XML / XSL

8-Jun-09 23:10

One thing I see a problem with immediately - the <a> tag won't be included in the text for <field> - it's a nested tag.

Aside form that, this is roughly what you need:

<xsl:stylesheet version="1.0" 
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform">


   <xsl:template match="/">
      <xsl:apply-templates select="//field"/>
   </xsl:template>

   <!-- 
      Template for a field element. Apply template for each of its text nodes.
   -->
   <xsl:template match="field">
      <ul>
         <xsl:apply-templates select="text()"/>
      </ul>
   </xsl:template>

   <!-- 
      Template for text() in a field element. Apply the line-splitting algorithm to the text
   -->
   <xsl:template match="/field/text()">
      <xsl:call-template name="t">
         <xsl:with-param name="s" select="."/>
      </xsl:call-template>
   </xsl:template>
  
   <!-- 
      Template t takes one string parameter s. It takes off the first line
      then recurses, terminating when the input parameter is the empty string
   -->
   <xsl:template name="t">
      <xsl:param name="s"/>
      <!--
         Is there any text to process?
      -->
      <xsl:if test="string-length($s)>0">
         <!--
            first = text before the first line break (or null if it's the last, unterminated line)
         -->
         <xsl:variable name="first" select="substring-before($s, '&#xA;')"/>
         <xsl:choose>
            <!--
               Output first if it has content
            -->
            <xsl:when test="string-length($first)>0">
               <li><xsl:value-of select="$first"/></li>
            </xsl:when>
            <xsl:otherwise>
               <!--
                  Output the input string if it doesn't contain a line break
               -->
               <xsl:if test="substring($s,1,1)!='&#xA;'">
                  <li><xsl:value-of select="$s"/></li>
               </xsl:if>
            </xsl:otherwise>
         </xsl:choose>
         <!--
            Recurse, passing in the text after the first line break
         -->
         <xsl:call-template name="t">
            <xsl:with-param name="s" select="substring-after($s, '&#xA;')"/>
         </xsl:call-template>
      </xsl:if>
   </xsl:template>
   
</xsl:stylesheet>

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

CSS...

Jamal Abdul Nasir8-Jun-09 3:11

Jamal Abdul Nasir

8-Jun-09 3:11

--> can v apply CSS on XML sheet?

--> if ya den how?

J A Nasir K

Christian Graus8-Jun-09 18:08

8-Jun-09 18:08

This is a stupid question. Even if it was written in English, it would still be stupid. How is a plain XML document displayed, that it would interact with CSS ?

Christian Graus

Driven to the arms of OSX by Vista.

"! i don't exactly like or do programming and it only gives me a headache." - spotted in VB forums.

I can do things with my brain that I can't even google. I can flex the front part of my brain instantly anytime I want. It can be exhausting and it even causes me vision problems for some reason. - CaptainSeeSharp

Stuart Dootson8-Jun-09 22:27

8-Jun-09 22:27

Christian Graus wrote:
How is a plain XML document displayed, that it would interact with CSS ?

Not such a stupid question, Christian - see this example[^]...

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Christian Graus9-Jun-09 11:17

9-Jun-09 11:17

OK, so the XML basically gets treated as XHTML ? I'd have expected to use XSLT to create that sort of a view.

Stuart Dootson9-Jun-09 13:02

9-Jun-09 13:02

Christian Graus wrote:
OK, so the XML basically gets treated as XHTML

Something like that - I guess each of the elements gets treated a bit like a DIV in HTML, i.e. no implied display semantics.

The 'original' stylesheet language was DSSSL, which was designed for use with SGML (ancestor of XML and HTML). Looking at this DSSSL sample[^], you can (kind of) see a resemblance to CSS. So, I guess there's a parallel link between SGML -> XML/XHTML/HTML on the one side and DSSSL -> CSS on t'other.

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Stuart Dootson8-Jun-09 22:25

8-Jun-09 22:25

Yes - see here[^]

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Another boring XML question. URGENT

Jamal Abdul Nasir2-Feb-10 7:58

Jamal Abdul Nasir

2-Feb-10 7:58

thanks alot sir...

J A Nasir K

Etienne_1237-Jun-09 6:15

7-Jun-09 6:15

Hi
I have 2 XML files. The one looks like the following at the top:

  <?xml version="1.0" standalone="yes" ?> 
- <NewDataSet>

And the other one as follows:

  <?xml version="1.0" standalone="yes" ?> 
- <dsParrys_offoice_furniture xmlns="http://tempuri.org/dsParrys_offoice_furniture.xsd">

I use the following in my XAML to bind to them:

<!--XmlDataProvider used to populate treeview-->
        <XmlDataProvider x:Key="dpNewDS" XPath="*"/>

        <!--ProductRange template-->
        <HierarchicalDataTemplate DataType="ProductRange" ItemsSource="{Binding XPath=*}">
            <WrapPanel>
                <TextBlock Text="{Binding XPath=ProductRangeDesc}"/>
            </WrapPanel>
        </HierarchicalDataTemplate>

        <!--Products template-->
        <HierarchicalDataTemplate DataType="Products" ItemsSource ="{Binding XPath=Products}">
            <WrapPanel>
                <TextBlock Text="{Binding XPath=Product}"/>
            </WrapPanel>
        </HierarchicalDataTemplate>

As well as these lines in the code behind file:
NOTE: I specify the path to the XML files using an OpenFileDialog

System.Windows.Forms.OpenFileDialog ofd = new System.Windows.Forms.OpenFileDialog();
            if (ofd.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                string fileName = ofd.FileName;
                XmlDataProvider dp = this.FindResource("dpNewDS") as XmlDataProvider;
                dp.Source = new Uri(fileName);
            }

Now when I select the FIRST XML file my treeview binds fine, but when I select to the 2nd one (the one with the XMLNS defined), it does not bind correctly.
Is there something I should change in my XAML when binding to the 2nd XML file?

