replace a String with XSLT - xslt

Question

I have a wsdl (that I get from a Web Service) where I have to replace the current address String to something else , The Idea was to use XSLT to do that. There is just one problem , I have never done anything with XSLT so i have no idea how to do that. I have found an simple example of how to do that but I dot get how do i Get the old string out of the wsdl so I can replace it.
Here is the Example
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:inm="http://www.inmagic.com/webpublisher/query" version='1.0'>
<xsl:output method="text" encoding="UTF-8"/>
<xsl:preserve-space elements="*"/>
<xsl:template match="text()"></xsl:template>
<xsl:template match="test">
<xsl:apply-templates/>
<xsl:for-each select="testObj">
'Notes or subject' <xsl:call-template name="rem-html"><xsl:with-param name="text" select="SBS_ABSTRACT"/></xsl:call-template>
</xsl:for-each>
</xsl:template>
<xsl:template name="rem-html">
<xsl:param name="text"/>
<xsl:variable name="newtext" select="translate($text,'a','b')"/>
</xsl:template>
</xsl:stylesheet>
UPDATE :
this is what i have now :
<soap:address location="http://localhost:4434/miniwebservice"/>
this is what i want to get :
<soap:address location="http://localhost:4433/miniwebservice"/>
I just replaced the number of the Port from 4434 to 4433

Answer 1

<xsl:template match="soap:address/#location">
<xsl:attribute name="location">
<xsl:call-template name="string-replace">
<xsl:with-param name="haystack" select="current()"/>
<xsl:with-param name="search">:4434/</xsl:with-param>
<xsl:with-param name="replace">:4433/</xsl:with-param>
</xsl:call-template>
</xsl:attribute>
</xsl:template>
<xsl:template match="#* | node()">
<xsl:copy>
<xsl:apply-templates select="#* | node()"/>
</xsl:copy>
</xsl:template>
Note that there are no built-in string replace function in XSLT, you'll need to take it somewhere else (e.g. http://symphony-cms.com/download/xslt-utilities/view/26418/ was used when writing this stylesheet).

Answer 2

Note that with XSLT 2.0 you have an easier way to proceed using regular expressions :
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:soap="..."
version="2.0">
<xsl:param name="newPort">4433</xsl:param>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="soap:address/#location">
<xsl:attribute name="location">
<xsl:value-of select="replace(.,
'^(http://[^/]*:)[0-9]{4}/',
concat('$1',$newPort,'/'))"/>
</xsl:attribute>
</xsl:template>
</xsl:stylesheet>
To make it work, you just have to change the namespace URI in xmlns:soap="..." to the soap namespace uri (i'm not sure of it) and use an XSLT 2.0 processor (e.g. : saxon).