Converting XML format using XSLT - xslt

Question

My requirement is to convert the following array
<array>
<value>755</value>
<value>5861</value>
<value>4328</value>
</array>
to this array.
<array>
<int>755</int>
<int>5861</int>
<int>4328</int>
</array>
Following is my XSLT code to do the transformation & it works. Is it the correct way because in one post I saw the use of "identity template". but I haven't used it.
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/array">
<xsl:element name="array">
<xsl:for-each select="value">
<xsl:element name="int">
<xsl:value-of select="." />
</xsl:element>
</xsl:for-each>
</xsl:element>
</xsl:template>
</xsl:stylesheet>

Answer 1

Your current method works. It is more of a "pull" style stylesheet. The "push" style uses apply-templates.
You could shorten it a bit by using element literals, which makes it a little easier to read:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/array">
<array>
<xsl:for-each select="value">
<int>
<xsl:value-of select="." />
</int>
</xsl:for-each>
</array>
</xsl:template>
</xsl:stylesheet>
A solution using the identity template and a custom template for the value element:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" indent="yes"/>
<!--identity template, which copies every attribute and
node(element, text, comment, and processing instruction)
that it matches and then applies templates to all of it's
attributes and child nodes (if there are any) -->
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<!--Specialized template that matches the value element.
Because it is more specific than the identity template above it has
a higher priority and will match when the value element is encountered.
It creates an int element and then applies templates to any attributes
and child nodes of the value element -->
<xsl:template match="value">
<int>
<xsl:apply-templates select="#*|node()"/>
</int>
</xsl:template>
</xsl:stylesheet>

Answer 2

You can do:
<xsl:template match="/array">
<array>
<xsl:for-each select="value">
<int>
<xsl:value-of select="." />
</int>
</xsl:for-each>
</array>
</xsl:template>

Answer 3

Here is a quote from the accepted answer of your link:
XSL cannot replace anything. The best you can do is to copy the parts you want to keep, then output the parts you want to change instead of the parts you don't want to keep.
That is where the identity template comes into play: it copies everything not targetted by another matching template. The upshot is, that if your base XML contains other content than just the array, then you should also include the identity template in your xslt. But if you are sure that your xml will contain no other content, then you don't need it.