Hi,

I have a need to replace all GUID (or UID consist of 32 character) in an installation project file (xml format). I wrote a xsl which replaces all uid elements with the one I specify in <uid>. Since UID must be unique, I must come up a way to insert individual UID. I can generate the UID in Java code. I am thinking (not sure it's possible) two ways of handling this: one is to generate a list of UID in an xml file, and import each UID to each <uid> element. Second is to invoke the function that returns the UID directly in xsl.

I am new to xsl, so I appreciate your help in this issue. Thanks!

Billy



<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml"/>
    <xsl:template match="*|@*">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="section[@name='Product']/productTree//key/uid">
        <uid>12345678901234567890123456789012</uid>
    </xsl:template>
</xsl:stylesheet>

Both techniques are possible. Writing a file containing the mappings from old to new UIDS is probably cleaner, because calling Java extension functions is processor-dependent.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Thanks Michael,

Could you point me to some reference online or books that talk about the mapping technique? Or perhaps, you can provide me with a sample code? I am planning to have the guid.xml for my GUID entries:

<guids>
  <guid>12345678900987654321123456789098</guid>
  <guid>abcdec78900987654321123456789098</guid>
  <guid>123456789adeca980a00123456789098</guid>
  <guid>403ab678900987654321123456789098</guid>
  <guid>12afes78900987654321123456789098</guid>
</guilds>

How do I import this xml into the xsl, and how do I reference each guid in <gui> element? Thanks!

Billy

You use the document() function to read the XML (or you can pass it as a parameter to your stylesheet, and then you locate the elements within it using path expressions just as in your primary document.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Hi Michael,

With the document() function, I was able to retrieve the data from an external xml file. However, It writes all UID with the first GUID. I know that somehow I need to use a loop or some reference to the next element in guid.xml, but just not sure where. could you shine some light?

Thanks!

Billy


<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml"/>
    <xsl:template match="*|@*">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="section[@name='Product']/productTree//key/uid">
        <uid><xsl:value-of select="document('guid.xml')/guids/guid"/></uid>
    </xsl:template>
</xsl:stylesheet>

I wrote the following code, but still it's not going through the guid.xml in increasing order:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml"/>
    <xsl:template match="*|@*">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>


    <xsl:template name="replacement">
        <xsl:param name="n"/>
        <uid><xsl:value-of select="document('guid.xml')/guids/guid[$n]"/></uid>
    </xsl:template>


    <xsl:template match="section[@name='Product']/productTree//key">
        <xsl:param name="n" select="1"/>
        <xsl:for-each select="uid">
                <xsl:call-template name="replacement">
                    <xsl:with-param name="n" select="$n"/>
                </xsl:call-template>
                <xsl:param name="n" select="$n+1"/>
        </xsl:for-each>
    </xsl:template>

</xsl:stylesheet>

You're getting close. This should give you a compile error:

        <xsl:param name="n" select="1"/>
        <xsl:for-each select="uid">
                <xsl:call-template name="replacement">
                    <xsl:with-param name="n" select="$n"/>
                </xsl:call-template>
                <xsl:param name="n" select="$n+1"/>
        </xsl:for-each>

because you can't use xsl:param here. (If you don't get an error, it's a bug in your XSLT processor). However, what you are trying to do is basically flawed, because XSLT, as a functional programming language, doesn't have updatable variables. You need to think of this as a function: calculate the variable $n as a function of something in your input. It might be as simple as position(), or something more complicated involving xsl:number.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference