Learning XPath with XSLT using ASP.NET 2.0 - Retrieving all values of all elements at all levels of hierarchies (or paths)
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In previous sections, we simply searched for a particular element throughout the document. Now, I would like to parse through every element (regardless of its path) within the document and display the values.
The following is the first approach for accomplishing the same (using a separate template):
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<b>
<xsl:apply-templates select ="//*"/>
</b>
</xsl:template>
<xsl:template match="*">
<xsl:value-of select ="text()"/>
<br/>
</xsl:template>
</xsl:stylesheet>
Within the above code, you can observe that I am using “//*”, which stands for “any element at any level”! The following is the second approach for accomplishing the same (using “xsl:for-each” construct).
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<b>
<xsl:for-each select="//*">
<xsl:value-of select="text()"/>
<br/>
</xsl:for-each>
</b>
</xsl:template>
</xsl:stylesheet>
In any of the above cases, when the XSLT gets executed, the transformation looks something like the following (I have added comments for understanding):
<b>
<br /> (for EmployeeDetails)
<br /> (for Department)
<br /> (for Employee)
1001<br /> (for Empno)
Jagadish<br /> (For Ename)
3400<br /> (For Sal)
<br /> (for Employee)
1002<br /> (for Empno)
Chatarji<br /> (for Ename)
2500<br /> (for Sal)
<br /> (for Employee)
.
.
.
</b>
In the upcoming articles, we shall look further into XPath expressions. I hope you enjoyed the article and any comments, suggestions, feedback, bugs, errors, enhancements etc. are highly appreciated at http://jagchat.spaces.live.com
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