XML / XSL

Re: Space in attribute name

9-Nov-09 23:01

If you have spaces in element names or in attribute names, then what you have **IS NOT XML**. XML doesn't allow spaces in element or attribute names. See the specification for names[^] (which is used for element and attribute names) in the XML spec.

So therefore, your question can only be answered logically with this statement: get rid of the spaces form your attribute names.

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

tassilo10-Nov-09 23:06

tassilo

10-Nov-09 23:06

Aniruddha,

Just check the spec, there is no such thing as attribute names with spaces. Validation of a corresponding document should fail.

Best regrads,

Stefan

How can I traverse and read values in an XML using VBScript?

Curtis Rowell6-Nov-09 1:03

Curtis Rowell

6-Nov-09 1:03

Given the xml snippet below, I am trying to read the Company value. I have other code which will obtain the Network Policy Zone and the Default Gateway of a given system. Using these two values, zone first, then dfgw, I need to get the Company value which correspondes so it can be set as an environment variable in a remote windows system.

Here's my mess of code, and I'm afraid it is a mess because I've been trying different things:

Sub ReadXML()
	Dim attrValue, child1, child2, gChild1
	getZone()
	WScript.Echo "Zone is: " & zone
		
	Set xDoc = CreateObject("Microsoft.XMLDOM")  'CreateObject("Msxml.DOMDocument") CreateObject("Microsoft.XMLDOM") 
	xDoc.async = False
	xDoc.load ("company.xml")  'Load XML document
	Set Root = xDoc.documentElement
	Set NodeList = Root.getElementsByTagName(zone)
	
	Set child1 = xDoc.documentElement.firstChild 
	Set child2 = child1.nextSibling 
	Set gChild1 = child1.firstChild
	
	Do Until attrValue <> dfgw
	attrValue  = xDoc.GetAttrValue("DFGW")
	WScript.Echo("DFGW Attribute Value is: " & attrValue)
	xDoc.nextSibling
	Loop
	
	company = gChild1.nodeValue("Company")
	WScript.Echo("Company Attribute Value is: " & attrValue)
	
	
	'Set ElemList = xDoc.getElementsByTagName("Zone0")
	'dfgw = ElemList.item(0).getAttribute("DFGW")
	
	WScript.Echo("Default Gateway: " & dfgw)
	'company = "BMC Software, Inc."
	SetCompany(company)
		
End Sub

Here's the XML I'm trying to read:

<br />
  <?xml version="1.0" ?> <br />
- <ZoneList><br />
- <Zone0><br />
- <DFGW><br />
  "192.168.25.1" <br />
  <Company>"CPC"</Company> <br />
  </DFGW><br />
- <DFGW><br />
  "142.101.230.1" <br />
  <Company>"Test Company"</Company> <br />
  </DFGW><br />
- <DFGW><br />
  "" <br />
  <Company>""</Company> <br />
  </DFGW><br />
- <DFGW><br />
  "" <br />
  <Company>""</Company> <br />
  </DFGW><br />
- <DFGW><br />
  "" <br />
  <Company>""</Company> <br />
  </DFGW><br />
  </Zone0><br />
- <Zone1><br />

Re: How can I traverse and read values in an XML using VBScript?

Stuart Dootson9-Nov-09 4:08

XSD Unique Implementation and XPath

9-Nov-09 4:08

XPath?

Here's some VBscript that uses XPath to select the nodes you want

Dim attrValue, child1, child2, gChild1
zone = "Zone0"
dfgw = """192.168.25.1"""
WScript.Echo "Zone is: " & zone
WScript.Echo "dfgw is: " & dfgw
	
Set xDoc = CreateObject("Microsoft.XMLDOM")  'CreateObject("Msxml.DOMDocument") CreateObject("Microsoft.XMLDOM") 
xDoc.async = False
xDoc.load ("a.xml")  'Load XML document

xpathString = "//" & zone & "/DFGW[text() = '" & dfgw & "']/Company"
WScript.Echo xpathString

Set CompanyNodeList = xDoc.selectNodes(xpathString)
WScript.Echo CompanyNodeList.Length

You just need to be sure that the string values are correctly escaped when building the query.

