获取SQL Server 2012中xml元素的位置

Question

如何使用SQL Server 2012获取行的行号？

这是xml

<Rows>
   <Row>Coating</Row>
   <Row>Drying</Row>
   <Row>Waxing</Row>
</Rows>

我需要像这样返回的数据

RowLabel  RowNumber
-------------------
Coating    1
Drying     2
Waxing     3

Answer 1

您可以使用一些有关SQL Server如何实现XML碎化的内部知识,并使用row_number()这样的方法.

declare @XML xml = 
'<Rows>
   <Row>Coating</Row>
   <Row>Drying</Row>
   <Row>Waxing</Row>
</Rows>'

select T.X.value('text()[1]', 'nvarchar(100)') as RowLabel,
       row_number() over(order by T.X) as RowNumber
from @XML.nodes('/Rows/Row') as T(X)

参考:使用DENSE_RANK唯一识别XML节点

或者你可以"安全地玩"并使用数字表.

select T.X.value('text()[1]', 'nvarchar(100)') as RowLabel,
       N.Number as RowNumber
from Numbers as N
  cross apply @XML.nodes('/Rows/Row[sql:column("N.Number")]') as T(X)
where N.Number between 1 and @XML.value('count(/Rows/Row)', 'int')