XML / XSL

Tools for merging multiple xsd files

25-Sep-09 2:40

Why does it make any difference? Semantically, both forms are exactly the same.

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

cateyes9922-Sep-09 19:03

cateyes99

22-Sep-09 19:03

Hi, I've got a XML schema which is split to multiple XSD files. Now we need to merge them into one big XSD file. Anyone knows a good tools I can use to do it? Manually merging would be too cumbersome, and error-prone.

Thanks very much.

Stuart Dootson24-Sep-09 10:04

24-Sep-09 10:04

How about the XML Schema include tag[^]?

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

cateyes9924-Sep-09 16:38

cateyes99

24-Sep-09 16:38

I don't think this include tag will solve my problem. What I'm facing is that
- Case 1
I've got WSDL which reference multiple xsd files, I use soapUI to generate Web service, but soapUI cannot handle those multiple xsd references because the refrence URLs in all files are still referencing to web sites which are not available, unless I manually hack those URLs to reference local xsd files. But this is tedious, each time I have to do it when WSDL changes. So I need a tools to put all XSD file's content inside the WSDL file. So soapUI will accept it.

- Case 2
I also initially had a big XSD file, but client asked to split a particular path in the main XSD to a seperate XSD file, and removed this from the main XSD then, for example.

the initial Main XSD's struct:

-root node - 1st level node 1 - 2nd level node 1
                              - 2nd level node 2
           - 1st level node 2 - 2nd level node 3
                              - 2nd level node 4
                              - 2nd level node 5

now, after split, it become two xsd files：

-root node - 1st level node 1 - 2nd level node 1
                              - 2nd level node 2
           - 1st level node 2 - 2nd level node 5

-root node - 1st level node 2 - 2nd level node 3
                              - 2nd level node 4

the problem is that I use .net xsd.exe tool to generate classes from those xsd files. After splitting to 2 xsd files which are compensatory, when I use xsd.exe to generate classes again, it will get conflict classes for the root level and 1st level Frown | :(

so I need a tool to merge these XSD files into a single one again for generating classes.

Any ideas? Smile | :)

Member 248888319-Nov-09 0:30

Member 2488883

19-Nov-09 0:30

Did you get any tool, if you have any solution please share with me. i am also facing similar problem.

Thanks in advance

jepskiluv14-Jan-10 17:30

jepskiluv

14-Jan-10 17:30

Also eager to hear any solution to this..

Use XML code into MS Word 2007 document

cateyes9919-Jan-10 15:14

cateyes99

19-Jan-10 15:14

It's extremely unfortunate.
Obviously, there is severe lack of such a wonderful tool Frown | :-(

I wish XML Spy could provide this kind of feature ...

Jim Krueger19-Sep-09 6:28

Jim Krueger

19-Sep-09 6:28

I am a newbie to xml data manipulation.
I have Pocket PC Creations software and have created an inspection application.
I can export in XML, HTML, Excel as well. (here is a link to the product specifics):

http://www.pocketpccreations.com/skills/lecture.asp?CourseID=4&LectureID=4&Page=7

I am trying to figure out a way to create a paragraph word report with the information gathered by the pocket pc.

Is it possible to incorporate the data from the exported data into an MS Word document? If it is, can you point me in a direction for help?

Jimdandy1956

Reading into XML section and getting same value again

Gareth H19-Sep-09 0:41

Gareth H

19-Sep-09 0:41

I am reading in some XML using the below code. The problem is when i attempt to get out the values for each Item, i get the same values again, ie: Debug prints out

Code = A123
Quantity = 10
---
Code = A123
Quantity = 10

using (TextReader textReader = new StringReader(xml))
using (XmlReader xmlReader = new XmlTextReader(textReader))
{
    XPathDocument xPathDocument = new XPathDocument(xmlReader);

    //Create navigator to get nodes out
    XPathNavigator xPathNavigator = xPathDocument.CreateNavigator();
    XmlNamespaceManager xmlNamespaceManager = new XmlNamespaceManager(xPathNavigator.NameTable);
    xmlNamespaceManager.AddNamespace("ns", "OrderSchema");

    //Get the order items
    XPathExpression xPathExpressionItems = xPathNavigator.Compile("//ns:Order//ns:Items");
    xPathExpressionItems.SetContext(xmlNamespaceManager);

