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http://issues.apache.org/bugzilla/show_bug.cgi?id=27275
xsl:sort not sorting correctly
joerg.heinicke@gmx.de changed:
What |Removed |Added
----------------------------------------------------------------------------
Status|NEW |RESOLVED
Resolution| |WORKSFORME
------- Additional Comments From joerg.heinicke@gmx.de 2004-03-09 00:21 -------
Hello Khanh Vo,
I tested exactly your stylesheet and it works for me, I get the output
<Document>
<p class="MsoBodyText">
<span style="font-family:"Times New Roman"">one</span>
</p>
<p class="MsoBodyText">
<span style="font-family:"Times New Roman"">two</span>
</p>
<p class="MsoBodyText">
<span style="font-family:"Times New Roman"">three</span>
</p>
</Document>
with the one and only template in the stylesheet:
<xsl:template match="/*">
<xsl:copy>
<xsl:for-each select="Para">
<xsl:sort data-type="number" select="substring-before(current(),':')"/>
<p class="MsoBodyText">
<span style="font-family:"Times New Roman"">
<xsl:value-of select="substring-after(current(),':')"/>
</span>
</p>
</xsl:for-each>
</xsl:copy>
</xsl:template>
Now I can only start with wild guessing: Are you sure the for-each is executed?
If you get exactly the output below I guess the template containing the for-each
is not accessed at all. Maybe you can solve this issue on a XSLT list like
Mulberry's: http://www.mulberrytech.com/xsl/xsl-list/.
Joerg
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