小编Ubu*_*ead的帖子

我在这里查了很多答案,我想我差不多了.有一件事困扰着我(由于某些原因,我的同伴需要它)如下:

我有以下输入XML:

<?xml version="1.0" encoding="utf-8"?>
<MyRoot>
  <MyRequest CompletionCode="0" CustomerID="9999999999"/>
  <List TotalList="1">
    <Order CustomerID="999999999" OrderNo="0000000001" Status="Shipped">
      <BillToAddress ZipCode="22221"/>
      <ShipToAddress ZipCode="22222"/>
      <Totals Tax="0.50" SubTotal="10.00" Shipping="4.95"/>
    </Order>
  </List>
  <Errors/>
</MyRoot>

我被要求制作这个:

<ns:MyNewRoot xmlns:ns="http://schemas.foo.com/response"  
xmlns:N1="http://schemas.foo.com/request"  
xmlns:N2="http://schemas.foo.com/details">
    <N1:MyRequest CompletionCode="0" CustomerID="9999999999"/>
    <ns:List TotalList="1">
            <N2:Order CustomerID="999999999" Level="Preferred" Status="Shipped">
                    <N2:BillToAddress ZipCode="22221"/>
                    <N2:ShipToAddress ZipCode="22222"/>
                    <N2:Totals Tax="0.50" SubTotal="10.00" Shipping="4.95"/>
            </N2:Order>
    </ns:List>
    <ns:Errors/>
</ns:MyNewRoot>

注意N2:Order的子节点还需要N2:前缀以及其余元素的ns:前缀.

我使用下面的XSL转换:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

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


<xsl:template match="/MyRoot">
 <MyNewRoot xmlns="http://schemas.foo.com/response"
   xmlns:N1="http://schemas.foo.com/request"
   xmlns:N2="http://schemas.foo.com/details">
     <xsl:apply-templates/>
 </MyNewRoot> …