In XSL, how can I print only the first line in for-each block?

Question

I made a list of books in XML. The following is the format of my XML file; there are some more <book> blocks, of course.

<data>
    <book>
        <title>Encyclopaedia Britannica</title>
        <category>encyclopedia</category>
        <language>English</language>
        <author>Encyclopaedia Britannica Editorial</author>
        <year>1768</year>
        <price>49.99</price>
    </book>
</data>

And I want to print the title of the most expensive one. I tried as follows in the .xsl file:

<p style = "display: block;">
    The most expensive book is "
    <xsl:for-each select="data/book">
        <xsl:sort select="price" order="descending"/>
        <!-- <xsl:value-of select="title"/> -->
    </xsl:for-each>
        <xsl:value-of select="data/book/title"/>
    "
</p>

When the <xsl:value-of> block in <xsl:for-each>(which I set as comment) is executed it prints all the books sorted in descending order. Of course, <xsl:value-of select="data/book/title"/> makes the first book of the original table get printed.

So I want to print the most expensive one by stopping printing the other books but the first one when the table is sorted.

I know that there's no such thing like break, so I thought using <xsl:if> block to check if the row is the first or not. Is this idea possible? Or can I do this in another way?

Answer 1

so I thought using <xsl:if> block to check if the row is the first or not.

Yes, this is the correct idea:

<xsl:for-each select="/data/book">
    <xsl:sort select="price" data-type="number" order="descending"/>
    <xsl:if test="position()=1">
        <xsl:value-of select="title"/>
    </xsl:if>
</xsl:for-each>

Note that this selects only the first most expensive title in case of a tie.

---

Or can I do this in another way?

That depends on which XSLT processor you use. In XSLT 2.0, you can use the max() function. And several XSLT 1.0 processors support the math:max() and math:highest() extension functions.