Working with XML & XSL

Question

FIRST EDIT
I'm fetching the Child 1 tag into a DropDownList in a C# form, Plz suggest the best practise code (C#) for deleting the Parent tag & all it's child tags in an XML file. Example:

    <Parents>
      <Parent>
        <Child 1>Something</Child 1>
        <Child 2>Something</Child 2>
        <Child 3>Something</Child 3>
        <Child 4>Something</Child 4>
      </Parent>
      <Parent>
        <Child 1>Something 1</Child 1>
        <Child 2>Something 1</Child 2>
        <Child 3>Something 1</Child 3>
        <Child 4>Something 1</Child 4>
      </Parent>
    </Parents>

--- Previous Question ---
How can I insert the following stylesheet tag into my new xml file which is created using C# code????

<?xml-stylesheet type="text/xsl" href="issuez.xsl"?>

C# code to create the xml file:-

new XDocument(
                        new XElement("issues",
                            new XElement("issue",
                            new XElement("cat", comboBox1.Text),
                            new XElement("desc", richTextBox1.Text),
                            new XElement("end", dateTimePicker1.Text),
                            new XElement("att", textBox2.Text)
                            )
                        )
                        ).Save(path);

Answer 1

This is the sort of thing that the back end producing the XML should handle. XSLT isn't the best place for lots of logic. Better to embed all that in the XML after you query for the news items. Just send them to the client in the proper form so they don't have to work so hard.

Answer 2

If you use the XSLT processor from PHP, you can use PHP-functions inside your XSLT script. All you need is call registerPhpFunctions before transformation. The result value in the right order can be used for sorting.

Answer 3

An XSLT stylesheet allows global parameters that can be set before a transformation is run. So with XSLT 1.0 and .NET's XslCompiledTransform if you need the current date in your stylesheet you can define a global parameter

  <xsl:param name="current-date"/>

and set that before running the transformation by creating an XsltArgumentList, setting the parameter to a value and format you want/need and then pass that XsltArgumentList as the second argument to the Transform method. Then in your stylesheet you can compare the date in an XML input element or attribute to the parameter.

As you use .NET a different option is to use XSLT 2.0; Microsoft does not support that but with Saxon 9 there is a third party solution. XSLT/XPath 2.0 have a function named current-date, that way you don't need a parameter.

Answer 4

First, make sure that dates in your XML are represented in the canonical YYYY-MM-DD format, and times as HH:MM:SS, so that XSLT (which, in 1.0, doesn't have date or time datatypes) can compare and sort them.

Second, use Steve Muench's technique for grouping. You generate a key on the items' dates, using xsl:key. The key() function can then be used to find a list of all items on a given date.

Using that key, you can build a list of the distinct dates that appear in the items. This is the Muenchian technique: find each item that's the first item in the list that key() returns for that item's date. This technique guarantees that you're always get one and only one item for each distinct date value. You then sort those items, and use their dates to drive the actual production of your output.

A minimal example:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;

  <xsl:key name="dates" match="/data/newsitem" use="@date"/>

  <xsl:template match="/">
    <output>
      <!-- find exactly one newsitem node for each distinct @date in the document -->
      <xsl:for-each select="/data/newsitem[generate-id() = generate-id(key('dates', @date)[1])]">
        <xsl:sort select="@date" order="descending"/>
        <xsl:call-template name="newsitems_for_date">
          <xsl:with-param name="date" select="@date"/>
        </xsl:call-template>
      </xsl:for-each>
    </output>
  </xsl:template>

  <xsl:template name="newsitems_for_date">
    <xsl:param name="date"/>
    <h1>
      <xsl:value-of select="$date"/>
    </h1>
    <xsl:apply-templates select="/data/newsitem[@date=$date]">
       <xsl:sort select="@time" order="descending"/>
    </xsl:apply-templates>
  </xsl:template>

  <xsl:template match="newsitem">
    <p>
      newsitem for <xsl:value-of select="@date"/>
    </p>
  </xsl:template>

</xsl:stylesheet>