Sorting XML having xsd format using xslt

Question

Sort XML with xsd format in ascending order

Because of xsd format in XML my xslt solution was not working ,what must be used in case of xsd?

Here is my XML Input:

4 1 7 What I tried:

<xsl:stylesheet version="1.0" xmlns:xsl="w3.org/1999/XSL/Transform"> <xsl:template match="/*"> xsl:copy xsl:apply-templates <xsl:sort select="number"/> </xsl:apply-templates> </xsl:copy> </xsl:template>

<xsl:template match="@*|node()">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>

</xsl:stylesheet> What i expect as Output:

1 4 7

Sort XML with xsd format in ascending order Because of xsd format in XML my xslt solution was not working ,what must be used in case of xsd? Here is my XML Input: 4 1 7 — Harshi, Jul 23 '20 at 17:29
Don't add information as comments please. Edit your question to improve it. — music2myear, Jul 24 '20 at 15:40

score 0 · Answer 1 · answered Jul 23 '20 at 19:47

0

I think you want something close to this (leaving out the namespace and other declarations around):

<xsl:template match="/">
  <xsl:element name="test">
    <xsl:for-each select="//number">
      <xsl:sort select="text(.)" data-type="number"/>
      <xsl:copy-of select="." />
    </xsl:for-each>
  </xsl:element>
</xsl:template>

answered Jul 23 '20 at 19:47

fratester

364
1
3

This din't work – Harshi Jul 24 '20 at 06:55
Sorry, as I said: I left out declaration etc. Here is the complete stylesheet: ``` ``` – fratester Jul 24 '20 at 07:04

Sorting XML having xsd format using xslt

1 Answers1