大家好，我是空空star，本篇带大家了解一道简单的力扣sql练习题。

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前言

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一、题目：1050. 合作过至少三次的演员和导演

ActorDirector 表：

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| actor_id    | int     |
| director_id | int     |
| timestamp   | int     |
+-------------+---------+
timestamp 是这张表的主键.

写一条SQL查询语句获取合作过至少三次的演员和导演的 id 对 (actor_id, director_id)

ActorDirector 表：
+-------------+-------------+-------------+
| actor_id    | director_id | timestamp   |
+-------------+-------------+-------------+
| 1           | 1           | 0           |
| 1           | 1           | 1           |
| 1           | 1           | 2           |
| 1           | 2           | 3           |
| 1           | 2           | 4           |
| 2           | 1           | 5           |
| 2           | 1           | 6           |
+-------------+-------------+-------------+Result 表：
+-------------+-------------+
| actor_id    | director_id |
+-------------+-------------+
| 1           | 1           |
+-------------+-------------+
唯一的 id 对是 (1, 1)，他们恰好合作了 3 次。

二、解题

1.正确示范①

提交SQL

select actor_id,director_id
from ActorDirector 
group by actor_id,director_id
having count(1)>=3;

运行结果

2.正确示范②

提交SQL

select actor_id,director_id
from ActorDirector 
group by actor_id,director_id
having count(*)>=3;

运行结果

3.正确示范③

提交SQL

select actor_id,director_id
from ActorDirector 
group by actor_id,director_id
having count(timestamp)>=3;

运行结果

4.正确示范④

提交SQL

select actor_id,director_id from(select actor_id,director_id,count(1) numfrom ActorDirector group by actor_id,director_id
) u 
where num>=3;

运行结果

5.其他

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总结

正确示范①思路：
按照演员和导演id分组group by actor_id,director_id
筛选合作次数大于等于3次的having count(1)>=3；
正确示范②思路：
按照演员和导演id分组group by actor_id,director_id
筛选合作次数大于等于3次的having count(*)>=3；
正确示范③思路：
按照演员和导演id分组group by actor_id,director_id
筛选合作次数大于等于3次的having count(timestamp)>=3；
正确示范④思路：
按照演员和导演id分组group by actor_id,director_id，
计算出次数count(1) as num，
然后在最外层限定num>=3。