XML Parsing - SQL Server

血红的双手。 提交于 2019-12-04 20:31:37

There is 1:n data at <Margin> and again 1:n data at <MarginRevenue>. You need to use .nodes() twice via APPLY.

declare @xml xml = '<Margins>
    <Margin type="type1" currencyCode="currencyCode1">
      <MarginRevenue>1.1</MarginRevenue>
      <MarginRevenue>1.2</MarginRevenue>
      <MarginRevenue>1.3</MarginRevenue>
    </Margin>
    <Margin type="type2" currencyCode="currencyCode2">
      <MarginRevenue>1.4</MarginRevenue>
      <MarginRevenue>1.5</MarginRevenue>
      <MarginRevenue>1.6</MarginRevenue>
    </Margin>
    <Margin type="type3" currencyCode="currencyCode3">
      <MarginRevenue>1.7</MarginRevenue>
      <MarginRevenue>1.8</MarginRevenue>
      <MarginRevenue>1.9</MarginRevenue>
      </Margin>
    </Margins>'

SELECT
 [Margin_Type]         = Marg.value('@type', 'varchar(100)') 
,[Margin_currencyCode] = Marg.value('@currencyCode', 'varchar(100)')
,[Revenue_Value]       = Rev.value('text()[1]','decimal(15,5)') 
FROM
@xml.nodes('Margins/Margin') AS A(Marg)
OUTER APPLY Marg.nodes('MarginRevenue') B(Rev);

The result

Type    currencyCode    Revenue_Value
-------------------------------------
type1   currencyCode1   1.10000
type1   currencyCode1   1.20000
type1   currencyCode1   1.30000
type2   currencyCode2   1.40000
type2   currencyCode2   1.50000
type2   currencyCode2   1.60000
type3   currencyCode3   1.70000
type3   currencyCode3   1.80000
type3   currencyCode3   1.90000

Actually, required result is not clear for me so please check below query for your required output and let me know if any changes required.

SELECT
[Margin_Revenue] = N.value('.', 'decimal(15,5)')
FROM
@xml.nodes('Margins/Margin/MarginRevenue') AS a(N)
SELECT
[Margin_Revenue] = N.value('(text())[1]', 'decimal(15,5)')
FROM
@xml.nodes('Margins/Margin/MarginRevenue') AS X(N)
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