Also - why the quotes around the string values?

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

Xml File Problems

md_refay4-Nov-09 4:19

md_refay

4-Nov-09 4:19

i have xml file
iam having c# application
i need to
1-read xml file in c# abd pass it as parameter to sql server store procedure (parameter type xml)
2- use procedure have xml parameter to insert data into table

md_refay

Re: Xml File Problems

The Man from U.N.C.L.E.11-Nov-09 11:45

The Man from U.N.C.L.E.

11-Nov-09 11:45

Not to put to fine a point on it, what is stopping you?
I'm not seeing a question or a problem here, just a task you have to do.

If you are having problems with your code post the problem code and errors and we can help.

If you have knowledge, let others light their candles at it.
Margaret Fuller (1810 - 1850)
www.JacksonSoft.co.uk

Skippums2-Nov-09 12:33

Skippums

2-Nov-09 12:33

Firstly, if anyone could send me a link that explains everything that is allowed in XSD xpath expressions, I would be very grateful. I have searched for hours, and still have no idea what subset of xpath is allowed for either the selector or the field nodes. Great thanks to anyone who knows where I can find some documentation on this feature of xsd.

On to my problem... I am attempting to have an element, "parent", that contains a key attribute called "key". Every such element may have 0 or more child elements, called "Child", such that they contain a keyref attribute called "key" as well. I am wondering if there is a way to enforce through the XSD that the value of "key" for the child elements cannot be equivalent to the value of "key" for the parent node. So, for example:

<xsd:element name="Example" type="ExampleType">
    <xsd:key name="ParentKey">
        <xsd:selector xpath="Parent"/>
        <xsd:field    xpath="@key"  />
    </xsd:key>
    <xsd:keyref name="ChildKeyRef" refer="ParentKey">
        <xsd:selector xpath="Parent/Child"/>
        <xsd:field    xpath="@key"        />
    </xsd:keyref>
</xsd:element>

<xsd:complexType name="ExampleType">
    <xsd:sequence>
        <xsd:element name="Parent" type="ParentType" maxOccurs="unbounded">
            <!-- This unique portion is where I need the help.  It almost works in that it prevents me from
                 giving a child element the parent's key, but it also has the unfortunate side effect of not
                 allowing me to have duplicate key references in the child elements.  How can this be done? -->
            <xsd:unique name="UniqueKeyRef">
                <xsd:selector xpath=".|Child"/>
                <xsd:field    xpath="@key"   />
            </xsd:unique>
        </xsd:element>
    </xsd:sequence>
</xsd:complexType>

<xsd:complexType name="ParentType">
    <xsd:attribute name="key" type="xsd:String" use="required"/>
    <xsd:sequence>
        <xsd:element name="Child" type="ChildType" minOccurs="0" maxOccurs="unbounded"/>
    </xsd:sequence>
</xsd:complexType>

<xsd:complexType name="ChildType">
    <xsd:attribute name="key" type="xsd:String" use="required"/>
</xsd:complexType>

<-- XML File -->

<Example>
    <Parent key="A">
        <Child key="B"/>
        <Child key="B"/> <!-- This is allowed, since it references a different element than it's parent -->
                         <!-- This duplicate is currently not allowed even though I want it to be.      -->
        <Child key="A"/> <!-- This is NOT allowed, because this element is a child of Parent "A"        -->
    </Parent>
    <Parent key="B"/>
</Example>

Hopefully this example has clearly illustrated my desire. Any number of children of Parent "A" can reference any other parent, including multiple children referencing the same parent (such as the two child elements referencing parent "B" in my example). Is there some way to rework my unique constraint, or by adding another constraint, such that these duplicates are allowed? Thanks,

Sounds like somebody's got a case of the Mondays

-Jeff

modified on Tuesday, November 3, 2009 10:19 AM

Christian Graus3-Nov-09 11:27

3-Nov-09 11:27

Not sure if this answers your question ( I didn't read it all ), but I've always used this[^] when I need to do XSLT/XPath stuff.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Stuart Dootson3-Nov-09 22:54

3-Nov-09 22:54

The Essential XML Quick Reference[^] is an excellent (and free) XML (+ XSD + XPath + XSLT) reference.