Christian Graus8-Jun-09 18:07

8-Jun-09 18:07

Don't say it's urgent. It's not, we really don't care. Did you cross post this, or did you fail to post it in the WPF forum ?

Etienne_1238-Jun-09 23:00

8-Jun-09 23:00

If you don't have an answer to post, why bother to reply? Seriously. And don't refer to "we" when its just you replying, please.
This HAS got to do with XML by the way, as well as WPF. When I post this sort of question in the WPF forums I get told that I should post it in the XML forum, now when I post it here I get told I should post it in the WPF forum.

Christian Graus9-Jun-09 14:54

9-Jun-09 14:54

Well, I'm sorry that someone told you that, because it seems to me that it has to do with how WPF parses XML, not just XML. So, you'll have more luck in the WPF forum. And I suggest that entirely to try to help you find an answer.

I said 'we' because I'm referring to community standards. It's rude to tell us that your question is urgent. If you can't wait for a free answer, find a way to pay for one.

How do I specify a new Uri path?

kamagra10-Jun-09 2:10

kamagra

10-Jun-09 2:10

Can you please specify the path for your XML files. So that we can look into it in broader way

Etienne_1234-Jun-09 4:56

Binding TreeView Header to XML Nodes

4-Jun-09 4:56

Hi guys

I use the following to specify the Uri path for the XmlDataProvider dp:

System.Windows.Forms.OpenFileDialog ofd = new System.Windows.Forms.OpenFileDialog();
            if (ofd.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                string fileName = ofd.FileName;
                XmlDataProvider dp = this.FindResource("dpNewDS") as XmlDataProvider;
                dp.Source = new Uri(fileName);
            }

Now I want to now, what if I don't want to use an OpenFileDialog to specify the path, but instead I want to specify it myself. I know typing in the folder address e.g C:\xmlfile.xml works fine, but I don't want to hard code the path. I want to keep the file in the "bin" folder, how do I specify the path to this folder?
I tried various methods but none seemed to work.

Etienne_1234-Jun-09 0:49

Re: Binding TreeView Header to XML Nodes

4-Jun-09 0:49

Hi everybody

I recently got it working to bind an XML file to a treeview using a XmlDataProvider, but my problem is that my XML file does not contain attributes which can be easily bound to. Here's a sample of what my XML looks like:

<Product_x0020_range>
     <Product_x0020_range_x0020_id>1</Product_x0020_range_x0020_id> 
     <Product_x0020_Range>Specials</Product_x0020_Range> 
     <Products>
          <quote_x0020_id>270</quote_x0020_id> 
          <date>2008-10-14T00:00:00+02:00</date> 
          <x_x0020_hide_x0020_options_x0020_id>12</x_x0020_hide_x0020_options_x0020_id> 
          <Product_x0020_range_x0020_id>1</Product_x0020_range_x0020_id> 
          <Product>QQ CORE 2200 X 800 X 1500 X 600</Product> 
          <height>0</height> 
          <breadth_x002F_depth>0</breadth_x002F_depth> 
          <length_x002F_width>0</length_x002F_width> 
          <Description_x0020_1>CORE 2200 X 800 X 1500 X 600</Description_x0020_1> 
          <Description_x0020_2>CORE 2200 X 800 X 1500 X 600</Description_x0020_2> 
          <wood_x0020_options_x0020_1>1</wood_x0020_options_x0020_1> 
          <wood_x0020_options_x0020_2>2</wood_x0020_options_x0020_2> 
          <wood_x0020_options_x0020_3>3</wood_x0020_options_x0020_3> 
          <frozen_x0020_price>2030</frozen_x0020_price> 
          <Frozen_x0020_hours>1</Frozen_x0020_hours> 
          <Prepaired_x0020_for>GILLIAN</Prepaired_x0020_for> 
          <Percent_x0020_mark_x0020_up>2.5</Percent_x0020_mark_x0020_up> 
          <Price_x0020_list>false</Price_x0020_list> 
     </Products>
</Product_x0020_range>

As you can see, ProductRangeID and the ProductRange description are not specified as attributes, but instead as a values between 2 tags each.

The code I'm using in my XAML to bind these are as follows:

<HierarchicalDataTemplate DataType="Product_x0020_range" ItemsSource="{Binding XPath=*}">
            <WrapPanel>
                <TextBlock Text="{Binding Path=Name}"/>
            </WrapPanel>
</HierarchicalDataTemplate>

        <HierarchicalDataTemplate DataType="Products" ItemsSource ="{Binding XPath=*}">
            <WrapPanel>
                    <TextBlock Text="{Binding XPath=Name}"/>
            </WrapPanel>
</HierarchicalDataTemplate>

This does not give me the output I want though. It lists all the Product Ranges in the treeview, but it displays Product_x0020_range as the header for the root treeview items, then when I expand this item, it then displays the actual ID and description of the product range e.g. 1, Specials.
How can I do my bindings so that "Specials" is displayed as the header for the root treeview items?

Etienne_1234-Jun-09 2:06

Re: Binding TreeView Header to XML Nodes

4-Jun-09 2:06

So what I'm asking is, how do I bind to the text contained within XML elements, instead of attributes?

Etienne_1234-Jun-09 2:10