PHP:如何制作电影和流派列表(例子) - MySQL加入多对多(想要?)和PHP?

时间:2010-09-07 15:16:51

标签: php mysql many-to-many movie

如何制作电影和流派列表 - MySQL加入多对多(我应该?)和PHP? 或者做其他方式?

示例:
在MySQL中:

Table 1:
movie_id | movie_title
1______  | title_movie_1
2______  | title_movie_2
3______  | title_movie_3

Table 2:
genre_id | genre_title
1______  | title_genre_1
2______  | title_genre_2
3______  | title_genre_3

Table 3:
id | movie_id | genre_1
1_ | __1____ | _1
2_ | __1____ | _2
3_ | __2____ | _1
4_ | __2____ | _3
5_ | __3____ | _1
6_ | __3____ | _3
7_ | __3____ | _2

结果PHP:

Movie: <a href="movie/1 (id_movie)">title_movie_1 _/a>
Genre: 
<a href="genre/1 (id_genre)">title_genre_1 _/a>, 
<a href="genre/2">title_genre_2 _/a>

Movie: <a href="movie/2 (id_movie)">title_movie_2 _/a>
Genre: 
<a href="genre/1">title_genre_1 _/a>,
<a href="genre/3">title_genre_3 _/a>

Movie: <a href="movie/3 (id_movie)">title_movie_3 _/a>
Genre: 
<a href="genre/1">title_genre_1 _/a>
<a href="genre/2">title_genre_2 _/a>
<a href="genre/3">title_genre_3 _/a>

我找不到... 请帮忙!我在等!

真诚的,Areku

2 个答案:

答案 0 :(得分:2)

SELECT
    T1.movie_id,
    T1.movie_title,
    T2.genre_id,
    T2.genre_title
FROM
    Table3 AS T3
    INNER JOIN Table1 AS T1 ON T3.movie_id = T1.movie_id
    INNER JOIN Table2 AS T2 ON T3.genre_1 = T2.genre_id
ORDER BY
    T1.movie_id

这将返回类似的内容:

movie_id | movie_title   | genre_id | genre_title
1        | title_movie_1 | 1        | title_genre_1
1        | title_movie_1 | 2        | title_genre_2
2        | title_movie_2 | 1        | title_genre_1
2        | title_movie_2 | 3        | title_genre_3
3        | title_movie_3 | 1        | title_genre_1
3        | title_movie_3 | 3        | title_genre_3
3        | title_movie_3 | 2        | title_genre_2

之后,您可以循环浏览所有记录;如果movie_id发生了变化,则打印movie_title,然后打印genre_title。

$currentMovie = 0;
while ($row = mysql_fetch_row($result))
{
    if($row['movie_id'] != $current_movie)
    {
        $currentMovie = $row['movie_id'];
        echo "Movie: <a href=\"movie/" + $row['movie_id'] + "\">" + $row['movie_title'] + "</a><br />Genre:<br />"
    }
    echo "<a href=\"genre/" + $row['genre_id'] + "\">" + $row['genre_id'] + "</a><br />"
}

答案 1 :(得分:1)

执行多对多SQL查询:

SELECT m.movie_id, m.movie_title, g.genre_id, g.genre_title
FROM movies m 
INNER JOIN movie_genres mg ON m.movie_id = mg.movie_id
INNER JOIN genres g ON g.genre_id = mg.genre_id

然后按电影ID循环结果和分组。