Print parameter or variable depending on a file's path only once

Question

Sorry in advance if the title is too vague.

I have to process several XML files referenced by each other with XSLT and search for certain errors.

My XMLs typically look like this:

<topic>
    ... some elements ...
    <topicref @href="path-to-another-file"/>
    ... some other elements ...
    <figure> ... </figure>
</topic>

And my desired output is:

path-to-a-file:
    Errors found

path-to-another-file:
    Other errors found

I get the paths from href attributes and I'd like to print a path if there's an error in the corresponding file.

The important parts of my XSLT:

<!-- ingress referenced files -->
<xsl:template match="//*[@href]">
    <xsl:apply-templates select="document(@href)/*">
        <xsl:with-param name="path" select="@href"/>
    </xsl:apply-templates>
    <xsl:apply-templates select="./*[@href]">
    </xsl:apply-templates>            
</xsl:template>

<!-- matching topic elements to check -->
<xsl:template match="topic">
    <xsl:param name="path"/>
    <xsl:if test=".//figure">
        <!-- Right now this is where I print the path of the current file -->
        <xsl:value-of select="$path"/>
    </xsl:if>
    <xsl:apply-templates select="figure">
        <xsl:with-param name="path" select="$path"/>
    </xsl:apply-templates>
</xsl:template>

<!-- check if there's any error -->
<xsl:template match="figure">
    <xsl:param name="path"/>

    <xsl:if test="...">
        <xsl:call-template name="printError">
            <xsl:with-param name="errorText" select="'...'"/>
            <xsl:with-param name="filePath" select="..."/>
            <xsl:with-param name="elementId" select="..."/>
        </xsl:call-template>
    </xsl:if>  
    <xsl:if test="...">
        <xsl:call-template name="printError">
            <xsl:with-param name="errorText" select="'...'"/>
            <xsl:with-param name="filePath" select="..."/>
            <xsl:with-param name="elementId" select="..."/>
        </xsl:call-template>
    </xsl:if>
</xsl:template>

<!-- print error message -->
<xsl:template name="printError">
    <xsl:param name="errorText"/>
    <xsl:param name="filePath"/>
    <xsl:param name="elementId"/>

    ... print out some stuff ...

</xsl:template>

Where should I print out the path of a file? With this transformation I always write it out if the file has a figure element even when it doesn't contain any mistakes. Like this:

path-to-a-file:
    Errors found

path-to-file-with-no-errors:

path-to-another-file:
    Other errors found

If I put the part in question elsewhere (i.e. in the error checking or printing template) it gets printed after every figure elements examined or errors printed.

I assume the problem could be solved with variables but I'm new to XSLT so I'm not sure how to go about it.

EDIT:

I need to display the path of the file when I find an error in it but only once, after the first error was found. Errors can be found only in files with figure elements.

I hope this clarifies my problem.

After reading your question, still not understood what do you want to do? — Smile, Oct 03 '13 at 10:57
I edited the question, I hope the problem is clear this way. — gombost, Oct 03 '13 at 11:21
Yes, but it's not what I need. This way the path gets printed out even when the file does not contain errors (I pointed this out in my post). — gombost, Oct 03 '13 at 11:36

Hobbes · Answer 1 · 2013-10-03T13:40:23.733

You can test if the current element matches a previous element in the XML, using something like:

<variable name='currentlocation' select="@href"/>
<xsl:if test="not(preceding-sibling::topicref[contains(@href, $currentlocation)])">

 (process this only if there is no preceding sibling with the same @href as the current location)

edit: this test would solve only half the problem. You need another test to check if there is an error in the other file.

Print parameter or variable depending on a file's path only once

1 Answers1