Re: [xsl] Decimal to hexadecimal in one function (XSLT/XPath 2)

[Michael Müller-Hillebrand <mmh@xxxxxxxxxxxxx>]

Am 17.02.2009 um 12:22 schrieb Yves Forkl:

> I want to convert the decimal integers returned by the XPath 2
> function fn:string-to-codepoints into a sequence of strings that use
> the conventional "U+..." hexadecimal notation of the Unicode standard.
>
> Neither the XPath 2 spec nor the XSLT FAQ offered a complete
> solution to this problem, so I developed my own. It is working, but
> I wonder how to integrate these two functions, which convert a
> decimal integer into its hexadecimal equivalent, into a single
> function which still returns just one string:
>
> <xsl:function name="my:decimal-to-hex" as="xs:string">
>  <xsl:param name="decimalNumber"/>
>  <xsl:sequence select="string-join(my:decimal-to-hex-
> internal($decimalNumber),'')"/>
> </xsl:function>
>
> <xsl:function name="my:decimal-to-hex-internal" as="xs:string+">
>  <xsl:param name="decimalNumber"/>
>  <xsl:variable name="hexDigits" select="'0123456789ABCDEF'"/>
>  <xsl:if test="$decimalNumber &gt;= 16">
>    <xsl:sequence
>      select="my:decimal-to-hex-internal(floor($decimalNumber div
> 16))"/>
>  </xsl:if>
>  <xsl:sequence
>    select="substring($hexDigits, ($decimalNumber mod 16) + 1, 1)"/>
> </xsl:function>


I don't think it is elegant at all, but you can collect the recursive
digits in a variable and then do this:

<xsl:function name="my:decimal-to-hex" as="xs:string">
  <xsl:param name="decimalNumber"/>
  <xsl:variable name="hexDigits" select="'0123456789ABCDEF'"/>
  <xsl:variable name="upperDigits">
    <xsl:if test="$decimalNumber &gt;= 16">
      <xsl:sequence select="string-join(my:decimal-to-
hex(floor($decimalNumber div 16)), '')"/>
    </xsl:if>
  </xsl:variable>
  <xsl:sequence select="string-join(($upperDigits,
substring($hexDigits, ($decimalNumber mod 16) + 1, 1)), '')"/>
</xsl:function>

- Michael


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