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Re: Display a image in picturebox

kpuneeth7

0:53 24 Feb '10

Thanks for the code.... BUt its giving indexoutofrange exception...... HElp
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Re: Display a image in picturebox

Stuart Dootson

1:08 24 Feb '10

You need to give a command line argument that specifies the XML file...the code comments DO say that.... Java, Basic, who cares - it's all a bunch of tree-hugging hippy crp* CodeProject MVP for 2010 - who'd'a thunk it!
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Re: Display a image in picturebox

kpuneeth7

2:26 24 Feb '10

Thanks dude..... I got it Thank u very much
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Multiple group using Muenchian Method [modified]

ONeil Tomlinson

5:15 23 Feb '10

I have the following input XML. I want to group all unique Code/ID combination.

<List>
 <Section>
  	<Code>AA</Code>
  	<ID>11</ID>
 <Section>
 <Section>
  	<Code>AA</Code>
  	<ID>11</ID>
 <Section>
 <Section>
  	<Code>CC</Code>
  	<ID>11</ID>
 <Section>
 <Section>
  	<Code>AA</Code>
  	<ID>22</ID>
 <Section>
 <Section>
  	<Code>AA</Code>
  	<ID>11</ID>
 <Section>
<List>

So result should be as shown below.

<Code>AA</Code>
<ID>11</ID>
 
<Code>CC</Code>
<ID>11</ID>
 
<Code>AA</Code>
<ID>22</ID>

Im using the Muenchian Method. Currely my XSL (below) is grouping by Code (and not Code/ID combination)

<?xml version="1.0" encoding="UTF-16"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" xmlns:var="http://schemas.microsoft.com/BizTalk/2003/var" exclude-result-prefixes="msxsl var" version="1.0" xmlns:ns0="http://A4C.Interface.HRP.Payroll.Schemas">
	<xsl:output omit-xml-declaration="yes" method="xml" version="1.0" />
 
	<xsl:key name="CCids" match="/List/Section" use="Code"/>
 
	<xsl:template match="/">
		<xsl:apply-templates select="/ns0:Payroll" />
	</xsl:template>
	<xsl:template match="/ns0:Payroll">
		<ns0:List>
			<xsl:for-each select="/List/Section [generate-id(.)=generate-id(key('CCids',Code))]">
				<Section>
					<Code>
						<xsl:value-of select="code/text()" />
					</Code>
					<ID>
						<xsl:value-of select="id/text()" />
					</ID>
				</Section>
			</xsl:for-each>
		</ns0:List>
	</xsl:template>
</xsl:stylesheet>

How do i get this to use a Code/ID combination? I tried nested for-each loop but no luck. Thanks

modified on Tuesday, February 23, 2010 11:40 AM

Re: Multiple group using Muenchian Method

Stuart Dootson

21:47 23 Feb '10

Firstly, your key needs to have an expression which reflects what you want to retrieve:

   <xsl:key name="CCids" match="/List/Section" use="concat(Code,'|',ID)"/>

And then you need to use that same expression when accessing items referenced by the key.

Oh - and in addition, I would suggest not using a for loop - it's not idiomatic XSL. Here's the file I'd use:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" xmlns:var="http://schemas.microsoft.com/BizTalk/2003/var" exclude-result-prefixes="msxsl var" version="1.0" xmlns:ns0="http://A4C.Interface.HRP.Payroll.Schemas">
   <xsl:output omit-xml-declaration="yes" method="xml" version="1.0" encoding="utf-8"/>
 
   <xsl:key name="CCids" match="/List/Section" use="concat(Code,'|',ID)"/>
 
   <xsl:template match="/">
      <ns0:List>
         <xsl:apply-templates select="/List/Section [generate-id(.)=generate-id(key('CCids',concat(Code,'|',ID)))]"/>
      </ns0:List>
   </xsl:template>
            
   <xsl:template match="Section">
      <Section>
         <Code>
            <xsl:value-of select="Code/text()" />
         </Code>
         <ID>
            <xsl:value-of select="ID/text()" />
         </ID>
      </Section>
   </xsl:template>
</xsl:stylesheet>

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

CodeProject MVP for 2010 - who'd'a thunk it!

