如何使用xslt转换此xml

Question

I have an XML in the following format:

<?xml version="1.0" encoding="utf-8"?>
<dataset  xmlns="http://developer.cognos.com/schemas/xmldata/1/"  xmlns:xs="http://www.w3.org/2001/XMLSchema-instance">
    <metadata>
          <item name="Col1" type="xs:string" length="14"/>
          <item name="Col2" type="xs:string" length="92"/>
          <item name="Col3" type="xs:string" length="66"/>
          <item name="Col4 With Space" type="xs:string" length="32"/>
    </metadata>
    <data>
        <row>
            <value>SomeVal1</value>
            <value>SomeVal2</value>
            <value>SomeVal3</value>
            <value>SomeVal4</value>
        </row>
        <row>
            <value>SomeVal11</value>
            <value>SomeVal22</value>
            <value>SomeVal33</value>
            <value>SomeVal44</value>
        </row>
    </data>
</dataset>

I want to transform it to this format:

<?xml version="1.0" encoding="utf-8"?>
<dataset>
    <data>
        <row>
            <Col1 type="xs:string" length="14">SomeVal1</Col1>
            <Col2 type="xs:string" length="92">SomeVal2</Col2>
            <Col3 type="xs:string" length="66">SomeVal3</Col3>
            <Col4_With_Space type="xs:string" length="32">SomeVal4</Col4_With_Space>
        </row>
        <row>
            <Col1 type="xs:string" length="14">SomeVal11</Col1>
            <Col2 type="xs:string" length="92">SomeVal22</Col2>
            <Col3 type="xs:string" length="66">SomeVal33</Col3>
            <Col4_With_Space type="xs:string" length="32">SomeVal44</Col4_With_Space>
        </row>
    </data>
</dataset>

I've never used xslt before and I am out of my depth. I tried something like this (which doesn't work) but I'm stuck.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns="http://www.w3.org/1999/xhtml" version="1.0">
<xsl:output encoding="UTF-8" indent="yes" method="xml" standalone="no" omit-xml-declaration="no"/>
    <xsl:template match="dataset">
        <xsl:for-each select="data/row">
            <Col1 select="1"/>
            <Col2 select="2"/>
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

The reason for wanting to transform is SSIS "Data Flow" "XML Source" doesn't recognise the first format. I'm trying to import data from an XML into a DB table. "XML Source" works with the 2nd format but not the first.

I think I should add, there is no guarantee XML metadata will always be the same, so transform needs to take the column names and types from the metadata, instead of hard-coding.

Answer 1

I've never used xslt before and I am out of my depth.

This is certainly not a task suitable for a beginner.

As I mentioned in the comment to your question, your first problem is that your source XML places all its elements in a namespace. You need to declare the same namespace in your stylesheet, assign it a prefix and use that prefix to address the elements in the source XML.

The other issue with your attempt is that you're not actually getting any data from the source XML.

<Col1 select="1"/>

will output the same static element for each row in your XML.

Now, the real complication here is getting the columns names and attributes from the metadata section and pairing them with the corresponding cell value from the current row. This could be accomplished as follows:

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:cog="http://developer.cognos.com/schemas/xmldata/1/" 
exclude-result-prefixes="cog">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>

<xsl:template match="/cog:dataset">
    <xsl:variable name="cols" select="cog:metadata/cog:item" />
    <dataset>
        <data>
            <xsl:for-each select="cog:data/cog:row">
                <xsl:variable name="curr-row-cells" select="cog:value" />
                <row>
                    <xsl:for-each select="$cols">
                        <xsl:variable name="i" select="position()" />
                        <xsl:element name="{translate(@name, ' ', '_')}">
                            <xsl:copy-of select="@type | @length"/>
                            <xsl:value-of select="$curr-row-cells[$i]" />
                        </xsl:element>
                    </xsl:for-each>
                </row>
            </xsl:for-each>
        </data>
    </dataset>
</xsl:template>

</xsl:stylesheet>

Result

<?xml version="1.0" encoding="UTF-8"?>
<dataset>
  <data>
    <row>
      <Col1 type="xs:string" length="14">SomeVal1</Col1>
      <Col2 type="xs:string" length="92">SomeVal2</Col2>
      <Col3 type="xs:string" length="66">SomeVal3</Col3>
      <Col4_With_Space type="xs:string" length="32">SomeVal4</Col4_With_Space>
    </row>
    <row>
      <Col1 type="xs:string" length="14">SomeVal11</Col1>
      <Col2 type="xs:string" length="92">SomeVal22</Col2>
      <Col3 type="xs:string" length="66">SomeVal33</Col3>
      <Col4_With_Space type="xs:string" length="32">SomeVal44</Col4_With_Space>
    </row>
  </data>
</dataset>

---

Important:

The supplied column names are not necessarily valid XML element names. In this example, "Col4 With Space" contains spaces which I have translated to underscores - but there are many other potential pitfalls that may cause a name to be unusable and generate a fatal error.