XML / XSL

22-Jan-07 19:07

I have downloaded from a website xml menu code.This includes a Horizantal menu and Vertical Menu of XML and XSL Files respectively. There is one Menu java script file and a Css file . The Download folder also includes code, default and help aspx files.How to display the XML menu . There is know html file loaded with all this files into it.

//Floder
Code.aspx
Default.aspx
Help.aspx
Menu.css
menu.js
vmenu.xml
vmenu.xsl
hmenu.xml
hmenu.xsl

Praveen

Re: XML Menu

Stefan Troschuetz22-Jan-07 22:14

Stefan Troschuetz

22-Jan-07 22:14

No html file needed, as the menus are probably used by the aspx files. Those are part of the ASP.NET technology and similar to php files allow you to create dynamic pages. Therefor they need to be processed by a special web server which then returns a html page to the client. Such a web server that allows you to run the aspx pages on your local systems is included in the free of charge Visual Web Developer 2005 Express Edition[^]

XML to HTML page

B.L.Praveen22-Jan-07 3:03

22-Jan-07 3:03

I trie to display a greeting Message in an html by redaing the words from xml.This was a my first simple progrma on xml.Use XMLDOM to load xml and xsl to memory.Transfomed into html. The html page dispplay only words of greeting.It doesn't display the text read from xml.
This my xsl code:

<body> words of greeting:

    xsl:value-of select="greeting"   
</body>

The Xml code

Hello world.

Praveen

Dustin Metzgar22-Jan-07 3:37

22-Jan-07 3:37

You have to change your <'s to <'s and >'s to >'s. We can't see your XML text.

B.L.Praveen22-Jan-07 18:07

22-Jan-07 18:07

The Xsl stylesheet File used by xml to display html page

<small><?xml version="1.0"?><!--hellohtm.xsl-->
<html xmlns:xsl="http://www.w3.org/1999/xsl/transform" xsl:version="1.0">
<head><title>Greeting</title></head>
<body><p> words of greeting:<br/>
    <b><i><u><xsl:value-of select="greeting"/></u></i></b>
   </p>
</body>
</html></small>

The Xml file hodliing greeting message to be printed

<small><?xml version="1.0"?>
<?xml-stylesheet type="type/xsl" href="Hello.xsl"?>
<greeting>Hello world.</greeting></small>

The Html page should display the greeting message read from the xml file:" The page displays only the word of greeting on the page it must also display the xml greeting message Hello World."

<small><html>
<body>
<script language="javascript">
// Load XML 
var xml = new ActiveXObject("Microsoft.XMLDOM")
xml.async = false
xml.load("Hello.xml")

// Load the XSL
var xsl = new ActiveXObject("Microsoft.XMLDOM")
xsl.async = false
xsl.load("Hello.xsl")

// Transform
document.write(xml.transformNode(xsl))
</script>

</body>
</html> </small>

Praveen

Dustin Metzgar22-Jan-07 18:23

22-Jan-07 18:23

Excellent, that helps a lot. You have to divide the XML from the XSL. Here's what the XML should look like:

<?xml version="1.0" encoding="utf-8"?>
<greeting>Hello world.</greeting>

And here's an example XSL:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
    <html>
    <body>
      <xsl:value-of select="greeting"/>
    </body>
    </html>
</xsl:template>
</xsl:stylesheet>

You had the XSL tags in the XML source document. The XML source document should be free-form, meaning that it can be pretty much anything as long as it's legal XML. The XSL is the transform and is pretty specific to the input and output expected. Check out zvon[^] or w3schools[^] for some tutorials on XSL.

Logifusion[^]

B.L.Praveen22-Jan-07 18:29

22-Jan-07 18:29

Thanks, it is displaying

Praveen

XML Xpath Xquery

Capatain galaxy20-Jan-07 9:59

Capatain galaxy

20-Jan-07 9:59

Hi Guys

This a ticklish one and espite hunting around the web I cannot find an answer.

Basically what I would like to do is to extract a single instance of any node. For example imagine an XMl file which contains loads of books, authors and keywords for each book. What I would like to do is to extract a single intance of each author.