On page 349, it defines the available XPath selectors (I'd copy them here, but it disallows copying - grrrr).

However, I don't think you can do what you want with the 'unique' relation. The correctness 'uniqueness' set would be '..|.' using the Child element as the context node. But you can't do that with XML Schemas.

Using the Parent as context node, there's no way to define the uniqueness set, I don't think, as there are n sets, where n==number of Child nodes.

However - using Schematron[^], you can validate this rule using this Schema:

<schema xmlns="http://www.ascc.net/xml/schematron" > 
     <pattern name="Check CHild/Parent keys"> 
          <rule context="Child"> 
               <assert test="@key!=../@key">Child key should differ from Parent key.</assert> 
               <report test="@name">Child key should differ from Parent key.</report> 
          </rule> 
     </pattern> 
</schema>

This should be used in addition to a schema describing the basic structure of the XML. Using xmllint (which comes with libxml2), I can confirm that this Schematron schema meets your requirements.

XML files can be validated against Schematron schemas using an XSLT processor, so even if your XML parser doesn't have Schematron support, you can still use Schematron.

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

Skippums4-Nov-09 2:28

Skippums

4-Nov-09 2:28

Yeah, that is what I was afraid of... I know I could do it using XSD 1.1 or, as you suggested, Schematron, but validating against either of those options isn't well supported (at least as far as I can tell) in Java or C. I actually tried what you suggested with the parent element, saw it wasn't supported, and figured that there must just be some other well-hidden way to go about it. I don't want to have a second definition file, so embedded Schematron may be the answer right now (if I can figure out how to validate against it), or I could just wait until next April when the XSD 1.1 specification is finalized. Anyway, thank you very much for the help!

Sounds like somebody's got a case of the Mondays

-Jeff

Stuart Dootson4-Nov-09 2:52

Sorted xml element while serialising

4-Nov-09 2:52

Schematron is supported by libxml2, so (therefore) is accessible to C or C++ or anything else that exposes that part of libxml2 through a binding.

The Schematron website describes the steps needed to perform Schaematron validation:

xslt -stylesheet iso_dsdl_include.xsl  theSchema.sch > theSchema1.sch
xslt -stylesheet iso_abstract_expand.xsl  theSchema1.sch > theSchema2.sch
xslt -stylesheet iso_svrl_for_xsltn.xsl  theSchema2.sch > theSchema.xsl
xslt -stylesheet theSchema.xsl  myDocument.xml > myResult.xml

Basically, generate an XSLT file from your Schematron schema (the first three steps). Then process the input XML file using your XSLT processor and the generated XSLT file to output something that tells you if the XML is valid per the Schematron schema.

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

Soundman32.22-Nov-09 1:28

Soundman32.2

2-Nov-09 1:28

I am using c# serialisation to write some xml. For some reason the output is not necessarily written in the order it appears in the code.

How can I order the elements?

For example:

[Serializable]
public class Request
{
  [XmlElement("Type")]
  public string Type
  {
    get;
    set;
  }

  [XmlElement("Id")]
  public int Id
  {
    get;
    set;
  }
}

Would write

<Request>
 <Id>1</Id>
 <Type>Get</Type>
</Request>

(Note this example is very simplified and doesn't actually write it wrongly).

Re: Sorted xml element while serialising

Soundman32.22-Nov-09 3:58

Soundman32.2

2-Nov-09 3:58

Fixed it.

