Question

我有以下数据模型：

`title`

- id
- name

`version`

- id
- name
- title_id

`version_price`

- id
- version_id
- store
- price

这是一个数据示例：

`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

我想构建一个查询，它将为我提供所有在iTunes上具有version_price而不在Google上的标题。我该怎么做？以下是我到目前为止的情况：

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

这给了我一个group_concat，它向我展示了我拥有的东西：

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

但是我如何构建一个查询来包含该项目是否在Google上（使用案例陈述或其他所需的内容）

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

它基本上是在做group_concat LIKE '%google%'而不是普通的where子句。

这是我当前查询的SQL小提琴的链接：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

在将count(case when store = 'google' then 1 end) >= 1分配给其中包含1的行后，

google会计算给定标题的所有行。（否则他们将被分配null而count将忽略空值。）此后，if检查count并对标题进行分类（如果标题至少为google {{1}} 1}}存储在它上面。

Answer 2

这将为您提供不在谷歌上的版本价格数量以及谷歌上的数量。（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

或类似于vkp的其他解决方案：

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

注意：ELSE NULL可以省略，因为如果没有提供ELSE，则暗示;但为了清楚起见，我加入了。

Answer 3

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;

如果需要过滤记录，请在查询中添加HAVING：

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;

Answer 4

我会像下面这样做

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

语法可能只是一点点，因为我现在无法测试它，但我相信这是你想要的精神。交叉申请也是一种非常有效的联接，总是有用的！

Answer 5

这可能是上述答案中效率最低的，但以下子查询可以使用%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

如何在group_concat中执行子查询

5 个答案: