Question

我在查询此问题时遇到了困难。

假设我们有3个表格，如下所示：

company
    id int IDENTITY,
    name nvarchar(255)

location
    id int IDENTITY,
    name nvarchar(255),
    company int

address
    id int IDENTITY,
    name nvarchar(255),
    location int

请忽略此用例的键和索引。

我想返回一个公司列表，其所有子位置都有一个指定的地址，该公司所有位置的地址都相同。

示例工作数据：

select c.name, l.name, a.name
from company as c
join location as l on l.company = c.id
left join address as a on a.location = l.id



company     location          address
----------  ----------------  -----------------
3-DOM       Ankh-Morpork      Street 1
Adrenalize  Cabot Cove        Avenue 2
Adrenalize  Mayberry          Road 49
InterBlurb  Hogsmeade         NULL
InterBlurb  Kakariko Village  Alleyway 13
Octocore    Nibelheim         Blueberry Creek 2
Octocore    Mos Eisley        Blueberry Creek 2

示例结果：

company   
----------
3-DOM     
Octocore

我将如何实现这一结果？我已尝试按l.company分组并选择count(distinct a.name)和其他内容，但我似乎无法掌握它。我非常感谢您对查询的解释，我需要理解这一点。

Answer 1

您可以尝试按公司名称进行分组，并仅保留所有位置地址相同的公司（这是内部查询的作用）：

WITH the_cte (companyName, locationName, address)
AS
(
    SELECT c.name, l.name, a.name
    FROM company AS c
    INNER JOIN location AS l
        ON l.company = c.id
    LEFT JOIN address AS a
        ON a.location = l.id
)

SELECT companyName, locationName, address
FROM the_cte
INNER JOIN
(
    SELECT companyName, COUNT(DISTINCT CASE WHEN address IS NULL THEN 0 ELSE 1 END)
    FROM the_cte
    GROUP BY companyName
    HAVING COUNT(DISTINCT CASE WHEN address IS NULL THEN 0 ELSE 1 END) = 1
) t
    ON the_cte.companyName = t.companyName

Answer 2

SELECT Name 
FROM (
                SELECT        company.name
                FROM            company INNER JOIN
                                         location ON company.id = location.company LEFT OUTER JOIN
                                         address ON location.id = address.location
                GROUP BY company.name
                )sub

WHERE Name NOT IN (
                SELECT        company.name
                FROM            company INNER JOIN
                                         location ON company.id = location.company LEFT OUTER JOIN
                                         address ON location.id = address.location
                WHERE        (address.location IS NULL)
                GROUP BY company.name

                UNION

                SELECT        company.name
                FROM            location INNER JOIN
                                         address ON location.id = address.location INNER JOIN
                                         company ON location.company = company.id
                GROUP BY company.name
                HAVING        (COUNT(DISTINCT address.name) >1 )

                )

使用外部选择查询，我们结合了3个sql查询
第一个查询获取所有公司名称
第二个查询获取没有地址的公司
第三个查询获得具有多个不同位置的公司

找到所有孩子都具有相同价值的父母

2 个答案: