虽然M.Ali's answer 将为您提供结果，但由于您使用的是 SQL Server 2012，因此我将取消旋转 name 和 address 列略有不同以获得最终结果。
由于您使用的是 SQL Server 2012，因此您可以使用 CROSS APPLY 和 VALUES 将这些多列取消透视为多行。但在您这样做之前，我会使用row_number() 来获取您将拥有的新列的总数。
使用 CROSS APPLY 来“取消透视”数据的代码如下所示：
select d.loanid,
col = c.col + cast(seq as varchar(10)),
c.value
from
(
select loanid, name, address,
row_number() over(partition by loanid
order by loanid) seq
from yourtable
) d
cross apply
(
values
('name', name),
('address', address)
) c(col, value);
见SQL Fiddle with Demo。这将使您的数据格式类似于：
| LOANID | COL | VALUE |
|--------|----------|----------|
| 1 | name1 | John |
| 1 | address1 | New York |
| 1 | name2 | Carl |
| 1 | address2 | New York |
| 1 | name3 | Henry |
| 1 | address3 | Boston |
您现在拥有一列 COL，其中包含所有新列名称，并且关联的值也位于一列中。现在，新列名称的末尾有一个数字（1、2、3 等），具体取决于每个 loanid 的总条目数。现在您可以申请 PIVOT：
select loanid,
name1, address1, name2, address2,
name3, address3
from
(
select d.loanid,
col = c.col + cast(seq as varchar(10)),
c.value
from
(
select loanid, name, address,
row_number() over(partition by loanid
order by loanid) seq
from yourtable
) d
cross apply
(
values
('name', name),
('address', address)
) c(col, value)
) src
pivot
(
max(value)
for col in (name1, address1, name2, address2,
name3, address3)
) piv;
见SQL Fiddle with Demo。最后，如果您不知道您将拥有多少对 Name 和 Address，那么您可以使用动态 SQL：
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX)
select @cols = STUFF((SELECT ',' + QUOTENAME(col+cast(seq as varchar(10)))
from
(
select row_number() over(partition by loanid
order by loanid) seq
from yourtable
) d
cross apply
(
select 'Name', 1 union all
select 'Address', 2
) c (col, so)
group by seq, col, so
order by seq, so
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query = 'SELECT loanid,' + @cols + '
from
(
select d.loanid,
col = c.col + cast(seq as varchar(10)),
c.value
from
(
select loanid, name, address,
row_number() over(partition by loanid
order by loanid) seq
from yourtable
) d
cross apply
(
values
(''name'', name),
(''address'', address)
) c(col, value)
) x
pivot
(
max(value)
for col in (' + @cols + ')
) p '
exec sp_executesql @query;
见SQL Fiddle with Demo。两个版本都给出了结果：
| LOANID | NAME1 | ADDRESS1 | NAME2 | ADDRESS2 | NAME3 | ADDRESS3 |
|--------|--------|----------|--------|----------|--------|----------|
| 1 | John | New York | Carl | New York | Henry | Boston |
| 2 | Robert | Chicago | (null) | (null) | (null) | (null) |
| 3 | Joanne | LA | Chris | LA | (null) | (null) |