如何做一个子查询GROUP_CONCAT

Question

，我有以下数据模型：

`title` 

- id 
- name 

`version` 

- id 
- name 
- title_id 

`version_price` 

- id 
- version_id 
- store 
- price 

这里是数据的一个例子：

`title` 

- id=1, name=titanic 
- id=2, name=avatar 

`version` 

- id=1, name="titanic (dubbed in spanish)", title_id=1 
- id=2, name="avatar directors cut", title_id=2 
- id=3, name="avatar theatrical", title_id=2 

`version_price` 

- id=1, version_id=1, store=itunes, price=$4.99 
- id=1, version_id=1, store=google, price=$4.99 
- id=1, version_id=2, store=itunes, price=$5.99 
- id=1, version_id=3, store=itunes, price=$5.99 

我想建立一个查询，这将使我所有在iTunes上有version_price但在Google上没有的标题。我将如何做到这一点？以下是我迄今为止：

select 
    title.id, title.name, group_concat(distinct store order by store) 
from 
    version inner join title on version.title_id=title.id inner join version_price on version_price.version_id=version.id 
group by 
    title_id 

这给了我GROUP_CONCAT这说明我是什么，我有：

id name group_concat(distinct store order by store) 
1 Titanic Google,iTunes        
2 Avatar iTunes          

但我怎么构建查询到包括该项目是否是在谷歌（使用CASE语句，Or的任何需要的）

id name group_concat(distinct store order by store) on_google 
1 Titanic Google,iTunes        true 
2 Avatar iTunes          false 

这将基本上是做了group_concat LIKE '%google%'，而不是一个正常的where子句。

下面是当前查询的SQL拨弄一个链接，我有：http://sqlfiddle.com/#!9/e52b53/1/0

Answer 1

使用条件聚合来确定标题是否在指定的商店中。

select title.id, title.name, group_concat(distinct version_price.store order by store), 
if(count(case when store = 'google' then 1 end) >= 1,'true','false') as on_google 
from version 
inner join title on version.title_id=title.id 
inner join version_price on version_price.version_id=version.id 
group by title.id, title.name 

count(case when store = 'google' then 1 end) >= 1计数所有行对于给定标题分配1到在他们google行之后。 （否则他们将被分配null并且count忽略空值。）之后，if检查count，并且如果标题上至少有一个google存储，则将其分类。

Answer 2

我会做类似下面

SELECT 
    * 
FROM 
    title t 
INNER JOIN 
    version v ON 
     v.title_id = t.id 
CROSS APPLY (
    SELECT 
     * 
    FROM 
     version_price vp 
    WHERE 
     vp.store <> 'google' 
) c ON c.version_id == v.id 

语法可能只是有点过，因为我不能测试它现在，但我相信这是你想要的精神。交叉申请也是一个非常有效的加入，它总是有帮助的！

Answer 3

这会给你不是谷歌版本价格的数量，而是谷歌上的数字。 （COUNT不计算空值。）

SELECT t.id, t.name 
    , COUNT(DISTINCT vpNotG.id) > 0 AS onOtherThanGoogle 
    , COUNT(DISTINCT vpG.id) > 0 AS onGoogle 
FROM title AS t 
    INNER JOIN version AS v ON t.id=v.title_id 
    LEFT JOIN version_price AS vpNotG 
     ON v.id=vpNotG.version_id 
     AND vpNotG.store <> 'Google' 
    LEFT JOIN version_price AS vpG 
     ON v.id=vpG.version_id 
     AND vpG.store = 'Google' 
GROUP BY t.id 

或类似VKP的另一个解决方案：

SELECT t.id, t.name 
    , COUNT(DISTINCT CASE WHEN store = 'Google' THEN vp.id ELSE NULL END) AS googlePriceCount 
    , COUNT(DISTINCT CASE WHEN store = 'iTunes' THEN vp.id ELSE NULL END) AS iTunesPriceCount 
    , COUNT(DISTINCT CASE WHEN store <> 'Google' THEN vp.id ELSE NULL END) AS nonGooglePriceCount 
FROM title AS t 
    INNER JOIN version AS v ON t.id = v.title_id 
    INNER JOIN version_price AS vp ON v.id = vp.version_id 
GROUP BY t.id 

注：ELSE NULL可以省略，因为如果没有提供ELSE它被暗示;但为了清楚起见，我包括在内

Answer 4

这可能是最没有效率的上述答案的，但下面的子查询会的工作，使用%like%条件：

select *, case when stores like '%google%' then 1 else 0 end on_google 
from (select title.id, title.name, group_concat(distinct store order by store) stores 
from version inner join title on version.title_id=title.id inner join version_price 
on version_price.version_id=version.id group by title_id) x 

id name stores on_google 
1 Titanic Google,iTunes 1 
2 Avatar iTunes 0 

Answer 5

http://sqlfiddle.com/#!9/b8706/2

你可以：

SELECT 
    title.id, 
    title.name, 
    group_concat(distinct version_price.store), 
    MAX(IF(version_price.store='google',1,0)) on_google 
FROM version 
INNER JOIN title 
ON version.title_id=title.id 
INNER JOIN version_price 
ON version_price.version_id=version.id 
GROUP BY title_id; 

并在需要过滤的记录中添加HAVING到查询：

SELECT 
    title.id, 
    title.name, 
    group_concat(distinct version_price.store), 
    MAX(IF(version_price.store='google',1,0)) on_google 
FROM version 
INNER JOIN title 
ON version.title_id=title.id 
INNER JOIN version_price 
ON version_price.version_id=version.id 
GROUP BY title_id 
HAVING on_google;