mysql - 是否可以创建一个表，然后在一个查询中从中进行选择？

Question

我很好奇是否有办法在表格上进行选择，该表格本身就是选择的产物。类似的东西（伪）

    select a,b,c from 
(select id as a, name as b, phone as c, date as d from some_table) 
where  d = now();

因此，实际上括号将返回一个表，其列由定义，as然后外部select可以去查询该结果。

如果不完全是这样，我很想知道我可以使用什么模糊的类似方法。

我需要这个的原因是我有一个相当广泛的数据透视查询，我在我的用户数据上运行，我希望能够从相当大的结果中进行选择。

Answer 1

您可能会遇到类似“派生表需要别名”的错误消息

以下作品：

select a,b,c 
from (
     select id as a, 
            name as b, 
            phone as c, 
            some_date as d 
     from some_table
) as t 
where  d <= now();

别名t定义了一个所谓的派生表。

SQLFiddle：http ://sqlfiddle.com/#!2/05fd6/2

Answer 2

对的，这是可能的。尝试这个：

select A.a,A.b,A.c from 
(select id as a, name as b, phone as c, date as d from some_table where  d = now()) A

它被称为派生表。

将子查询移到WHERE外部：

select A.a,A.b,A.c, A.d from 
(select id as a, name as b, phone as c, date as d from some_table) A
where A.d = now()