我需要查询以下xml:
declare @xml xml = '<root>
<level1>
<property1>Value1</property1>
<property2>Value2</property2>
<level2List>
<level2>Child1</level2>
<level2>Child2</level2>
</level2List>
</level1>
<level1>
<property1>Value3</property1>
<property2>Value4</property2>
<level2List>
<level2>Child3</level2>
<level2>Child4</level2>
</level2List>
</level1>
</root>'
我需要以下结果:
Property1 Property2 Child
Value1 Value2 Child1
Value1 Value2 Child2
Value3 Value4 Child3
Value3 Value4 Child4
我有这个查询:
select col.value('Property1','varchar(100)') Property1,
col.value('Property2','varchar(100)') Property2
from @xml.nodes('//root/level1') as tab(col)
但是我不知道如何将父节点与其子节点合并。有想法吗?
答案 0 :(得分:2)
我将使用以下内容:
SELECT
X.value('(../../property1)[1]', 'varchar(20)') Property1,
X.value('(../../property2)[1]', 'varchar(20)') Property2,
X.value('.', 'varchar(20)') Child
FROM @xml.nodes('//level2') N(X)
答案 1 :(得分:2)
您可以从底部开始搜索:
DECLARE @xml AS XML = '<root>
<level1>
<property1>Value1</property1>
<property2>Value2</property2>
<level2List>
<level2>Child1</level2>
<level2>Child2</level2>
</level2List>
</level1>
<level1>
<property1>Value3</property1>
<property2>Value4</property2>
<level2List>
<level2>Child3</level2>
<level2>Child4</level2>
</level2List>
</level1>
</root>';
SELECT
n.value('../../property1[1]','varchar(100)') property1,
n.value('../../property2[1]','varchar(100)') property2,
n.value('.','varchar(100)') Child
FROM @xml.nodes('//level1//level2') AS x(n)
或通过扩展您的原始尝试:
FROM @xml.nodes('//level1') AS n1(l1)
CROSS APPLY l1.nodes('.//level2') AS n2(l2)
答案 2 :(得分:1)
您也可以使用APPLY
:
SELECT col.value('property1[1]', 'varchar(100)') AS property1,
col.value('property2[1]', 'varchar(100)') AS property2,
col1.value('text()[1]','varchar(100)') AS Child
FROM @xml.nodes('root/level1') AS tab(col) OUTER APPLY
col.nodes('level2List/level2') tab1(col1);