CODEWITHSUNDEEP
c#xmlxslt
Dmitry Peredera
Dmitry Peredera

Reputation: 23

Get filename from path using XSLT 1.0 (c# .Net Core)

Please help me. How can I get filename using XSLT 1.0 from string like this?

filestore:/722601940006/2018/02/09/file.jpg

A file can have a different name and extension. A path can have a different depth.

I tried to use regular expression and function tokenize():

<xsl:value-of select="tokenize('$file/@fh:fileRef','.*/(.*?)$')"/>

But I found out that .Net Core doesn't support XSLT 2.0

Upvotes: 2

Views: 1376

Answers (1)

zx485
zx485

Reputation: 29062

Because you are restricted to XSLT-1.0 a recursive template is necessary. So one XSLT-1.0 solution to avoid waiting for Microsoft to respond would be 

<xsl:template name="fileName">
  <xsl:param name="str" />
  <xsl:choose>
    <xsl:when test="normalize-space(substring-after($str,'/'))">
      <xsl:call-template name="fileName">
        <xsl:with-param name="str" select="substring-after($str,'/')" />
      </xsl:call-template>  
    </xsl:when>
    <xsl:otherwise>
      <xsl:value-of select="$str" />
    </xsl:otherwise>
  </xsl:choose>
</xsl:template>

Calling it with 

<xsl:call-template name="fileName">
  <xsl:with-param name="str" select="yourFileNameString" />   <!-- insert string here -->
</xsl:call-template>

emits the string

file.jpg

Upvotes: 4

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