querystring using xslt

Question

is it possible to access a query string using xslt?

i have a url e.g

www.example.com/page.aspx?k=aa&lang=en

I want to do something like

if lang = en

<div>displaly stuff</div>

else

 <div>display other stuff</div>

can you show me how to do this using xslt?

Answer 1

Not in native XSLT no but you can write extension objects that handle complex functions that are outside of the scope of native XSLT so for example in your XSLT you can insert a namespace into the stylesheet such as

xmlns:ex="my:Qs"

and then call

<xsl:variable name="qs" select="my:Qs('parameterName')"/>

Answer 2

No, because XSLT is not compiled code that runs on its own. It is a technology for transforming XML data and it is invoked by some XSLT processor.

So, it is up to you to provide the necessary variable data as parameters to your XSLT processor.

Answer 3

is it possible to access a query string using xslt?

Yes, if the query string is passed as a parameter.

The code below shows that no extension function is required to access a query-string. It can be passed as a (global) parameter. This is to be preferred as it reduces the need for extensions and results in cleaner and more readable code.

Then one can perform tokenization (with the tokenize() function in XSLT 2.0 or in XSLT 1.0 using the str-split-to-words template of FXSL 1.x or a self-written recursive tokenization template.)

XSLT 1.0 solution:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:ext="http://exslt.org/common"
>

   <xsl:import href="strSplit-to-Words.xsl"/>

   <xsl:output indent="yes" omit-xml-declaration="yes"/>
     <xsl:param name="pQString" select=
     "'?login=userId&amp;tag=XSLT&amp;lang=en&amp;level=expert'"
     />


    <xsl:template match="/">
    <xsl:variable name="vwordNodes">
      <xsl:call-template name="str-split-to-words">
        <xsl:with-param name="pStr" select="$pQString"/>
        <xsl:with-param name="pDelimiters" 
                  select="'?&amp;'"/>
      </xsl:call-template>
    </xsl:variable>

    <xsl:variable name="vLang" select=
      "substring-after(ext:node-set($vwordNodes)/*
                             [starts-with(.,'lang=')]
                               [last()],
                       'lang='
                      )
      "/>

      <xsl:value-of select="concat('lang = ', $vLang)"/>
    </xsl:template>
</xsl:stylesheet>

when the above transformation is applied on any XML document (will not be used), the wanted result is produced:

lang = en

Do note the use of the FXSL 1.x str-split-to-words template and the use of the EXSLT ext:node-set() extension function.

XSLT 2.0 solution:

<xsl:stylesheet version="2.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"&gt;

   <xsl:output indent="yes" omit-xml-declaration="yes"/>

     <xsl:param name="pQString" as="xs:string" select=
     "'?login=userId&amp;tag=XSLT&amp;lang=en&amp;level=expert'"
     />

    <xsl:template match="/">
      <xsl:variable name="vLang" as="xs:string" select=
      "substring-after(
                       tokenize($pQString, '\?|&amp;')
                                 [starts-with(.,'lang=')]
                                    [last()],

                       'lang='
                       )
      "/>

      lang = "<xsl:sequence select='$vLang'/>"
    </xsl:template>
</xsl:stylesheet>

When the above XSLT 2.0 transformation is performed, it produces the correct result:

  lang = "en"