Mysql 如何使用视图执行复杂的sql查询，而不是依赖中间（具体）表

SQL粉丝：

我正在尝试通过使用以下用例来挖掘我的一些生疏的sql技能：

假设我们有一家有线电视公司，有跟踪以下内容的db表：

• 电视节目,
• 观看我们节目的客户，以及
• 观看活动（特定客户观看特定节目的日期）

比如说，我们想要一份关于每个节目的浏览量和观众平均年龄的报告 观众。我在这里意识到的关键是，如果同一个人在不同的日期看X台节目两次 我们不能让这个人的年龄对“X秀观众的平均年龄”的计算贡献两倍

首先，我定义了我的表并在其中粘贴了一些数据（这是mysql语法，b.t.w）：

现在，我创建了一个查询，将表连接起来，并给出每个客户的showname和年龄 谁看了一个节目

     select date, shows.showid, cust.custid, showname, age  from 
        ->                       watched 
        ->                           inner join
        ->                       cust
        ->                           on cust.custid = watched.custid
        ->                               inner join 
        ->                                   shows
        ->                               on shows.showid = watched.showid    ;
    +------+--------+--------+----------+-----+
    | date | showid | custid | showname | age |
    +------+--------+--------+----------+-----+
    |    1 |      1 |      1 | bingo    |  20 |
    |    2 |      1 |      1 | bingo    |  20 |
    |    3 |      1 |      1 | bingo    |  20 |
    |    4 |      1 |      1 | bingo    |  20 |
    |    1 |      1 |      2 | bingo    |  30 |
    |    1 |      1 |      3 | bingo    |  40 |
    |    1 |      1 |      4 | bingo    |  20 |
    |    2 |      2 |      2 | animals  |  30 |
    +------+--------+--------+----------+-----+
    8 rows in set (0.00 sec)

但请注意，customer id 1作为观察者多次出现 “宾果游戏”的表演，我希望他只算一次

因此，我创建了一个查询，列出了节目和观看过节目的客户， 但每个客户只统计一次

    mysql>  select age, showname, showid, custid  from 
        ->              ( select date, shows.showid, cust.custid, showname, age  from 
        ->                  watched 
        ->                      inner join
        ->                  cust
        ->                      on cust.custid = watched.custid
        ->                          inner join 
        ->                              shows
        ->                          on shows.showid = watched.showid    
        ->              ) as VIEWS
        ->          group by custid, showname;
    +-----+----------+--------+--------+
    | age | showname | showid | custid |
    +-----+----------+--------+--------+
    |  20 | bingo    |      1 |      1 |
    |  30 | animals  |      2 |      2 |
    |  30 | bingo    |      1 |      2 |
    |  40 | bingo    |      1 |      3 |
    |  20 | bingo    |      1 |      4 |
    +-----+----------+--------+--------+
    5 rows in set (0.00 sec)

接下来——还有（我希望你能给我一点建议）。。。 我试着创建一个视图，给我看的每一个节目的名字，看这个节目的人的平均年龄，以及节目id。 我计划加入一个查询，让我知道每个节目的浏览量。但视图创建失败，如图所示：

    mysql> create view viewages as 
        -> select showname, avg(age), showid
        ->       from 
        ->          (select age, showname, showid, custid  from 
        ->              ( select date, shows.showid, cust.custid, showname, age  from 
        ->                  watched 
        ->                      inner join
        ->                  cust
        ->                      on cust.custid = watched.custid
        ->                          inner join 
        ->                              shows
        ->                          on shows.showid = watched.showid    
        ->              ) as VIEWS
        ->          group by custid, showname)
        ->              as DISTINCT_CUST_VIEWS
        ->          group by showname;
    ERROR 1349 (HY000): View's SELECT contains a subquery in the FROM clause

好的。。所以那没用。我让它工作了，但我做的方式看起来很俗气。我使用中间表

有没有一位sql摇滚明星可以在没有桌子的情况下给我展示一种更好的方法。。也许有一个视角，就像我试过的那样 创造，或者更好的东西

以下是我的蹩脚解决方案：

drop table if exists viewage ;

create table viewage (
    showname varchar(256) not null,
    avg_age   float  not null,
    showid    int    not null
);

insert into viewage 
select showname, avg(age), showid
      from 
         (select age, showname, showid, custid  from 
             ( select date, shows.showid, cust.custid, showname, age  from 
                 watched 
                     inner join
                 cust
                     on cust.custid = watched.custid
                         inner join 
                             shows
                         on shows.showid = watched.showid    
             ) as VIEWS
         group by custid, showname)
             as DISTINCT_CUST_VIEWS
         group by showname;


##  Finally join the table with average age for each show with a query that does the count of views for each show:



drop table if exists viewage ;

create table viewage (
    showname varchar(256) not null,
    avg_age   float  not null,
    showid    int    not null
);

insert into viewage 
select showname, avg(age), showid
      from 
         (select age, showname, showid, custid  from 
             ( select date, shows.showid, cust.custid, showname, age  from 
                 watched 
                     inner join
                 cust
                     on cust.custid = watched.custid
                         inner join 
                             shows
                         on shows.showid = watched.showid    
             ) as VIEWS
         group by custid, showname)
             as DISTINCT_CUST_VIEWS
         group by showname;



select count(*), showname, avg_age from  
  watched 
    inner join 
  viewage
    on viewage.showid = watched.showid 
    group by showname;




+----------+----------+---------+
| count(*) | showname | avg_age |
+----------+----------+---------+
|        1 | animals  |      30 |
|        7 | bingo    |    27.5 |
+----------+----------+---------+
2 rows in set (0.00 sec)

提前感谢您的帮助

-克里斯

我试着创建一个视图，给我每个被观看的节目的名称， 观看该节目者的平均年龄，以及该节目的id

错误消息很清楚--

这里有一种解决这个特殊问题的方法

-- Age of each show's customers.
create or replace view show_cust_ages as 
select distinct watched.showid, cust.custid, cust.age
from watched 
inner join cust  on cust.custid = watched.custid;

-- Average age of show's customers. This queries the previous view.
create or replace view show_avg_ages as
select showid, avg(age) avg_age
from show_cust_ages
group by showid;

-- Your goal.
create or replace view show_name_avg_ages as
select t1.showid, t2.showname, t1.avg_age
from show_avg_ages t1
inner join shows t2 on t2.showid = t1.showid;

在制作过程中，我会花更多的时间思考事物的名称，而不是在这里

---

您应该知道，在MySQL中，基于视图的视图在大表上的性能可能很差。

感谢CREATE TABLE和INSERT语句。

-- Age of each show's customers.
create or replace view show_cust_ages as 
select distinct watched.showid, cust.custid, cust.age
from watched 
inner join cust  on cust.custid = watched.custid;

-- Average age of show's customers. This queries the previous view.
create or replace view show_avg_ages as
select showid, avg(age) avg_age
from show_cust_ages
group by showid;

-- Your goal.
create or replace view show_name_avg_ages as
select t1.showid, t2.showname, t1.avg_age
from show_avg_ages t1
inner join shows t2 on t2.showid = t1.showid;