Question

我正在寻找一种方法，在开头添加此节点“LoanSecondaryStatusDates”，并将相应的结束标记“LoanSecondaryStatusDates”添加到结尾。我在SQL服务器中使用“FOR XML”生成了以下内容，但无法弄清楚如何添加开始和结束标记。如果使用“FOR XML”可以做到这一点，那么一个例子会很棒，不过C＃就没问题了。谢谢！

目前：

<Loans>
    <Loan>
      <GUID></GUID>
      <AgentCompanyName></AgentCompanyName>
        <LoanSecondaryStatus>
            <StatusName>Name</StatusName>
            <StatusDate>Date</StatusDate>
        </LoanSecondaryStatus>
        <LoanSecondaryStatus>
            <StatusName>Name</StatusName>
            <StatusDate>Date</StatusDate>
        </LoanSecondaryStatus>
    </Loan>
</Loans>

最终结果应为：

 <Loans>
   <Loan>
   <GUID></GUID>
   <AgentCompanyName></AgentCompanyName>
    <LoanSecondaryStatusDates>
        <LoanSecondaryStatus>
            <StatusName>Name</StatusName>
            <StatusDate>Date</StatusDate>
    </LoanSecondaryStatus>
        <LoanSecondaryStatus>
            <StatusName>Name</StatusName>
            <StatusDate>Date</StatusDate>
        </LoanSecondaryStatus>
    </LoanSecondaryStatusDates>
</Loan>
</Loans>

FOR XML查询

SELECT 
   [GUID] 
  ,[AgentCompanyName],
   (
   SELECT 
    'Borrower Docs Sent/Req' as 'StatusName',
    CASE WHEN t.BorrowerDocsSent IS NOT NULL THEN t.BorrowerDocsSent ELSE '' END as 'StatusDate'

FROM Encompass_loanData as t
WHERE t.[GUID] = E.[GUID]
FOR XML PATH('LoanSecondaryStatus'), TYPE
  ),
  (
   SELECT 
    t.BorrowerCity as 'StatusName',
    t.[GUID] as 'StatusDate'

FROM Encompass_loanData as t
WHERE t.[GUID] = E.[GUID]
FOR XML PATH('LoanSecondaryStatus'), TYPE
  )

From Encompass_loanData E
FOR XML PATH ('Loan'), type, root('Loans')

SQL小提琴：http://sqlfiddle.com/#!6/d672a/2/0

Answer 1

您发布的xml没有有效的root。所以我添加了<xml>作为根。

这也可以在C＃中轻松完成。但这是一种使用xpath的方法。

  <xml>
  <LoanSecondaryStatus>
    <StatusName>Name</StatusName>
    <StatusDate>Date</StatusDate>
  </LoanSecondaryStatus>
  <LoanSecondaryStatus>
    <StatusName>Name</StatusName>
    <StatusDate>Date</StatusDate>
  </LoanSecondaryStatus>
  </xml>

这是XSL。

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="/">
    <LoanSecondaryStatusDates>     
      <xsl:for-each select="xml/LoanSecondaryStatus">
        <LoanSecondaryStatus>
          <StatusName>
            <xsl:value-of select="StatusName"/>
          </StatusName>
          <StatusDate>
            <xsl:value-of select="StatusDate"/>
          </StatusDate>
        </LoanSecondaryStatus>
      </xsl:for-each>      
    </LoanSecondaryStatusDates>
  </xsl:template>
</xsl:stylesheet>

输出：

<?xml version="1.0" encoding="utf-8"?>
<LoanSecondaryStatusDates>
  <LoanSecondaryStatus>
    <StatusName>Name</StatusName>
    <StatusDate>Date</StatusDate>
  </LoanSecondaryStatus>
  <LoanSecondaryStatus>
    <StatusName>Name</StatusName>
    <StatusDate>Date</StatusDate>
  </LoanSecondaryStatus>
</LoanSecondaryStatusDates>

更新了SQL查询

SELECT 
   [GUID] 
  ,[AgentCompanyName],
   (
   SELECT NULL,   
            (
                SELECT 
                        'Borrower Docs Sent/Req' as 'StatusName',
                        CASE WHEN t.BorrowerDocsSent IS NOT NULL THEN t.BorrowerDocsSent ELSE '' END as 'StatusDate'
                FROM Encompass_loanData as t
                WHERE t.[GUID] = E.[GUID]
                FOR XML PATH('LoanSecondaryStatus'), TYPE
            ),
            (   
                SELECT NULL AS X
                FOR XML PATH('LoanSecondaryStatusDates'), TYPE
            ),
            (
                SELECT 
                        t.BorrowerCity as 'StatusName',
                        t.[GUID] as 'StatusDate'
                FROM Encompass_loanData as t
                WHERE t.[GUID] = E.[GUID]
                FOR XML PATH('LoanSecondaryStatus'), TYPE
            ),
            NULL
    FOR XML PATH('LoanSecondaryStatusDates'),TYPE
    )
    FROM Encompass_loanData E
    WHERE [LASTMODIFIED] >= '20160121'
    FOR XML PATH ('Loan'), type, root('Loans')

Answer 2

看看XElement类，你可以轻松地修改xml树。

https://msdn.microsoft.com/en-us/library/system.xml.linq.xelement(v=vs.110).aspx

http://www.dotnetperls.com/xelement

Answer 3

我能够用这个XSL包装子元素

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output indent="yes"/>
    <xsl:strip-space elements="*"/>

    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="Loan">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()[not(self::LoanSecondaryStatus)]"/>
            <LoanSecondaryStatusDates>
                <xsl:apply-templates select="LoanSecondaryStatus"/>
            </LoanSecondaryStatusDates>
        </xsl:copy>
    </xsl:template>

</xsl:stylesheet>

如何在节点集

3 个答案: