如何使用 MySQL 的 INNER JOIN 匹配多行？

Question

使用 INNER JOIN 选择标签为“动作”和“戏剧”的电影的正确方法是什么？

我试过这个查询，结果一定是“Taken, The Godfather”，但是没有返回结果。

SELECT 
    f.film_guid,
    f.film_name
FROM
    films as f
INNER JOIN
    film_labels as l ON l.film_guid = f.film_guid
WHERE
    l.label = 'Action' AND l.label = 'Drama'

表格：电影

+------------+----------------+
| film_guid  | film_name      |
+------------+----------------+
| filmguid_1 | Taken          |
| filmguid_2 | Matrix         |
| filmguid_3 | The Godfather  |
+------------+----------------+

表格：film_labels

+------------+----------------+
| film_guid  | label          |
+------------+----------------+
| filmguid_1 | Action         |
| filmguid_1 | Drama          |
| filmguid_1 | Family         |
| filmguid_2 | Action         |
| filmguid_3 | Action         |
| filmguid_3 | Drama          |
+------------+----------------+

Answer 1

您正在film_labels 中查找包含动作和戏剧的行，这不可能发生。您需要在与给定电影相对应的标签中搜索，这表明聚合：

SELECT f.film_guid, f.film_name
FROM films as f
INNER JOIN film_labels as l ON l.film_guid = f.film_guid
WHERE l.label IN ('Action', 'Drama')  -- either one, or the other
GROUP BY f.film_guid, f.film_name
HAVING COUNT(*) = 2                   -- both match

请注意，您也可以将exists 与相关子查询一起使用。键入的时间有点长，但 可能 更有效（使用正确的索引），因为它避免了聚合的需要：

SELECT f.*
FROM films as f
WHERE
    EXISTS (SELECT 1 FROM film_labels l WHERE l.film_guid = f.film_guid AND l.label = 'Action')
    AND EXISTS (SELECT 1 FROM film_labels l WHERE l.film_guid = f.film_guid AND l.label = 'Drama')

为了第二个查询的性能，您需要在film_labels(film_guid , label) 上建立索引。