具有多对多表的数据透视表答案

【问题标题】：Pivot Table with many to many table具有多对多表的数据透视表
【发布时间】：2013-09-10 18:00:16
【问题描述】：

我的 SQL Fiddle 在这里：http://sqlfiddle.com/#!3/d5c60

CREATE TABLE customer 
    (
     id int identity primary key, 
     name varchar(20), 
    );

CREATE TABLE warehouse 
    (
     id int identity primary key, 
     name varchar(20), 
    );

CREATE TABLE customerwarehouse 
    (
     id int identity primary key, 
     customerid int,
      warehouseid int
    );

INSERT INTO customer (name) 
VALUES
('CustA'),
('CustB'),
('CustC');

INSERT INTO warehouse (name) 
VALUES
('wh01'),
('wh02'),
('wh03');

INSERT INTO customerwarehouse (customerid, warehouseid)
VALUES
(1,1),
(2,1),
(2,2),
(3,1),
(3,2),
(3,3);

我想编写一个查询，以以下格式返回客户/仓库数据：

Customer    WH1    WH2    WH3
CustA       wh01    
CustB       wh01   wh02
CustC       wh01   wh02   wh03

我这样做的尝试为所有仓库返回 null。

如何构造查询以返回所需格式的数据？

【问题讨论】：

标签： sql sql-server pivot

【解决方案1】：

为了获得结果，您需要连接表并应用 PIVOT 函数。我还建议使用 row_number() 窗口函数来获取每个客户的仓库数量 - 这将是用作新列标题的值。

select customername, wh1, wh2, wh3
from
(
  select w.name warehousename,
    c.name customername,
    'wh'+cast(row_number() over(partition by c.id
                                order by w.id) as varchar(10)) seq
  from customer c
  inner join customerwarehouse cw
    on c.id = cw.customerid
  inner join warehouse w
    on cw.warehouseid = w.id
) d
pivot
(
  max(warehousename)
  for seq in (wh1, wh2, wh3)
) piv;

见SQL Fiddle with Demo。如果您有未知数量的值，那么您将需要使用动态 SQL 来获取结果：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT distinct ',' + QUOTENAME('wh'+cast(row_number() over(partition by customerid
                                                                                  order by warehouseid) as varchar(10))) 
                    from customerwarehouse
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT customername, ' + @cols + ' 
            from 
            (
                select w.name warehousename,
                  c.name customername,
                  ''wh''+cast(row_number() over(partition by c.id
                                              order by w.id) as varchar(10)) seq
                from customer c
                inner join customerwarehouse cw
                  on c.id = cw.customerid
                inner join warehouse w
                  on cw.warehouseid = w.id
            ) x
            pivot 
            (
                max(warehousename)
                for seq in (' + @cols + ')
            ) p '

execute sp_executesql @query;

见SQL Fiddle with Demo。两者都给出结果：

| CUSTOMERNAME |  WH1 |    WH2 |    WH3 |
|        CustA | wh01 | (null) | (null) |
|        CustB | wh01 |   wh02 | (null) |
|        CustC | wh01 |   wh02 |   wh03 |

【讨论】：

结果正好相反（CustC 与所有仓库链接，CustA 仅与 WH1 链接）。无论如何，我确信这是一个小错误，可以轻松修复。 +1 动态方法；这非常有用。
@w0lf 已修复，我对连接列进行了粗指/反转。谢谢

【解决方案2】：

这是我在查看 this MSDN page 上的复杂 PIVOT 示例后得出的结论：

SELECT
  CustomerName,
  case when [wh01] is null then null else 'wh01' end,
  case when [wh02] is null then null else 'wh02' end,
  case when [wh03] is null then null else 'wh03' end
FROM (
  SELECT
    c.Name AS CustomerName,
    cw.id AS cwid,
    w.name AS WarehouseName
  FROM Customer c
  JOIN CustomerWarehouse cw
    ON c.id = cw.customerId
  JOIN Warehouse w
    ON w.id = cw.warehouseId
) AS SourceTable
pivot (
  max(cwid)
  FOR WarehouseName IN (
      [wh01], [wh02], [wh03]
    )
) AS PivotTable

在 SQLFiddle 上：http://sqlfiddle.com/#!3/d5c60/42

【讨论】：