Re: Multiple group using Muenchian Method

ONeil Tomlinson

1:05 24 Feb '10

Thanks..Really helpful
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How Do I deserialize the following Xml string? [modified]

ashishtango

0:05 23 Feb '10

Hi,
I am getting this error There is an error in XML document (1, 2)
When i Deserialize following Xml.

<EnvironmentConfigurations>
<EnvironmentConfiguration>
<EnvironmentName>ABC</EnvironmentName>
<CMSConfiguration>
<Address>net.tcp://localhost:9000/Manager</Address>
<Binding>netTcpBinding</Binding>
<Contract>Common.Interfaces.Server</Contract>
</CMSConfiguration>
</EnvironmentConfiguration>
<EnvironmentConfiguration>
<EnvironmentName>XYZ</EnvironmentName>
<CMSConfiguration>
<Address>net.tcp://localhost:9000/Manager</Address>
<Binding>netTcpBinding</Binding>
<Contract>Common.Interfaces.Server</Contract>
</CMSConfiguration>
</EnvironmentConfiguration>
</EnvironmentConfigurations>

modified on Tuesday, February 23, 2010 5:11 AM

Re: How Do I deserialize the following Xml string?

Stuart Dootson

0:54 23 Feb '10

Deserialize to what? The way in which it's deserialized depends on the type you think it is. Are you sure this was serialized from .NET? That's not the only way to read and write XML, you know - maybe you should be using an XMLReader and parsing the XML yourself? Java, Basic, who cares - it's all a bunch of tree-hugging hippy crp* CodeProject MVP for 2010 - who'd'a thunk it!
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Re: How Do I deserialize the following Xml string?

ashishtango

17:52 23 Feb '10

I have serialized using XmlSerializer. I have removed Xml version and namespace information from the XML. I have changed the root node from <ArrayOfEnvironmentConfigurations> to <EnvironmentConfigurations> I want to desrialize this xml. Thank you.
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Re: How Do I deserialize the following Xml string?

Stuart Dootson

21:27 23 Feb '10

The answer to your problem is very, very clear - don't mess with the XML that the serializer produces - all of the things you changed are in there for a reason!!!! Java, Basic, who cares - it's all a bunch of tree-hugging hippy crp* CodeProject MVP for 2010 - who'd'a thunk it!
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Re: How Do I deserialize the following Xml string?

ashishtango

19:35 24 Feb '10

The problem was with the root node element which i changed. I kept the root name as ArrayOfEnvironmentConfigurations and the problem was solved. Thanks for replying.
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Accessing a image from a xml file and display in visual studio picture box

kpuneeth7

22:15 22 Feb '10

Hi everyone, I came across this prob... I am a beginner in xml. I wanted to create a xml file which stores the location of the images. This xml file must be accessed in visual studio picture box and it must display each image wen i click the next button and back wen i press the previous button... Please help
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Re: Accessing a image from a xml file and display in visual studio picture box

Richard MacCutchan

23:31 22 Feb '10

kpuneeth7 wrote:
I am a beginner in xml.

