经过多次搜索，我找不到如何做到这一点。
我的网络搜索得到 Pivots 和 Concats 以及 Cases 和 Subqueries 等，但没有一个能很好地解决我的问题。多行到单行的问题对我没有帮助。
问题：
一张桌子上有个人。另一个表中有这些个人的地址（有时是多个）。我需要一个查询来将每个个人的多个地址放在一行（在适当的列中）。
这是一个带有表和查询的 MySQL Fiddle：
在那个 SQL Fiddle 中，结果有 6 个独特个体的 9 条记录：
Number | Name | EyeColor | HairColor | Street | City | State | Zip | Street2 | City2 | State2| Zip2 | Street3 | ...
1 | John Smith | blue | red | 100 Pine Street | New York | NY | 10019 | | | | 0 | | ...
2 | Nancy Jones | green | red | 200 Pine Street | New York | NY | 10018 | | | | 0 | | ...
3 | Bobby Joe | blue | black | 310 Oak Street | New York | NY | 10018 | | | | 0 | | ...
7 | Little Lebowski | green | blond | 100 Apple Street | New York | NY | 10018 | | | | 0 | | ...
7 | Little Lebowski | green | blond | 200 Hickory Street | New York | NY | 10018 | | | | 0 | | ...
7 | Little Lebowski | green | blond | 1234 Pineapple Street | New York | NY | 10018 | | | | 0 | | ...
2 | Nancy Jones | green | red | 230 Golden Street | New York | NY | 10018 | | | | 0 | | ...
8 | Sarah Shepard | brown | brown | (null) | (null) | (null) | (null)| (null) | (null)| (null)| (null)| (null) | ...
这是我需要最终结果的 SQL Fiddle：
最终结果 SQL Fiddle 总共有 6 个记录，用于 6 个唯一的个人及其在个人行的列中的多个地址：
Number | Name | EyeColor | HairColor | Street | City | State | Zip | Street2 | City2 | State2| Zip2 | Street3 | City3 | State3 | Zip3 | Street4 | City4 | State4 | Zip4 | Street5 | City5 | State5 | Zip5 | Street6 | City6 | State6 | Zip6
1 | John Smith | blue | red | 100 Pine Street | New York | NY | 10019 | | | | 0 | | | | 0 | | | | 0 | | | | 0 | | | | 0
2 | Nancy Jones | green | red | 200 Pine Street | New York | NY | 10018 | 230 Golden Street | New York | NY | 10018 | | | | 0 | | | | 0 | | | | 0 | | | | 0
3 | Bobby Joe | blue | black | 310 Oak Street | New York | NY | 10018 | | | | 0 | | | | 0 | | | | 0 | | | | 0 | | | | 0
7 | Little Lebowski | green | blond | 100 Apple Street | New York | NY | 10018 | 200 Hickory Street| New York | NY | 10018 | 1234 Pineapple Street | New York | NY | 10018 | | | | 0 | | | | 0 | | | | 0
8 | Sarah Shepard | brown | brown | | | | 0 | | | | 0 | | | | 0 | | | | 0 | | | | 0 | | | | 0
9 | Joe Profigliani | brown | brown | | | | 0 | | | | 0 | | | | 0 | | | | 0 | | | | 0 | | | | 0
顺便说一下，我正在使用我得到的 MySQL 表，尽管对为解决方案构建临时表的方法持开放态度，但问题是如何组合最终结果中显示的数据，而不是关于原始表。
可能有数千条记录，尽管我不希望任何个人拥有超过 6 个地址。（如果可用的地址字段多的话，让它优雅地失败会很方便，但这不是问题的核心。）
我希望这是一些简单的事情，我只是没有提出正确的问题。
您如何决定哪种（姓名、地址）组合是“正确的”组合？
好吧，我认为这可能是我问题的前 50%。所有地址都是“正确的”，因为它们都被写入个人的行。例如，IndividualNumber 7 是 LittleLebowski，他有 3 个地址，所有三个地址都应出现在最终结果中该个人的行中。我的假设是第一个“出现”的地址作为第一个地址。（具有自动递增值的临时表？）一旦我回答了有关记录顺序的问题，那么另外 50% 的问题就是将它们写入该个人行上的相应列？
该问题可以分为 3 个主要操作：
IndividualNumber（使用变量）对行进行分区NULL您可以在此处查看示例：SQL Fiddle
IndividualNumber：此查询的行为类似于 Oracle (>= 10g)、PostgreSQL (>= 8.4) 和 SQL Server (>= 2012) 中可用的 ROW_NUMBER() 窗口函数。
MySQL 没有实现它，它必须使用变量和CASE语句来完成：
SELECT @row := CASE WHEN inf.IndividualNumber = @id
THEN @row + 1 ELSE 1 END as row
, @id := inf.IndividualNumber as IndividualNumber
, inf.IndividualAddressStreet
, inf.IndividualCity
, inf.IndividualState
, inf.IndividualZip
FROM (SELECT @row := 0, @id := 0) v
