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Getting a substring AFTER the last occurrence of a character in XSLT

Tags:

c#

xml

xslt

I have a string in an XML file that looks similar to this:

M:Namespace.Class.Method(Something a, Something b)

The number of period (.) characters is abritrary, meaning it can be only 2 as in this example, but can be more.

I would like to use XSLT to get a substring of this string from the last '.' character, so that i will only be left with:

Method(Something a, Something b)

I could not achieve this using the standard substring/substring-after functions.

Is there an easy way to do this?

like image 856
lysergic-acid Avatar asked Jan 31 '12 11:01

lysergic-acid


2 Answers

In XSLT 1.0 you will need to use a recursive template, like this:

  <xsl:template name="substring-after-last">
    <xsl:param name="string" />
    <xsl:param name="delimiter" />
    <xsl:choose>
      <xsl:when test="contains($string, $delimiter)">
        <xsl:call-template name="substring-after-last">
          <xsl:with-param name="string"
            select="substring-after($string, $delimiter)" />
          <xsl:with-param name="delimiter" select="$delimiter" />
        </xsl:call-template>
      </xsl:when>
      <xsl:otherwise><xsl:value-of 
                  select="$string" /></xsl:otherwise>
    </xsl:choose>
  </xsl:template>

and invoke it like this:

<xsl:call-template name="substring-after-last">
  <xsl:with-param name="string" select="'M:Namespace.Class.Method(Something a, Something b)'" />
  <xsl:with-param name="delimiter" select="'.'" />
</xsl:call-template>

In XSLT 2.0, you can use the tokenize() function and simply select the last item in the sequence:

tokenize('M:Namespace.Class.Method(Something a, Something b)','\.')[last()]
like image 156
Mads Hansen Avatar answered Nov 07 '22 02:11

Mads Hansen


Here is a more efficient solution O(N) vs. O(N^2) for the accepted answer:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output method="text"/>

 <xsl:template match="text()" name="skipAfterDots">
  <xsl:param name="pTotalString" select="."/>
  <xsl:param name="pTotalLength" select="string-length(.)"/>
  <xsl:param name="pPosition" select="1"/>
  <xsl:param name="pLastFound" select="-1"/>

  <xsl:choose>
    <xsl:when test="$pPosition > $pTotalLength">
      <xsl:value-of select="substring($pTotalString, $pLastFound + 1)"/>
    </xsl:when>
    <xsl:otherwise>
      <xsl:variable name="vIsDot" select=
       "substring($pTotalString, $pPosition, 1) = '.'"/>

      <xsl:call-template name="skipAfterDots">
        <xsl:with-param name="pTotalString" select="$pTotalString"/>
        <xsl:with-param name="pTotalLength" select="$pTotalLength"/>
        <xsl:with-param name="pLastFound" select=
        "$pLastFound * not($vIsDot) + $pPosition * $vIsDot"/>
        <xsl:with-param name="pPosition" select="$pPosition+1"/>
      </xsl:call-template>
    </xsl:otherwise>
  </xsl:choose>
 </xsl:template>
</xsl:stylesheet>

When this transformation is applied on the following XML document:

<t>M:Namespace.Class.Method(Something a, Something b)</t>

the wanted, correct result is produced:

Method(Something a, Something b)

Explanation:

This solution doesn't contain any call to the substring-after() function. Instead, at each step only the one character of the string is compared for equality with the dot character. Because there are at most N characters, this is O(N) -- linear complexity.

On the contrary, the accepted answer calls the substring-after() function on every step. In the worst case there could be N dots and thus this would be O(N^N) -- quadratic complexity.

Note: We make the reasonable assumption that in both solutions locating the k-th character of a string is O(1).

like image 27
Dimitre Novatchev Avatar answered Nov 07 '22 02:11

Dimitre Novatchev