关于URL上的第一个问题:https://www.w3resource.com/sql-exercises/movie-database-exercise/subqueries-exercises-on-movie-database.php#SQLEDITOR“用SQL编写查询以列出在电影《安妮·霍尔》中扮演过的演员的所有信息。”,谁能解释一下为什么查询这个原因即使三个表中的每个表中都不存在重复的id列值,下面的结果也会导致重复的条目:actor
,movie_cast
和movie
?
SQL查询:
SELECT actor.act_id, actor.act_fname, actor.act_lname, actor.act_gender
FROM actor
INNER JOIN movie_cast ON actor.act_id = movie_cast.act_id
INNER JOIN movie ON movie_cast.mov_id = movie.mov_id
WHERE movie.mov_title = 'Annie Hall'
上述在w3resource上的查询的输出是:
act_id act_fname act_lname act_gender
111 Woody Allen M
111 Woody Allen M
3个表的示例:
演员:
act_id | act_fname | act_lname | act_gender
--------+----------------------+----------------------+------------
101 | James | Stewart | M
102 | Deborah | Kerr | F
103 | Peter | OToole | M
104 | Robert | De Niro | M
105 | F. Murray | Abraham | M
106 | Harrison | Ford | M
107 | Nicole | Kidman | F
108 | Stephen | Baldwin | M
109 | Jack | Nicholson | M
110 | Mark | Wahlberg | M
111 | Woody | Allen | M
112 | Claire | Danes | F
113 | Tim | Robbins | M
114 | Kevin | Spacey | M
115 | Kate | Winslet | F
116 | Robin | Williams | M
117 | Jon | Voight | M
118 | Ewan | McGregor | M
119 | Christian | Bale | M
120 | Maggie | Gyllenhaal | F
121 | Dev | Patel | M
122 | Sigourney | Weaver | F
123 | David | Aston | M
124 | Ali | Astin | F
电影播放:
act_id | mov_id | role
--------+--------+--------------------------------
101 | 901 | John Scottie Ferguson
102 | 902 | Miss Giddens
103 | 903 | T.E. Lawrence
104 | 904 | Michael
105 | 905 | Antonio Salieri
106 | 906 | Rick Deckard
107 | 907 | Alice Harford
108 | 908 | McManus
110 | 910 | Eddie Adams
111 | 911 | Alvy Singer
112 | 912 | San
113 | 913 | Andy Dufresne
114 | 914 | Lester Burnham
115 | 915 | Rose DeWitt Bukater
116 | 916 | Sean Maguire
117 | 917 | Ed
118 | 918 | Renton
120 | 920 | Elizabeth Darko
121 | 921 | Older Jamal
122 | 922 | Ripley
114 | 923 | Bobby Darin
109 | 909 | J.J. Gittes
119 | 919 | Alfred Borden
电影:
mov_id | mov_title | mov_year | mov_time | mov_lang | mov_dt_rel | mov_rel_country
--------+----------------------------------------------------+----------+----------+-----------------+------------+-----------------
901 | Vertigo | 1958 | 128 | English | 1958-08-24 | UK
902 | The Innocents | 1961 | 100 | English | 1962-02-19 | SW
903 | Lawrence of Arabia | 1962 | 216 | English | 1962-12-11 | UK
904 | The Deer Hunter | 1978 | 183 | English | 1979-03-08 | UK
905 | Amadeus | 1984 | 160 | English | 1985-01-07 | UK
906 | Blade Runner | 1982 | 117 | English | 1982-09-09 | UK
907 | Eyes Wide Shut | 1999 | 159 | English | | UK
908 | The Usual Suspects | 1995 | 106 | English | 1995-08-25 | UK
909 | Chinatown | 1974 | 130 | English | 1974-08-09 | UK
910 | Boogie Nights | 1997 | 155 | English | 1998-02-16 | UK
911 | Annie Hall | 1977 | 93 | English | 1977-04-20 | USA
912 | Princess Mononoke | 1997 | 134 | Japanese | 2001-10-19 | UK
913 | The Shawshank Redemption | 1994 | 142 | English | 1995-02-17 | UK
914 | American Beauty | 1999 | 122 | English | | UK
915 | Titanic | 1997 | 194 | English | 1998-01-23 | UK
916 | Good Will Hunting | 1997 | 126 | English | 1998-06-03 | UK
917 | Deliverance | 1972 | 109 | English | 1982-10-05 | UK
918 | Trainspotting | 1996 | 94 | English | 1996-02-23 | UK
919 | The Prestige | 2006 | 130 | English | 2006-11-10 | UK
920 | Donnie Darko | 2001 | 113 | English | | UK
921 | Slumdog Millionaire | 2008 | 120 | English | 2009-01-09 | UK
922 | Aliens | 1986 | 137 | English | 1986-08-29 | UK
923 | Beyond the Sea | 2004 | 118 | English | 2004-11-26 | UK
924 | Avatar | 2009 | 162 | English | 2009-12-17 | UK
926 | Seven Samurai | 1954 | 207 | Japanese | 1954-04-26 | JP
927 | Spirited Away | 2001 | 125 | Japanese | 2003-09-12 | UK
928 | Back to the Future | 1985 | 116 | English | 1985-12-04 | UK
925 | Braveheart | 1995 | 178 | English | 1995-09-08 | UK
您的movie_cast
表中有演员Woody Allen
的重复条目。您可以通过以下查询进行检查。您可以在下面的图像中看到输出,因此您在查询中获得了两条记录。如果要消除重复的条目,请在查询中尝试DISTINCT
。
SELECT *
FROM movie_cast
WHERE movie_cast.act_id = 111
查询不同的记录。
SELECT DISTINCT actor.act_id, actor.act_fname, actor.act_lname, actor.act_gender
FROM actor
INNER JOIN movie_cast ON actor.act_id = movie_cast.act_id
INNER JOIN movie ON movie_cast.mov_id = movie.mov_id
WHERE movie.mov_title = 'Annie Hall'
OP来自评论的问题。
为什么建议的解决方案不会导致任何重复?建议的解决方案:SELECT * FROM actor WHERE act_id IN(SELECT act_id FROM movie_cast WHERE mov_id IN(SELECT mov_id FROM movie WHERE mov_title ='Annie Hall'));
答案
因为inner query
将具有包含重复的mov_id
的列表。但是outer query
将仅对每个记录检查一次inner query
结果是否包含current
行的mov_id
。