It turns out, fields are serialised before properties and my code was actually like this:

[Serializable]
public class Request
{
  [XmlElement("Type")]
  public string Type
  {
    get;
    set;
  }

  [XmlElement("Id")]
  public int Id;   // <<<--- Note this is a field, not a property
}

attribute centric

Michael A. Klausmeyer1-Nov-09 8:27

Michael A. Klausmeyer

1-Nov-09 8:27

I have just been successfull at generating an element centric xml query using linq to my sqlserver database. Can someone show me an example how to do it to generate an attribute centric result?

I also am able to save the results to a file on my c drive but need to save it to an xml file in my project.

Thank you in advance.

How to Password Protect a XML SpreadSheet

ANSHUMAN44429-Oct-09 5:14

ANSHUMAN444

29-Oct-09 5:14

I am generating a Excel Sheet(XML SpraedSheet) through XSLT transform.How to password protect the XML SpreadSheet in XSLT.

Is there is any tag to password Protect?

Is there is any other solutions to password protect the XML Spreadsheet?

Best Regards

Re: How to Password Protect a XML SpreadSheet

Christian Graus1-Nov-09 9:30

C# XML datagridview delete row

1-Nov-09 9:30

It's as if you understand XSLT, and yet somehow have no idea what XML is. How is that possible ?

XML is a string. You can write external code to encode the entire string ( meaning it won't be XML anymore ), or to encode the contents of the nodes. You'd then need corresponding code to decode the information. You can invent a 'password protect tag', but it's plainly not going to alter the fact that all you have is a plain text file.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Wheels01227-Oct-09 3:38

Wheels012

27-Oct-09 3:38

Good morning.

I am populating a datagridview with a dataset from an XML file.

I now want to write some code to delete a row on the datagridview to simulate a user clicking on the row and hitting the Delete button.

Any ideas on how to accomplish this?

Thank you, WHEELS

C# Append XML File

Wheels01223-Oct-09 5:42

Wheels012

23-Oct-09 5:42

Good morning.

I have the following simple XML file.

<?xml version="1.0" standalone="yes"?>
<SHRPT_SITE>
  <RECORD>
    <Name>Apps (doc)</Name>
    <Address>http://unumtfsprod/sites/IServices/Apps/Forms/AllItems.aspx</Address>
  </RECORD>
  <RECORD>
    <Name>Internet Services Architecture (arch)</Name>
    <Address>http://teamspace/sites/iservices/architecture/default.aspx</Address>
  </RECORD>
  <RECORD>
    <Name>iServices Customer Self Service (staff)</Name>
    <Address>http://teamspace/sites/isgcrm/custselfsvc/default.aspx</Address>
  </RECORD>
</SHRPT_SITE>

I want the user to be able to click a (favorites) button and be able to add a site name and URL to the file. What would be the best code approach to accomplish this?

Thank you, WHEELS

Re: C# Append XML File

Christian Graus23-Oct-09 10:35

Is it possible to call a comand line program from within an Xml File Preprocessor

23-Oct-09 10:35

Use an XmlDocument to create a new record, and then append it to the shrpt_site node.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Re: C# Append XML File

Wheels01226-Oct-09 1:24

Wheels012

26-Oct-09 1:24

Thank you Christian.

ONeil Tomlinson23-Oct-09 1:09

ONeil Tomlinson

23-Oct-09 1:09

Is it possible to call a comand line program from within an Xml File Preprocessor. I have a connection string that is encrypted in a settings file and i want to decrypt it before assigning it to the conString value

<configuration>
  <appSettings>
    <!-- ifdef ${_xml_preprocess} -->
    <!-- <add key="conString" value="${connectionStringWithPassowrd}"/> -->
    <!-- else -->
    <add key="conString" value="connection string"/>
    <!-- endif -->
  </appSettings>
</configuration>

Re: Is it possible to call a comand line program from within an Xml File Preprocessor

Christian Graus23-Oct-09 10:44