    XPathNodeIterator xPathNodeIteratorItems = xPathNavigator.Select(xPathExpressionItems);

    do
    {
        string code = xPathNodeIteratorItems.Current.SelectSingleNode("//ns:Code", xmlNamespaceManager).ToString();
        string quantity = xPathNodeIteratorItems.Current.SelectSingleNode("//ns:Quantity", xmlNamespaceManager).ToString();

        System.Diagnostics.Debug.WriteLine("code == " + code);
        System.Diagnostics.Debug.WriteLine("quantity == " + quantity);
        System.Diagnostics.Debug.WriteLine("----");

    }
    while (xPathNodeIteratorItems.MoveNext());
}

Heres the xml i am reading in:

<?xml version="1.0" encoding="UTF-8" ?>
<Order>
	<Items>
		<Item>
			<Code>A123</Code>
			<Quantity>10</Quantity>
		</Item>
		<Item>
			<Code>A456</Code>
			<Quantity>5</Quantity>
		</Item>
	</Items>
</Order>

Any ideas what i am doing wrong?

Regards,
Gareth.

(FKA gareth111)

Re: Reading into XML section and getting same value again

Stuart Dootson19-Sep-09 22:11

Re: Reading into XML section and getting same value again

19-Sep-09 22:11

The only thing that jumps out is that you use Current before calling MoveNext the first time - the documentation[^] says you need to call MoveNext before Current.

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

Lakhan Pal Garg22-Sep-09 19:08

Lakhan Pal Garg

22-Sep-09 19:08

Hi Gareth-

Please check this method:

private void ReadXMLData()
    { 
        XmlDocument xDoc=new XmlDocument();
        xDoc.LoadXml("<?xml version=\"1.0\" encoding=\"UTF-8\" ?><Order><Items><Item><Code>A123</Code><Quantity>10</Quantity></Item><Item><Code>A456</Code><Quantity>5</Quantity></Item></Items></Order>");
        XmlNodeList xNodeList = xDoc.GetElementsByTagName("Item");
        for (Int16 iCount = 0; iCount < xNodeList.Count; iCount++)
        {
            Response.Write(xNodeList[iCount].ChildNodes.Item(0).InnerText+ "<br/>");
            Response.Write(xNodeList[iCount].ChildNodes.Item(1).InnerText + "<br/>");
            Response.Write("-----<br/>");
        }
    }

Thanks & Regards
Lakhan Pal Garg
Free Code Snippets
http://lakhangarg.blogspot.com[^]

XSD problem

gantww18-Sep-09 10:09

gantww

18-Sep-09 10:09

I've created the following XML Schema (not for production use, but to learn by doing).

Here's what is burning me at the moment:
The xs:element elements that refer to classes or properties are giving me warnings that "Type 'class' is not declared". I'm still learning this stuff, so I'd really appreciate it if someone could take a quick peek at it and advise me of the (probably obvious) problem in the code below.

Thanks,
Will

<?xml version="1.0" encoding="utf-8"?>
<xs:schema targetNamespace="http://gantsoftwaresystems.com/classspec.xsd"
    elementFormDefault="qualified"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
>
	<xs:element name="classDefs">
		<xs:complexType>
		  <xs:sequence minOccurs="1" maxOccurs="unbounded">
			<xs:element name="classes">
			  <xs:complexType>
				<xs:sequence minOccurs="1" maxOccurs="unbounded">					
					<xs:element name="class" type="class"/>
				</xs:sequence>            
			  </xs:complexType>
			</xs:element>
			<xs:element name="views">
			  <xs:complexType>
				<xs:sequence>
					<xs:element name="view" type="class"/>
				</xs:sequence>            
			  </xs:complexType>
			</xs:element>
		  </xs:sequence>      
		</xs:complexType>
	</xs:element>
	<xs:complexType name="class">
		<xs:sequence minOccurs="1" maxOccurs="unbounded">
			<xs:element name="property" type="property"/>
		</xs:sequence>
		<xs:attribute name="name" type="xs:string"/>
		<xs:attribute name="base" type="xs:string"/>
		<xs:attribute name="hasImageUrl" type="xs:boolean"/>
	</xs:complexType>
	<xs:complexType name="property">
		<xs:sequence minOccurs="0" maxOccurs="unbounded">
			<xs:element name="alsoTriggerPropertyChanged">
				<xs:complexType>
					<xs:attribute name="propertyName" type="xs:string"/>
				</xs:complexType>
			</xs:element>
		</xs:sequence>
		<xs:attribute name="name" type="xs:string"/>
		<xs:attribute name="type" type="xs:string"/>
		<xs:attribute name="changesImageUrl" type="xs:boolean"/>
		<xs:attribute name="imageUrlQStringParamName" type="xs:string"/>
		<xs:attribute name="imageUrlOverridePropertyName" type="xs:string"/>
	</xs:complexType>
</xs:schema>

