Question

我在SQL Server 2005中有一个XML列：

<treatmentDetails>
      <treatmentFactor>
        <treatmentFactorName>Varieties</treatmentFactorName>
        <plot>Whole</plot>
        <levels>
          <level>Var1</level>
          <level>Var2</level>
        </levels>
      </treatmentFactor>
      <treatmentFactor>
        <treatmentFactorName>Nitrogen</treatmentFactorName>
        <plot>Whole</plot>
        <levels>
          <level>N1</level>
          <level>N2</level>
          <level>N3</level> 
        </levels>
      </treatmentFactor>
</treatmentDetails>

由此，我想创建这样的输出：

Varieties Nitrogen    
Var1      N1         
var1      N2
var1      N3
var2      N1
var2      N2
var3      N3

我尝试使用xmlfield.query或.value，但我能够提取单个值（例如Var1）或XML标记，<level>Var1</level> ...任何帮助都会很棒帮助

Answer 1

试试这个：

DECLARE @Test TABLE (ID INT NOT NULL, XmlContent XML)

INSERT INTO @TEST VALUES(1, '<treatmentDetails>
      <treatmentFactor>
        <treatmentFactorName>Varieties</treatmentFactorName>
        <plot>Whole</plot>
        <levels>
          <level>Var1</level>
          <level>Var2</level>
        </levels>
      </treatmentFactor>
      <treatmentFactor>
        <treatmentFactorName>Nitrogen</treatmentFactorName>
        <plot>Whole</plot>
        <levels>
          <level>N1</level>
          <level>N2</level>
          <level>N3</level> 
        </levels>
      </treatmentFactor>
</treatmentDetails>')

SELECT
    Varieties = TDLevel.value('(.)[1]', 'varchar(20)'),
    Nitrogen = TDLevel2.value('(.)[1]', 'varchar(20)')
FROM
    @test
CROSS APPLY
    XmlContent.nodes('/treatmentDetails/treatmentFactor[1]/levels/level') AS Tbl(TDLevel)
CROSS APPLY
    XmlContent.nodes('/treatmentDetails/treatmentFactor[2]/levels/level') AS Tbl2(TDLevel2)

这会给我一个像你正在寻找的输出。

但是：我不清楚你是否总是在<treatmentFactor>下只有两个<treatmentDetails>个节点 - 我只是选择了第一个和第二个子节点 - 这可能需要根据需要进行调整关于你的数据。

Answer 2

试试这个

select
    TV.V.value('(.)[1]', 'varchar(20)'),
    TN.N.value('(.)[1]', 'varchar(20)')
from table1 as A
    outer apply Data.nodes('/treatmentDetails/treatmentFactor[treatmentFactorName = "Varieties"]/levels/level') as TV(V)
    outer apply Data.nodes('/treatmentDetails/treatmentFactor[treatmentFactorName = "Nitrogen"]/levels/level') as TN(N)

<强> sql fiddle demo

如何从具有元素序列的XML列中提取

2 个答案: