Question

选择一个组，其中一个是主要，例如5 分支。因此，地点总数为6。在每个6中，搜索3工作人员job_types LIKE "%C%"。如果在其中一个6个地方，3名工作人员具有给定参数，则查询必须获得所有6个地点的结果。

澄清： 3工作人员必须在同一主干/分支机构工作。

因为项目本身非常困难，所以使用 RAW 查询获得结果会更好：

商家信息表

id     |    mainorbranch    |    name
--------------------------------------
1           Main                 Apple
2           Branch               Apple London
3           Branch               Apple Manchester
4           Main                 IBM
5           Branch               IBM London
etc ...

关系

business_branches表

b_id     |    branch_id    |    id
--------------------------------------
1             1                 1
2             2                 1
3             3                 1
4             4                 4
5             5                 4
// etc

people_details表

d_id     |    id    |    job_types
--------------------------------------
1             1          C
2             3          D
3             2          F
4             4          C
5             5          C
// etc

people_branches表

pb_id     |    branch_id    |    id
--------------------------------------
1              1                 3
2              3                 2
3              4                 4
4              2                 5
5              1                 1
// etc

我需要得到什么：

Business id    |    Name    |    Postcode
-----------------------------------------
1                   Apple        postcode
2                   Apple 232    postcode
3                   Apple 323    postcode
// etc...

帮助程序的数据库结构 http://sqlfiddle.com/#!9/206733

Simplified, minified SQL file with total of 110k+ rows

更新

通过 @KikiTheOne 回答有点工作，但它只获得了一半的结果。另一半失踪了。

Answer 1

正如聊天中所讨论的那样。这是一个解决方案：

如果你需要公司信息......就像postcode那样得到@ t1.XXXX。

我改变了

"pb_id" "branch_id" "id"
"1" "1" "3"
"2" "3" "2"
"3" "1" "4"
"4" "1" "5"
"5" "1" "1"

所以我在1个分支中得到3个人

SELECT 
    t1.id as "Business id",
    t1.name as Name,
    'postcode' as "Postcode" 
FROM SO_business as t1 inner join 
(
    SELECT * FROM SO_busness_branches as t3 
    inner join 
    (
        SELECT 
            t5.branch_id as inner_branch,
            count(t5.branch_id) as workers_in,
            max(t6.job_types) as job_types,
            max(t7.id) as mainbranch
        FROM 
            SO_people_branches as t5
                inner join SO_people_details as t6
                    on t5.id = t6.id 
                inner join SO_busness_branches as t7 
                    on t5.branch_id = t7.branch_id 
        WHERE 
            t6.job_types LIKE '%C%' 
        GROUP BY 
            t5.branch_id
    ) as t4
        on t3.id = t4.inner_branch 
    WHERE t4.workers_in >= 3
) as t2 
    on t1.id = t2.branch_id

说明：

- 。1最内层SQL计数所有分支与工人（工作人员数量初始化）和Job_type =％c％并加入分支的主要ID。

- .2第二个SQL获取该信息，并且仅选择所有具有worker＆gt; = 3

的分支

- 。3外部SQL选择所有内部INFOS，并从Inner SQL返回所有branches / main和branchID-Main。 AND将它们连接到Business表，这样你就可以显示所有信息，如那里的信息码

Answer 2

希望这可行，因为解释与问题中列的名称的关系并不容易。

使用此脚本，您可以添加任何一组main / branch来搜索

SELECT b.id, b.name, 'postcode' as postcode 
FROM business b
INNER JOIN business_branches bb ON (bb.branch_id = b.id)
WHERE bb.id IN (
  SELECT bb1.id FROM people_details pd
  INNER JOIN people_branches pb ON (pb.id = pd.id)
  INNER JOIN business_branches bb1 ON (bb1.branch_id = pb.branch_id)
  INNER JOIN business b1 ON (b1.id = bb1.branch_id)
  WHERE
  pd.job_types like '%C%' AND
  bb1.id IN (1,4) -- You can add as many group of businesses (main/branch combinations) using the main branch key
  GROUP BY pb.branch_id HAVING count(pb.branch_id) >= 3
)

计算与分支机构的业务工人

2 个答案: