XSLT从以下类型的XML结构中除去除某些节点以外的所有节点

Question

<Rootnode>
    <Properties Attribute ="xxx">
        <Type>1</Type>
        <Size>10</Size>
    </Properties>
    <Other>
        <blah>h</blah>
    </Other>
    <Other2>
        <blah>h</blah>
    </Other2>
    <Properties Attribute ="xxx">
        <xType>5</xType>
        <xSize>10</xSize>
    </Properties>
    <Items>
       <Item4>8</Item4>
    </Items>
    <Items>
       <Item6>8</Item6>
    </Items>
    <Properties Attribute ="xxx">
        <zType>1</zType>
        <zSize>10</zSize>
    </Properties>
    <Items place="UK">
       <Item1>8</Item1>
    </Items>
 </Rootnode>

Now what I want is to ONLY include the Properties and Items. Preferably to join the Properties and Items groups together if the Attributes are the same name and value and to SORT both Properties and Items based on the Attributes and to sort the child nodes of both alphabetically. So far I've reached a blank ;(

Required output is pretty much as ABach has shown. Although one thing I forgot to mention is that there could be other attributes on each Properties or Items and that I know the name of the Attribute I wish to sort on. That I can amend easily enough.

ie, required output

<?xml version="1.0"?>
<Rootnode>
  <Properties Attribute="xxx">
    <Size>10</Size>
    <Type>1</Type>
    <xSize>10</xSize>
    <xType>5</xType>
    <zSize>10</zSize>
    <zType>1</zType>
  </Properties>
  <Items>
    <Item4>8</Item4>
    <Item6>8</Item6>
  </Items>
  <Items place="UK">
    <Item1>8</Item1>
  </Items>
</Rootnode>

And apologies for lack of my efforts so far...I ended up in a right mess and didn't think it would help much...I'm pretty new to this stuff :)

Answer 1

As already pointed out by @LarsH, by not showing us the expected output XML, we're left to guess at what you really want. That said, here's my attempt at an XSLT 1.0 solution.

When this XSLT:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
  <xsl:output omit-xml-declaration="no" indent="yes" />
  <xsl:strip-space elements="*" />

  <xsl:key
    name="PropertiesByAttributeNameVal"
    match="Properties"
    use="concat(name(@*[1]), '+', @*[1])" />

  <xsl:key
    name="ItemsByAttributeNameVal" 
    match="Items"
    use="concat(name(@*[1]), '+', @*[1])" />

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

  <xsl:template match="Rootnode">
    <Rootnode>
      <xsl:apply-templates 
        select="Properties[
          generate-id() = 
          generate-id(key(
            'PropertiesByAttributeNameVal',
             concat(name(@*[1]), '+', @*[1]))[1])]">
        <xsl:with-param name="pKeyName"
          select="'PropertiesByAttributeNameVal'" />
        <xsl:sort select="concat(name(@*[1]), '+', @*[1])" />
      </xsl:apply-templates>
      <xsl:apply-templates
        select="Items[
          generate-id() = 
          generate-id(key(
            'ItemsByAttributeNameVal',
            concat(name(@*[1]), '+', @*[1]))[1])]">
        <xsl:with-param name="pKeyName"
          select="'ItemsByAttributeNameVal'" />
        <xsl:sort select="concat(name(@*[1]), '+', @*[1])" />
      </xsl:apply-templates>
    </Rootnode>
  </xsl:template>

  <xsl:template match="Properties|Items">
    <xsl:param name="pKeyName" />
    <xsl:copy>
      <xsl:apply-templates select="@*" />
      <xsl:apply-templates
        select="key($pKeyName, concat(name(@*[1]), '+', @*[1]))/*">
        <xsl:sort select="name()" />
      </xsl:apply-templates>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

...the run against the provided XML:

<Rootnode>
  <Properties Attribute="xxx">
    <Type>1</Type>
    <Size>10</Size>
  </Properties>
  <Other>
    <blah>h</blah>
  </Other>
  <Other2>
    <blah>h</blah>
  </Other2>
  <Properties Attribute="xxx">
    <xType>5</xType>
    <xSize>10</xSize>
  </Properties>
  <Items>
    <Item4>8</Item4>
  </Items>
  <Items>
    <Item6>8</Item6>
  </Items>
  <Properties Attribute="xxx">
    <zType>1</zType>
    <zSize>10</zSize>
  </Properties>
  <Items place="UK">
    <Item1>8</Item1>
  </Items>
</Rootnode>

...what I guess is the correct output XML is produced:

<?xml version="1.0"?>
<Rootnode>
  <Properties Attribute="xxx">
    <Size>10</Size>
    <Type>1</Type>
    <xSize>10</xSize>
    <xType>5</xType>
    <zSize>10</zSize>
    <zType>1</zType>
  </Properties>
  <Items>
    <Item4>8</Item4>
    <Item6>8</Item6>
  </Items>
  <Items place="UK">
    <Item1>8</Item1>
  </Items>
</Rootnode>

Note that if this same XSLT is run against a slightly modified XML document (that has more groupings, etc.):

<?xml version="1.0" encoding="utf-8"?>
<Rootnode>
  <Properties Attribute="xxx">
    <Type>1</Type>
    <Size>10</Size>
  </Properties>
  <Other>
    <blah>h</blah>
  </Other>
  <Other2>
    <blah>h</blah>
  </Other2>
  <Properties Attribute="yyy">
    <xType>5</xType>
    <xSize>10</xSize>
  </Properties>
  <Items>
    <Item4>8</Item4>
  </Items>
  <Items place="US">
    <Item9>8</Item9>
  </Items>
  <Items>
    <Item1>8</Item1>
  </Items>
  <Properties Attribute2="xxx">
    <zType>1</zType>
    <zSize>10</zSize>
  </Properties>
  <Properties Attribute="xxx">
    <elephantType>5</elephantType>
    <elephantSize>15</elephantSize>
  </Properties>
  <Items place="UK">
    <Item1>8</Item1>
  </Items>
</Rootnode>

...again, what I assume is the correct answer is produced:

<?xml version="1.0"?>
<Rootnode>
  <Properties Attribute="xxx">
    <Size>10</Size>
    <Type>1</Type>
    <elephantSize>15</elephantSize>
    <elephantType>5</elephantType>
  </Properties>
  <Properties Attribute="yyy">
    <xSize>10</xSize>
    <xType>5</xType>
  </Properties>
  <Properties Attribute2="xxx">
    <zSize>10</zSize>
    <zType>1</zType>
  </Properties>
  <Items>
    <Item1>8</Item1>
    <Item4>8</Item4>
  </Items>
  <Items place="UK">
    <Item1>8</Item1>
  </Items>
  <Items place="US">
    <Item9>8</Item9>
  </Items>
</Rootnode>

Assumptions:

• I assume that each <Properties> and <Items> element has only one attribute and that it should be the grouping determiner.
• If the above is not true, at the very least, I assume that the first attribute of that element should be the grouping determiner.

Explanation:

1. Because this is an XSLT 1.0 solution, Muenchian Grouping is the order of the day when grouping nodes and attributes under unique selectors; therefore, we define two keys: one for <Properties> elements and one for <Items> elements.

2. The first template is the Identity Transform - its job is to output all nodes and attributes from the source document to the result document as-is.

3. The second template matches the <Rootnode> element. It is instructed to apply templates to only those <Properties> and <Items> elements who are the first to appear in their respective keys; this has the intended effect of only processing unique elements (based upon the name and value of their first attribute).

   When the <xsl:apply-templates> element is specified, note that in both instances, it is instructed to sort the results by that same attribute name/value pairing.

   Do note that each <xsl:apply-templates> element is given a parameter (via <xsl:with-param> ). As you'll see, the code to process both <Properties> and <Items> elements is nearly identical; the only difference is the key that we grab results from. Because of this, I chose to consolidate that logic into the third template and account for variability via this parameter.

4. The third template matches both <Properties> and <Items> elements. For each, the original node is copied (as are its attributes). Finally, templates are applied against all child elements of this element (with appropriate sorting occuring [this time, based on the name of the child element itself]).

Answer 2

You will be more likely to receive help if you already have something rather than nothing. To get started, create a template that matches Properties and Items ( match="Properties | Items" ) whose content just copies the matched element: <xsl:copy-of select="." /> <xsl:copy-of select="." /> .

That will give you some working code to show.

The next step I would suggest is posting a sample of the desired output, and the actual output given by your XSLT code.

That will provide a much smaller gap for people to bridge in answering your question.