Then you should study some of the tutorials here and on MSDN on the use of XML. From the remainder of your question it appears that you are asking for a complete image viewer solution. If you have no idea how to start on this problem then you have a lot of studying still to complete.

txtspeak is the realm of 9 year old children, not developers. Christian Graus

xsd to entity [modified]

pokiri

18:45 22 Feb '10

<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema" id="BookEasy">

<xsd:element name="BookEasy">

<xsd:complexType>

<xsd:sequence>

<xsd:element name="Control">

<xsd:complexType>

<xsd:sequence>
<xsd:element name="ControlID" type="xsd:int"/>

<xsd:element name="ReservationRequest">

<xsd:complexType>

<xsd:sequence>
<xsd:element name="InDate" type="xsd:dateTime"/>
<xsd:element name="Period" type="xsd:int"/>
<xsd:element name="Adults" type="xsd:int"/>
<xsd:element name="Children" type="xsd:int"/>
<xsd:element name="Infants" type="xsd:int"/>
</xsd:sequence>
</xsd:complexType>
</xsd:element>

<xsd:element name="Operators">

<xsd:complexType>

<xsd:sequence>

<xsd:element name="Operator" minOccurs="0" maxOccurs="unbounded">

<xsd:complexType>

<xsd:sequence>
<xsd:element name="OperatorID" type="xsd:int"/>

<xsd:element name="Rooms">

<xsd:complexType>

<xsd:sequence>

<xsd:element name="Room" minOccurs="0" maxOccurs="unbounded">

<xsd:complexType>

<xsd:sequence>
<xsd:element name="RoomID" type="xsd:int"/>

<xsd:element name="Availability">

<xsd:complexType>

<xsd:sequence>
<xsd:element name="IsAvailable" type="xsd:boolean"/>

<xsd:element name="Days">

<xsd:complexType>

<xsd:sequence>

<xsd:element name="Day" maxOccurs="unbounded">

<xsd:complexType>

<xsd:sequence>
<xsd:element name="Date" type="xsd:dateTime"/>
<xsd:element name="IsAvailable" type="xsd:boolean"/>

<xsd:element name="Rates">

<xsd:complexType>

<xsd:sequence>
<xsd:element name="FullRate" type="xsd:decimal"/>
<xsd:element name="LastMinuteRate" type="xsd:decimal"/>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:sequence>
</xsd:complexType>
</xsd:element>
</xsd:schema>

modified on Tuesday, February 23, 2010 12:29 AM

Re: xsd to entity

Mark Nischalke

22:01 22 Feb '10

Do you have a question? I know the language. I've read a book. - _Madmatt
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How to delete duplicate nodes from XML throgh C#?

deadlyabbas

20:12 18 Feb '10

Hi, I want to delete the duplicate and some hard coded XML node in XMLDocument throgh C#. What is the method to do the same? I am new to the use of XML in C#. Please help me out.....!!!!!!!!!!! Thanks
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Problem with MSXML6 + namespace + XPath

Patrice.Espie

5:57 18 Feb '10

Hi All,

I try to use MSXML with xml files which contains namespaces. But I can't figure out how to do!

Here is an excerpt of my XML file:

<?xml version="1.0" encoding="utf-8"?>
<BENCH xmlns:VODPC_VOD="urn:VODPC:VOD:schema:20060809">
<VODPC_VOD:GRAPH state="running" currentFlowID="1" removeOnExit="FALSE">
</VODPC_VOD:GRAPH>
</BENCH>

When I load the file with MSXML6, no problem.
I try to access node BENCH with XPath "/BENCH" : no problem.
But when I try to access node VODPC_VOD:GRAPH with XPath "/BENCH/VODPC_VOD:GRAPH", I get an exception, which contain no valuable information:

First-chance exception at 0x75ac9617 in XXXXXX.exe: 0xE0000001: 0xe0000001.
First-chance exception at 0x75ac9617 in XXXXXX.exe: 0xE0000001: 0xe0000001.
First-chance exception at 0x75ac9617 in XXXXXX.exe: 0xE0000001: 0xe0000001.
First-chance exception at 0x75ac9617 in XXXXXX.exe: 0xE0000001: 0xe0000001.
First-chance exception at 0x75ac9617 in XXXXXX.exe: Microsoft C++ exception: _com_error at memory location 0x0025dac4..