Re: XSD problem

Stuart Dootson19-Sep-09 22:04

19-Sep-09 22:04

You need to namespace qualify the type references - you declare them in your target namespace, but when you reference them, don't mention that namespace. You can either use explicit namespace references or (and this is easier) use your target namespace as the default namespace for the schema.

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

sending xml document

aamirzada16-Sep-09 18:31

aamirzada

16-Sep-09 18:31

i need some help on xml. i want to send an xml document to an xml web service using get method. but i didn't find any article, helping material. i did not use xml before. please help me.

thanx

amirzada

Voted to remove message

N a v a n e e t h16-Sep-09 18:45

N a v a n e e t h

16-Sep-09 18:45

Please stick one forum.

Navaneeth

How to use google | Ask smart questions

Re: sending xml document

Stuart Dootson16-Sep-09 21:28

16-Sep-09 21:28

You want to 'send' an XML payload with a GET…an amusing concept. Look at the verbs we're using SEND, GET - they're not compatible. You want to use a POST method to send an XML document to a server.

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

Re: sending xml document

aamirzada25-Sep-09 18:40

aamirzada

25-Sep-09 18:40

thanx for your help but now i get an exception when i try create an xml document it show that the body and html tag is not closed how can i handle can you help me please.
thanx

amirzada

Re: sending xml document

Stuart Dootson25-Sep-09 19:04

25-Sep-09 19:04

More detail - where does the HTML come in - that's NOT XML.

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

XML Validation using XSD

scrapper15-Sep-09 4:00

scrapper

15-Sep-09 4:00

I have a tool called "Flat file to XML converter" to convert the text file information to XML file using the XSD template available. Text file contains employee details. Whenever a name with apostrphe(') comes the tool converts it as ' instead of '. So whenever the XML file is decoded to view, it is showing the name with '. Please help me in this.
To make any changes in the tool is not possible and no XSLT is available for the XML file. Only XSD is there. I need to validate this ' and replace it with ' Is it possible?

Only a change in XSD is allowed. No seperate application/tool to convert the XML is allowed.

Example:-
flat fle contains the name- O'NEIL, JOHN.
XML file after conversion - O 'NEIL, JOHN.
Output needed - O'NEIL, JOHN

Stuart Dootson15-Sep-09 7:12

15-Sep-09 7:12

Validation DOES NOT alter the input XML - it just tells you if it meets the Schema or not.

Why does it matter, anyway? Any conforming XML reader will convert the ' entity to an apostrophe when it reads it!

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

scrapper16-Sep-09 3:32

scrapper

16-Sep-09 3:32

But whenever a name with apostrophe comes the "flat file to XML" converter convert it to ' So the XML parser can't change it to apostrophe when it process it. What do i do to avoid it?

Stuart Dootson16-Sep-09 3:39

16-Sep-09 3:39

The XML parser will interpret the ' as an apostrophe and store it as such internally. The apostrophe is only converted to ' when written to the XML file.

The XML Standard[^] says that the apostrophe may be stored as ' to permit use in attribute values. By always translating an apostrophe to ' you make sure the apostrophe is always in a readable form.

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

scrapper16-Sep-09 3:44

scrapper

16-Sep-09 3:44

sorry friend, i meant whenever a name with apostrophe comes the "flat file to XML converter" converts it to &";apos instead of ' (please omit the double quotes). So the XML parser can't parse it to &apos.

Stuart Dootson16-Sep-09 21:27