英文:
Find the genres that have at least 2 same directors, then print common genres and director id?
问题
数据集如下：
g1 = {d1,d3,d5}g2 = {d1,d3,d2}g3 = {d4,d3,d5}
我有一个名为director_genre的表格，内容如下：
director_id | genred1 g1d1 g2d2 g2d3 g1d3 g2d3 g3d4 g3d5 g1d5 g3
我想要的输出是：
genre | director_idg1 d1,d3,d5g2 d1,d3g3 d3,d5
我考虑的查询是：
SELECT genreFROM director_genreGROUP BY genre;
之后，根据问题要求，我不会提供其他逻辑。请帮我完成这个任务。
英文:
The dataset is like this:
g1 = {d1,d3,d5}g2 = {d1,d3,d2}g3 = {d4,d3,d5}
I have the following table called director_genre:
director_id | genred1 g1d1 g2d2 g2d3 g1d3 g2d3 g3d4 g3d5 g1d5 g3
The output I want is
genre | director_idg1 d1,d3,d5g2 d1,d3g3 d3,d5
The query I thought of is
SELECT genreFROM director_genreGROUP BY genre;
After this, I am not to put the logic as asked in the question.
Please help me to get this done.
答案1
得分: 3
我建议使用以下查询：
SELECT d.genre, GROUP_CONCAT(director_id)FROM director_genre dWHERE EXISTS (SELECT *FROM director_genre d2WHERE d.director_id = d2.director_id AND d.genre <> d2.genre)GROUP BY d.genre;
嵌套的子查询将仅选择与至少两种类型相关联的导演记录（对于d中的每个记录，它会查看是否存在d2中具有相同director_id但不同genre的记录）。然后，它们将按genre分组。
英文:
I would suggest the following query:
SELECT d.genre, GROUP_CONCAT(director_id)FROM director_genre dWHERE EXISTS (SELECT *FROM director_genre d2WHERE d.director_id = d2.director_id AND d.genre <> d2.genre)GROUP BY d.genre;
The nested subquery will select only records of directors who are associated with at least 2 genres (for each record in d it looks whether there exists a record in d2 with the same director_id but different genre). Then they will be grouped by genre.
答案2
得分: 1
你想要一个（类型，导演）列表，其中导演还与另一种类型相关联。然后分组。可以这样做：
select genre, group_concat(director_id separator ', ') as directorsfrom dgwhere exists (select *from dg as xwhere x.director_id = dg.director_idand x.genre <> dg.genre)group by genre
英文:
You want a list of (genre, director) where the director is also associated with another genre. Then group. This could be done like so:
select genre, group_concat(director_id separator ', ') as directorsfrom dgwhere exists (select *from dg as xwhere x.director_id = dg.director_idand x.genre <> dg.genre)group by genre
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