I set these flags before to load:

m_pXMLDoc->async = VARIANT_FALSE;
m_pXMLDoc->preserveWhiteSpace = VARIANT_TRUE;
m_pXMLDoc->validateOnParse = VARIANT_TRUE;
m_pXMLDoc->resolveExternals = VARIANT_TRUE;

As you can understand, I work in C++. I'm an IT profesionnal since 18 year.

My problem is that I do not know what namespace is used in the file, so I don't want to have to tell MSXML6 what namespaces are used IN THE FILE. I guess it's its job, no ?

If anybody have an idea ...
Thanks a lot

Patrice

Re: Problem with MSXML6 + namespace + XPath

Stuart Dootson

1:10 23 Feb '10

You have to tell MSXML what namespace prefixes your XPath expressions use. When parsing XML, the parser remmebers the namespace URIs, NOT the prefixes used in the documents. That means that your code has to define the prefix to URI mappings used in your XPaths. Java, Basic, who cares - it's all a bunch of tree-hugging hippy crp* CodeProject MVP for 2010 - who'd'a thunk it!
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Problem with loading a xml to DataSet

alireza_shokoie

0:42 16 Feb '10

Hi all,

I've xml string and I can load it with XmlDataDocument.LoadXml successfully,
so that OuterXml shows the real xml string. But when I want to load in a dataset, the result dataset has no table(Tables.Count=0).
Typical code is as follows:

...
XmlDataDocument Doc = new XmlDataDocument();

/*
string Str = "<Result><DN>22800661</DN><ID>1</ID><Status><Line>Idle</Line></Status></Result>";
*/

Doc.LoadXml(Str);
DataSet datasetResult = Doc.DataSet;
if(datasetResult.Tables.Count == 0) return;
...

and it returns that means datasetResult.Tables.Count is always 0.

Can anyone help me, what is wrong?

Thanks in advance

Re: Problem with loading a xml to DataSet

daveyerwin

3:51 18 Feb '10

Maybe this is what you want ?

string Str = "<Result><DN>22800661</DN><ID>1</ID><Status><Line>Idle</Line></Status></Result>";
                  System.IO.StringReader sr = new System.IO.StringReader(Str);
                  XmlDataDocument Doc = new XmlDataDocument();
                  Doc.DataSet.ReadXml(sr);
                  DataSet datasetResult = Doc.DataSet;
                  if (datasetResult.Tables.Count == 0) return;

XSL help

Member 3997606

1:13 7 Feb '10

Hello all you wonderful people



 

<?xml version="1.0" encoding="unicode"?>

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

<_locDefinition>

      <_locDefault _loc="locData"/>

      <_locTag _locAttrData="pad-left,pad-right">p</_locTag>

      <_locTag _locAttrData="height,width">textarea</_locTag>

      <_locTag _locAttrData="minsize,noindent">select</_locTag>

      <_locTag _locAttrData="size">progressbar</_locTag>

      <_locTag _locAttrData="size,maxlen">input</_locTag>

      <_locTag _locAttrData="value">osxml:update</_locTag>

</_locDefinition>

<xsl:template match="fve-bad-pcr-registers">

<osxml:text-mode-ui>

  <head>

    <footer color="RGBIXXXX"><xsl:text> ENTER=Continue</xsl:text></footer>

    <meta key="ENTER" action="continue"/>

  </head>

  <body background-color="XXXX" foreground-color="RGBX">

    <font foreground-color="RGBI">

    Windows BitLocker Drive Encryption Information<br/>

    </font>

    <br/>

    <br/>

    The system boot information has changed since BitLocker was enabled.<br/>

    <br/>

    You must supply a BitLocker recovery key to start this system.<br/>

    <br/>

    Confirm that the changes to the system boot information are authorized.<br/>

    <br/>

    If the changes to the system boot information are trusted, then suspend and resume BitLocker. This will reset BitLocker to use the new boot information.<br/>

    <br/>

    Otherwise restore the system boot information.<br/>

    <br/>

  </body>

</osxml:text-mode-ui>

</xsl:template>

 

if User presses continue it goes to next line, if user press the "B" key then it goes to "progress-bar", otherwise quit.

