如何加快从两个表搜索的SQL查询？

Question

这是一个示例，因此我有表user和table city ，它们是这样的：

user列是（user_id，city_id，时间戳） [user_id和city_id是唯一的]

city列是（city_name，city_id） [city_id是唯一的]

我想从特定城市获取给定日期的用户数，因此基本上我这样做是这样的：

select city_id, city_name, 
    (select count(user.user_id) 
     from user, city 
     where DATE_FORMAT(user.timestamp, '%Y-%m-%d') = '2017-03-07' 
     and user.city_id = ct.city_id) as user_count
from city ct 
where (city_id = 20 or city_id = 30)

结果：

city_id, city_name, user_count
20       New York   100
30       LA         200

然后我意识到这比直接搜索要慢得多

select count(user.user_id) 
from user, city 
where DATE_FORMAT(user.timestamp, '%Y-%m-%d') = '2017-03-07' 
    and user.city_id = 20

为什么是这样？ 原始搜索中的ct.city_id是否已设置为20或30？ 如何优化搜索并以所需的表格格式获取结果？

Answer 1

您可以改进查询，避免子选择，而使用内部联接和分组依据

select city_id, city_name,    count(user.user_id) 
from user
inner join city on user.city_id = city.city_id
where DATE_FORMAT(user.timestamp, '%Y-%m-%d') = '2017-03-07' 
and city_id in (city_id = 20 or city_id = 30)
group by city_id, city_name

Answer 2

我假设MySQL选择在第一个查询中将派生表具体化为内部临时表，而不在第二个查询中做出选择。

对于派生表（FROM子句中的子查询），优化器具有以下选择：

• 将派生表合并到外部查询块中
• 将派生表具体化为内部临时表

来源： Mysql文档-8.2.2优化子查询，派生表和视图

Answer 3

尝试这个：

select city_id, city_name, count(user.user_id) as user_count
from city ct 
inner join user on user.city_id = ct.city_id
where (ct.city_id = 20 or ct.city_id = 30)
AND DATE_FORMAT(user.timestamp, '%Y-%m-%d') = '2017-03-07'