If you like you could call it an index.

any pointers would be immensley useful.

for the more SQL orientated the relevant code code would be select distinct author from books.

Thanks in advance

CQ

Re: XML Xpath Xquery

Dustin Metzgar22-Jan-07 5:18

VB.NET | XML Namespace parsing issues

22-Jan-07 5:18

In this case, XSL is a little weak. There is a distinct-values[^] function in XPath 2. Microsoft doesn't like XPath 2 and has decided not to support it. You can use another parser like XmlSpy and it will work fine. If you're using an XSLT, you can do a sort on a for-each so you could get all the like author names together, but that's about it.

cdnhype19-Jan-07 8:37

19-Jan-07 8:37

Hello,

This is my first delve into XML namespaces and I'm having a VERY difficult time parsing the XML because of the Namespace. If I remove the namespace then I get no problems. It would really be appreciated if anybody out there can cure my headache!

Here is my code:

Dim xmlDoc As New System.Xml.XmlDocument()
Dim resolver as XmlUrlResolver = new XmlUrlResolver()
Dim datasource as XmlDataSource = new XmlDataSource()

resolver.Credentials = New NetworkCredential("username", "password")
xmlDoc.XmlResolver = resolver
xmlDoc.Load("http://somedomain.com/webservicename")

Dim xsn as XmlNamespaceManager = new XmlNamespaceManager(xmlDoc.NameTable)
xsn.AddNamespace("t", "urn:ahw:itemlist")

xmlData.Document = xmlDoc
xmlData.TransformSource = "test.xsl"

Here is the XML:

<?xml version="1.0" encoding="UTF-8"?>
<itemlist xmlns="urn:ahw:itemlist">
<item name="Photos" friendly-name="Photos" />
<item name="Videos" friendly-name="Videos" />
<item name="OtherMedia" friendly-name="Other Media" />
</item-type-list>

Here is the xsl:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="itemlist">
<TABLE BORDER="2">
<TR>
<TD>Display Name</TD>
<TD>Name</TD>
</TR>
<xsl:for-each select="item">
<TR>
<TD><xsl:value-of select="@friendly-name"/></TD>
<TD><xsl:value-of select="@name"/></TD>
</TR>
</xsl:for-each>
</TABLE>
</xsl:template>
</xsl:stylesheet>

Here is my asp.net page (the main part where I want the XML to display):

<form runat="server">
<asp:Xml ID="xmlData" runat="server"/>
</form>

Thanks so much for your time!

Chris

Dustin Metzgar19-Jan-07 8:51

19-Jan-07 8:51

Well, first I gotta ask if there's a typo in your input XML. The end tag for the document element doesn't match. The next thing I'd like to know is what is the error you're getting? Do you get the error when you load the document, when you put it in the data document, or when you run the transform?

cdnhype19-Jan-07 8:53

19-Jan-07 8:53

Sorry. Here is the correct XML:

<?xml version="1.0" encoding="UTF-8"?>
<itemlist xmlns="urn:ahw:itemlist">
<item name="Photos" friendly-name="Photos" />
<item name="Videos" friendly-name="Videos" />
<item name="OtherMedia" friendly-name="Other Media" />
</itemlist>

I'm not getting an error. Its just nothing appears on page.

Stefan Troschuetz19-Jan-07 8:59

Stefan Troschuetz

19-Jan-07 8:59

Try putting two slashes before itemlist: <xsl:template match="//itemlist">

cdnhype19-Jan-07 9:02

19-Jan-07 9:02

sorry but that had no impact.

Thanks for the quick replies!

cdnhype19-Jan-07 9:14

19-Jan-07 9:14

I got it working now!!!

Heres is what I needed to change in the XSL:

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:t="urn:ahw:itemlist"
exclude-result-prefixes="t">
<xsl:template match="t:itemlist">
<TABLE BORDER="2">
<TR>
<TD>Display Name</TD>
<TD>Name</TD>
</TR>
<xsl:for-each select="t:item">
<TR>
<TD><xsl:value-of select="@friendly-name"/></TD>
<TD><xsl:value-of select="@name"/></TD>
</TR>
</xsl:for-each>
</TABLE>
</xsl:template>
</xsl:stylesheet>