 

<xsl:template match="fve-bad-code-id">

How do I prompt the user, trap a keypress and act upon the input in XML? What scripting language would you recommend to do this?
What do the loc lines do?

I am a newbie! John

I Need VB.net help

Re: XSL help

Stuart Dootson

2:09 7 Feb '10

XML is not a prgramming language - it's more a data format. So…why are you using XML and what are you trying to achieve? Java, Basic, who cares - it's all a bunch of tree-hugging hippy crp* CodeProject MVP for 2010 - who'd'a thunk it!
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Using MSXML2.ServerXMLHTTP.4.0 and ASP

tommywick65

5:49 6 Feb '10

I have an asp page that request data from a remote site (API) via XML.

My code is as follows:

DataToSend = "http://jc.tee.id.com/api/2.0/"& Request("resource") &"?apikey=a0d523ddnfjfjkkfkfkf32c&format=HTML&pret typrint=true&links=resource-path-encoded&request_handler_uri=http%3A%2F%2Fwww.mysit e.com%2Fitems&styles=enhanced"

set xmlhttp = Server.CreateObject("MSXML2.ServerXMLHTTP.4.0")

xmlhttp.Open "GET", DataToSend, false

xmlhttp.setRequestHeader "Content-Type", "application/x-www-form-urlencoded"

xmlhttp.send

Response.Write xmlhttp.ResponseText

This code works fine and returns data with no problem. However on page where the API has AJAX in it, the AJAX doesnt function.

For example, executing my code I pull up a table of contents page:

A | B | C | D | E

apple
adam
aligator
acorn

If I click in the title names, it takes me to the proper URL with no problem. if I click on a navigation letter (which is suppose to execute AJAX on the API side and pull in new titles), it just reloads the page.

I am assuming its because of MSXML2.ServerXMLHTTP.4.0.

I tried switching to 6.0 but this did nothing as well. Am i not able to execute AJAX or something through this remote site??

Any help or insight anyone could give me would be greatly appreciated.
Thanks

XSL import another XSL on Client-Side.

yeob

21:22 4 Feb '10

I try XSL Transfrom on client-side using javascript. My XSL file has a import tag in itself, but it couln't find the another XSL file. I have to excute this transform on Client-Side for
some reason. How can I put the import href content
Here goes my test code. this page written with .NET

Javascript code on the webpage.
function Transform() {
var xsl = new ActiveXObject("MSXML.DOMDocument");
var xml = new ActiveXObject("MSXML.DOMDocument");
xsl.async = false;
xsl.load(dataXmlPath);
xsl.async = false;
xsl.load(styleXslPath);

var data = xml.transformNode(xsl);
}

number.xml (XML Data file)-----------------------
<?xml version='1.0'?>
<data>
<circle>
<radius>12</radius>
</circle>
<circle>
<radius>37.5</radius>
</circle>
</data>

calc.xsl (XSL Style file)------------------------
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:include href="calc2.xsl"/>
</xsl:stylesheet>

calc2.xsl (XSL import file)------------------------
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="data">
<circles>
<xsl:for-each select="circle">
<circle>
<xsl:copy-of select="node()"/>
<circumference>
<xsl:value-of select="radius"/>
</circumference>
</circle>
</xsl:for-each>
</circles>
</xsl:template>
</xsl:stylesheet>

Re: XSL import another XSL on Client-Side.

daveyerwin

5:19 5 Feb '10

<script> function Transform() { var xsl = new ActiveXObject("MSXML.DOMDocument"); var xml = new ActiveXObject("MSXML.DOMDocument"); xml.async = false; xml.load("number.xml"); xsl.async = false; xsl.load("calc.xsl"); return xml.transformNode(xsl); } alert(Transform()); </script>
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