如何合并（覆盖）两个xml文档？

Question

Suppose I have an A document like this:

<document>
    <element>
        <value>1</value>
        <wom>bat</wom>
    </element>
    <bar>
        <baz />
        <baz />
        <baz />
    </bar>
</document>

and a B document like this:

<document>
    <element>
        <value>2</value>
    </element>
    <bar>

    </bar>
</document>

With the result which looks like this:

<document>
    <element>
        <value>2</value>
        <wom>bat</wom>
    </element>
    <bar>

    </bar>
</document>

So what I'd like to achieve is to overwrite values in a tag (like in element ) in document A with values provided from document B but leave the sibling values untouched. If the tag in B however is empty (leaf) I want its counterpart in A to be emptied as well. I've checked this question but it is merging not overwriting. How can I solve this problem?

Clarification: A and B documents have the same structure but B has less elements. I have to empty every element in A which is empty in B and I have to overwrite every inner element in an element if it is not empty (see my example).

Answer 1

One approach could be to navigate over DocumentA, but passing in a parameter set to the equivalent node in Document B.

To start with, match the document node of A, and start the matching off with the document node from B

   <xsl:template match="/">
      <xsl:apply-templates>
         <xsl:with-param name="parentB" select="document('DocB.xml')"/>
      </xsl:apply-templates>
   </xsl:template>

Then, you would have a template matching any element (in A) with the current (parent) node in B as the parameter

   <xsl:template match="*">
      <xsl:param name="parentB"/>

To find the equivalent 'child' node in B, you would first find the current position of the A node (should there be more than one child of the same name), and then check if such a child exists under the parent B node

<xsl:variable name="posA">
   <xsl:number  />
</xsl:variable>
<xsl:variable name="nodeB" select="$parentB/*[local-name() = local-name(current())][number($posA)]"/>

Then, it is just a case of determining whether to copy the A or B node. To copy the B node, the B node would have to exist, and not have any child elements (it might have child text nodes though, which would be copied

<xsl:when test="$nodeB and not($nodeB/*)">
   <xsl:copy-of select="$nodeB/node()"/>
</xsl:when>

Otherwise, continue processing the A node (passing in the current B node as a parameter).

Try this XSLT

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

   <xsl:template match="/">
      <xsl:apply-templates>
         <xsl:with-param name="parentB" select="document('DocB.xml')"/>
      </xsl:apply-templates>
   </xsl:template>

   <xsl:template match="*">
      <xsl:param name="parentB"/>
      <xsl:variable name="posA">
          <xsl:number  />
       </xsl:variable>
      <xsl:variable name="nodeB" select="$parentB/*[local-name() = local-name(current())][number($posA)]"/>
      <xsl:copy>
         <xsl:choose>
            <xsl:when test="$nodeB and not($nodeB/*)">
               <xsl:copy-of select="$nodeB/node()"/>
            </xsl:when>
            <xsl:otherwise>
               <xsl:apply-templates select="@*|node()">
                  <xsl:with-param name="parentB" select="$nodeB"/>
               </xsl:apply-templates>
            </xsl:otherwise>
         </xsl:choose>
      </xsl:copy>
   </xsl:template>

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