如何用两个相关的“许多”表展平一个表的结果？

我已经对数据库中的某些表进行了重组，以使其更加灵活，但是我不确定如何编写SQL从中提取有意义的数据。

我有以下表格（为简洁起见，略有缩写）：

CREATE TABLE Loans(
    Id int,
    SchemaId int,
    LoanNumber nvarchar(100)
);

CREATE TABLE SchemaFields(
    Id int,
    SchemaId int,
    FieldName nvarchar(255)
);

CREATE TABLE LoanFields(
    Id int,
    LoanId int,
    SchemaFieldId int,
    FieldValue nvarchar(4000)
);

带有以下数据：

INSERT INTO Loans (Id, SchemaId, LoanNumber) VALUES (1, 1, 'ABC123');

INSERT INTO SchemaFields (Id, SchemaId, FieldName) VALUES (1, 1, 'First Name');
INSERT INTO SchemaFields (Id, SchemaId, FieldName) VALUES (2, 1, 'Last Name');

INSERT INTO LoanFields (Id, LoanId, SchemaFieldId, FieldValue) VALUES (1, 1, 1, 'John');
INSERT INTO LoanFields (Id, LoanId, SchemaFieldId, FieldValue) VALUES (2, 1, 2, 'Doe');

目的是获得一个查询，该查询对于所有字段均适用于贷款。（在现实世界中，同一模式可能会有20-30个字段，但示例中只有2个）：

LoanNumber   First Name    Last Name
----------   -----------   ----------
ABC123       John          Doe

我无法使用引用“名字”和“姓氏”的数据透视表，因为我不知道实际上会出现什么。

我这里有一个已经准备好架构的SQL Fiddle。

如何获得理想的结果？

可以使用PIVOT函数完成此操作，但是由于听起来您想基于schemaId更改查询，因此您将需要使用动态SQL。

如果您具有已知数量的值或知道特定schemaID的列，则可以对查询进行硬编码。静态查询为：

select loannumber,
  [First Name], 
  [Middle Name], 
  [Last Name]
from
(
  select 
    l.loannumber,
    sf.fieldname,
    lf.fieldvalue
  from loans l
  left join loanfields lf
    on l.id = lf.loanid
  left join schemafields sf
    on lf.schemafieldid = sf.id
    and l.schemaid = sf.schemaid
) src
pivot
(
  max(fieldvalue)
  for fieldname in ([First Name], [Middle Name], [Last Name])
)piv;

请参阅带有演示的SQL Fiddle。

如果您有一个未知数，或者您希望根据SchemaId要传递给过程的来更改列，则将使用动态SQL生成SQL字符串：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX),
    @schemaId int = 1

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(FieldName) 
                    from SchemaFields 
                    where schemaid = @schemaid
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT loannumber,' + @cols + ' 
            from 
            (
              select 
                l.loannumber,
                sf.fieldname,
                lf.fieldvalue
              from loans l
              left join loanfields lf
                on l.id = lf.loanid
              left join schemafields sf
                on lf.schemafieldid = sf.id
                and l.schemaid = sf.schemaid
              where sf.schemaid = '+cast(@schemaid as varchar(10))+'
            ) x
            pivot 
            (
                max(fieldvalue)
                for fieldname in (' + @cols + ')
            ) p '

execute(@query);

请参阅带有演示的SQL Fiddle。这两个查询都将生成结果：

| LOANNUMBER | FIRST NAME | LAST NAME | MIDDLE NAME |
-----------------------------------------------------
|     ABC123 |       John |       Doe |      (null) |
|     XYZ789 |    Charles |     Smith |         Lee |

您可以使用此模式。要获取更多架构字段，只需在第二行到最后一行扩展所有架构字段名称即可。

select *
  from (
    select l.LoanNumber, s.FieldName, f.FieldValue
      from Loans l
      join Schemafields s on s.SchemaId = l.SchemaId
      join LoanFields f on f.LoanId = l.Id and f.SchemaFieldId = s.Id) p
pivot (
    max(FieldValue) for FieldName in ([First Name], [Last Name])
    ) v;

如果SchemaId = 1代表Loan类型的字段，那么您还可以动态生成schema field names使用如下所示内容的完整列表。

declare @sql nvarchar(max) =
    stuff((select ',' + quotename(FieldName)
      from SchemaFields
     where SchemaId = 1
       for xml path(''), type).value('/','nvarchar(max)'),1,1,'');
select @sql = '
select *
  from (
    select l.LoanNumber, s.FieldName, f.FieldValue
      from Loans l
      join Schemafields s on s.SchemaId = l.SchemaId
      join LoanFields f on f.LoanId = l.Id and f.SchemaFieldId = s.Id) p
pivot (
    max(FieldValue) for FieldName in (' + @sql + ')
    ) v';
exec (@sql);