, InfoaboutThemTable as inf
ORDER BY inf.IndividualNumber
它row为 中的每个IndividualNumber（分区）返回一个从 1 到 n的唯一值InfoaboutThemTable：
row IndividualNumber IndividualAddressStreet IndividualCity IndividualState IndividualZip
1 1 100 Pine Street New York NY 10019
1 2 200 Pine Street New York NY 10018
2 2 201 Pine Street New York NY 10018
3 2 230 Golden Street New York NY 10018
4 2 456 Golden Street New York NY 10018
1 3 310 Oak Street New York NY 10018
1 7 100 Apple Street New York NY 10018
2 7 200 Hickory Street New York NY 10018
3 7 1234 Pineapple Street New York NY 10018
因为您还不知道如何订购Address1, Address2, ... 它仅用于ORDER BY inf.IndividualNumber按IndividualNumber无特定顺序进行分区。
ORDER BY inf.IndividualNumber, inf.IndividualAddressStreet如果您要分区IndividualNumber并为分区的成员提供编号，则可以替换为IndividualAddressStreet。
一旦每个分区的每一行IndividualNumber都有一个唯一的row值，它就可以用于将行转置/旋转到列。
MySQL 没有实现该PIVOT运算符。数据可以使用 a 移动和旋转到 XGROUP BY d.IndividualNumber列，对于每个转置的列， aCASE WHEN row = X THEN ... END与聚合 ( MAX) 一起使用：
MAX(CASE WHEN row = 1 THEN d.IndividualAddressStreet END) AS Street1
MAX(CASE WHEN row = 1 THEN d.IndividualCity END) AS City1
...
MAX(CASE WHEN row = 2 THEN d.IndividualAddressStreet END) AS Street2
...
它只包含 3 个组，但您可以轻松地将其扩展到 6 个或 N 个组。
最后IndividualsTable是LEFT JOIN旋转子查询，以便将名称和颜色添加到所需的输出。
NULL使用 将值替换为空字符串COALESCE。
SELECT idt.IndividualNumber
, idt.Name
, idt.IndividualEyeColor
, idt.IndividualHairColor
, COALESCE(grp.Street1, '') as Street1
, COALESCE(grp.City1, '') as City1
, COALESCE(grp.State1, '') as State1
, COALESCE(grp.Zip1, '') as Zip1
, COALESCE(grp.Street2, '') as Street2
, COALESCE(grp.City2, '') as City2
, COALESCE(grp.State2, '') as State2
, COALESCE(grp.Zip2, '') as Zip2
, COALESCE(grp.Street3, '') as Street3
, COALESCE(grp.City3, '') as City3
, COALESCE(grp.State3, '') as State3
, COALESCE(grp.Zip3, '') as Zip3
FROM IndividualsTable idt
LEFT JOIN (
SELECT d.IndividualNumber as IndividualNumber
, MAX(CASE WHEN row = 1 THEN d.IndividualAddressStreet END) AS Street1
, MAX(CASE WHEN row = 1 THEN d.IndividualCity END) AS City1
, MAX(CASE WHEN row = 1 THEN d.IndividualState END) AS State1
, MAX(CASE WHEN row = 1 THEN d.IndividualZip END) AS Zip1
, MAX(CASE WHEN row = 2 THEN d.IndividualAddressStreet END) AS Street2
, MAX(CASE WHEN row = 2 THEN d.IndividualCity END) AS City2
, MAX(CASE WHEN row = 2 THEN d.IndividualState END) AS State2
, MAX(CASE WHEN row = 2 THEN d.IndividualZip END) AS Zip2
, MAX(CASE WHEN row = 3 THEN d.IndividualAddressStreet END) AS Street3
, MAX(CASE WHEN row = 3 THEN d.IndividualCity END) AS City3
, MAX(CASE WHEN row = 3 THEN d.IndividualState END) AS State3
, MAX(CASE WHEN row = 3 THEN d.IndividualZip END) AS Zip3
FROM
(
SELECT @row := CASE WHEN inf.IndividualNumber = @id
THEN @row + 1 ELSE 1 END as row
, @id := inf.IndividualNumber as IndividualNumber
, inf.IndividualAddressStreet
, inf.IndividualCity
, inf.IndividualState
, inf.IndividualZip
FROM (SELECT @row := 0, @id := 0) v
, InfoaboutThemTable as inf
ORDER BY inf.IndividualNumber
) d
GROUP BY d.IndividualNumber
) grp
ON grp.IndividualNumber = idt.IndividualNumber
;