Re: [xml] xml find and replace

[Piotr Sipika <piotreks optonline net>]

On 07/12/2012 09:53 AM, stuart shepherd wrote:
> Searching the web I've seen some examples in XSLT on how to
> do something like this, but I have never used XSLT. Does anyone know if
> there is a way to do this in XML.

XSLT is your best bet.

Here's a sample stylesheet which will:
 - change the name of all elements named 'old' to 'new'
 - change the name of all elements with an attribute named 'old' to 'new'

---- BEGIN stylesheet ----

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

  <!-- Template matches every node and attribute-->
  <xsl:template match="node()|@*">

    <xsl:choose>
      <!-- if name of element is 'old' or element contains an attribute
           named 'old', replace it with 'new' -->
      <xsl:when test="@old or name(.)='old'">
        <xsl:variable name="old_node" select="node()"/>
        <xsl:variable name="old_value" select="./text()"/>
        <xsl:variable name="attributes" select="@*"/>
        <xsl:variable name="children" select="./*"/>

        <!-- create a 'new' element with attributes and children from
             the 'old' node, with its old value -->
        <xsl:element name="new">
          <xsl:copy-of select="$attributes|$children"/>
          <xsl:value-of select="$old_value"/>
        </xsl:element>
      </xsl:when>

      <!-- else, just copy the node/attributes -->
      <xsl:otherwise>
        <xsl:copy>
          <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
      </xsl:otherwise>
    </xsl:choose>
  </xsl:template>

</xsl:stylesheet>

---- END stylesheet ----

Sample XML:
<?xml version="1.0" standalone="yes"?>
<swap>
  <old attr="attr1">This used to be inside the old node
    <old_child1>This is old_child1</old_child1>
  </old>
  <foo attr="attr2">This is just a middle element...</foo>
  <foo old="old attribute"/>
</swap>

Output of stylesheet processing using xsltproc (libxml 20708, libxslt
10126 and libexslt 815):

$ xsltproc swap.xsl swap.xml
<?xml version="1.0"?>
<swap>
  <new attr="attr1"><old_child1>This is old_child1</old_child1>This used
to be inside the old node
  </new>
  <foo attr="attr2">This is just a middle element...</foo>
  <new old="old attribute"/>
</swap>


You can modify the attribute match criteria (@old) to actually evaluate
its contents ([ old='old_parent']), etc...

Hope this helps.

Piotr