Adding new node to xml as a first child using xslt

Question

Hi,

I am new to xslt. Below is my Xml:

<root id=1 name=abc>

<child position=1>

...........

</child>

<child position=2>

-------

</child>

<child position=3>

-------

</child>

</root>

I would like to add a new node between <root> and <child position=1> as below using xslt.

<root id=1 name=abc>

<new node>

---------

</new node>

<child position=1>

...........

</child>

<child position=2>

-------

</child>

<child position=3>

-------

</child>

</root>

Can any body  help me?

Answer 1

Write a template for "root" that creates a shallow copy and copies the attribute node, inserts the new node and copies the child nodes e.g.

  <xsl:template match="root">

     <xsl:copy>

        <xsl:copy-of select="@*"/>

        <new_node>...</new_node>

        <xsl:copy-of select="node()"/>

    </xsl:copy>

  </xsl:template>

---

MVP Data Platform Development My blog

Answer 2

Thanks a lot..It's perfect code...

Answer 3

Hi,

In the above xml I have to remove the ab from name attribute.

the output should be

<root id=1 name=c>

<new node>

---------

</new node>

<child position=1>

...........

</child>

<child position=2>

-------

</child>

<child position=3>

-------

</child>

</root>

Can any body  help me?

Answer 4

Please, next time, for a new question start a new thread.

If you want to do more manipulations then it is best to write templates for the different tasks e.g.

<xsl:template match="@* | node()">

  <xsl:copy>

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

  </xsl:copy>

</xsl:template>

<xsl:template match="root">

  <xsl:copy>

     <xsl:apply-templates select="@*"/>

      <new_node>...</new_node>

      <xsl:apply-templates/>

  </xsl:copy>

</xsl:template>

<xsl:template match="root/@name">

  <xsl:attribute name="{name()}">

    <xsl:value-of select="translate(., 'ab', '')"/>

   </xsl:attribute>

</xsl:template>

---

MVP Data Platform Development My blog