2 | act1,act2,act3 |
| 2002 | dir1 | title3 | act4 |
查詢2:
select YT.year, YT.movietitle,
REPLACE(REGEXP_SUBSTR(YT.actorname||',','.*?,',1,lvl.lvl),',','') AS actorname
from films11 YT
join (select level as lvl
from dual
connect by level <= (select max(regexp_count(actorname,',')+1) from films11)
) lvl on lvl.lvl <= regexp_count(YT.actorname,',')+1
order by YT.year, YT.movietitle, actorname
有了一個很好的笛卡爾積:
Results:
| YEAR | MOVIETITLE | ACTORNAME |
|------|------------|-----------|
| 2000 | title1 | act1 |
| 2000 | title1 | act2 |
| 2001 | title2 | act1 |
| 2001 | title2 | act2 |
| 2001 | title2 | act3 |
| 2002 | title3 | act4 |
你運行它ONCE,並使用它的一切移動到一個標準化的DB
以下是完整的腳本到你的架構改變的東西更方便......
CREATE TABLE actors(
id_actor NUMBER GENERATED BY DEFAULT ON NULL AS IDENTITY,
act_name VARCHAR2(100)
)
;
CREATE TABLE directors(
id_director NUMBER GENERATED BY DEFAULT ON NULL AS IDENTITY,
dir_name VARCHAR2(100)
)
;
CREATE TABLE movies(
id_movie NUMBER GENERATED BY DEFAULT ON NULL AS IDENTITY,
mov_year NUMBER,
mov_name VARCHAR2(100),
director_id NUMBER
)
;
CREATE TABLE playedby(
movie_id NUMBER,
actor_id NUMBER
)
;
INSERT INTO directors (dir_name)
SELECT DISTINCT director dir_name
FROM films11
;
INSERT INTO movies (mov_year, mov_name, director_id)
SELECT year mov_year, movietitle mov_name, directors.id_director director_id
FROM films11
INNER JOIN directors ON directors.dir_name = films11.director
;
INSERT INTO actors (act_name)
SELECT DISTINCT t.actorname act_name
FROM (
SELECT YT.year, YT.movietitle,
REPLACE(REGEXP_SUBSTR(YT.actorname||',','.*?,',1,lvl.lvl),',','') AS actorname
FROM films11 YT
JOIN (SELECT level AS lvl
FROM dual
CONNECT BY level <= (SELECT MAX(REGEXP_COUNT(actorname,',')+1) FROM films11)
) lvl ON lvl.lvl <= REGEXP_COUNT(YT.actorname,',')+1
) t
;
INSERT INTO playedby (movie_id, actor_id)
SELECT movies.id_movie movie_id, actors.id_actor actor_id
FROM (
SELECT YT.year, YT.movietitle,
REPLACE(REGEXP_SUBSTR(YT.actorname||',','.*?,',1,lvl.lvl),',','') AS actorname
FROM films11 YT
JOIN (SELECT level AS lvl
FROM dual
CONNECT BY level <= (SELECT MAX(REGEXP_COUNT(actorname,',')+1) FROM films11)
) lvl ON lvl.lvl <= REGEXP_COUNT(YT.actorname,',')+1
) t
INNER JOIN actors ON t.actorname = actors.act_name
INNER JOIN movies ON t.year = movies.mov_year AND t.movietitle = movies.mov_name
;
這之後你可以做一個選擇這樣的:
查詢3:
SELECT mov_year, mov_name, dir_name, act_name
FROM movies
INNER JOIN directors ON directors.id_director = movies.director_id
INNER JOIN playedby ON movies.id_movie = playedby.movie_id
INNER JOIN actors ON playedby.actor_id = actors.id_actor
WHERE act_name like '%act2%'
order by mov_year asc
Results:
| MOV_YEAR | MOV_NAME | DIR_NAME | ACT_NAME |
|----------|----------|----------|----------|
| 2000 | title1 | dir1 | act2 |
| 2001 | title2 | dir2 | act2 |
橫向轉動顯示器 – Strawberry
屏幕截圖中已經有答案。您的新表只需要在ACTORNAME列中爲每個不同的Actor創建一行,如截圖的第三個表所示。或者,您可以使用@Strawberry的建議。 – JuveLeo1906
表有太多的行能夠做到這一點。我想知道是否有一個SQL查詢在逗號分隔'ACTORNAME'列中的名稱,就像你說到一個新的行... – 2milli