Re: Creating XPath from given node - XML / XSL Discussion Boards

Re: Fop problems

Dustin Metzgar18-Jul-06 11:47

Dustin Metzgar

18-Jul-06 11:47

Awesome! Glad to help out.

Logifusion[^]

Re: Fop problems

Support12311-Jul-06 5:49

Support123

11-Jul-06 5:49

Okay no, The xsl is to long now... Smile | :)

He who laughs last...

Crystal report exported xml

Naveed Kamboh10-Jul-06 3:06

Naveed Kamboh

10-Jul-06 3:06

Hi Friend,

I have exported a crystal file as an XML file.
Now I want to read that file using ASP.net.
Does any one help me how to read Crystal Exported XML file using ASP.NET.?

Thanks

Naveed Kamboh

Population of .NET property grid [modified]

tom groezer9-Jul-06 20:17

tom groezer

9-Jul-06 20:17

I'm trying to develop an application wherein I have an XML file. The contents of this XML file should be dynamically converted into a C# class(with properties). I learnt when the contents of a C# file object ie properties are given to Microsoft Property Grid in .NET, the fields in the grid are populated with the C# class properties.

Lets say for XML1 there is one property grid UI and for XML2 there is another property grid UI. The point that I'm trying to stress is that depending on the XML a particular UI comes up.

The whole idea behind the above design is to show XML data to the user and when the user edits the data from the grid the XML file gets updated.

I was exploring the idea of developing a wizard in VC8 that should address this situation. The wizard shall be similar to something like that in VC applications. On invoking it shall ask for the kind of XML file that the Property Grid should be generated for.Depeding on that somewhere there should be logic to convert this XML input to a C# file so that by the time the wizrd is finished wiht poicking the XML file the property grid is already populated.

I know of some of the tools that are available that address this situation. Any insight on this front shall be highly appreciated.

Thanks

-- modified at 2:17 Monday 10th July, 2006

Creating XPath from given node

mhadamji7-Jul-06 1:48

mhadamji

7-Jul-06 1:48

Hi guys,

I am validating my xml with schema.
the validation rotine return line number and line position, if the validation fails.

given the line number,line position i can get to node that fails the validation test using IXmlLineInfo.

see http://msdn2.microsoft.com/en-us/library/system.xml.ixmllineinfo.aspx

I need to build the xpath from that node

any ideas are helpfull

Thanks

Mustafa
http://www.hostukdomain.com

Re: Creating XPath from given node

Dustin Metzgar7-Jul-06 3:37

Dustin Metzgar

7-Jul-06 3:37

What I would suggest is that you take advantage of the position() function. The problem you'll be up against is that you may have a lot of similar nodes. So, if you have XML that looks like this:

<Foo>
   <Bar>9</Bar>
   <Bar>7</Bar>
   <Bar>asd</Bar>
</Foo>

Maybe the asd is the part that fails the schema, so you need an XPath that gets to the third Bar node. You could write something like this:
/Foo/Bar[position()=3]

If you know the node that fails, you should be able to get an XPath to it.

Logifusion[^]

Re: Creating XPath from given node [modified]

mhadamji7-Jul-06 5:29

mhadamji

7-Jul-06 5:29

<Thanks Dustin

Indeed a very similar position coz i could have multiple nodes of same name.
I am thinking abt ur solution but during runtime i might not know the node position that could fail.

sample xml

<root>
<level1>
<nest11</nest1>
<nest2>2</nest2>
</level1>
<level2>
<nest1>1</nest1>
<nest2>some bad value</nest2>
</level2>
</root>

then i should run some rotine to return /root/level2/nest2

i will try position() function but if can throw more light will be helpfull.

cheers

Mustafa
http://www.hostukdomain.com[^]

-- modified at 11:30 Friday 7th July, 2006

Re: Creating XPath from given node

Dustin Metzgar7-Jul-06 5:35

Dustin Metzgar

7-Jul-06 5:35

Well, first you have to change the <'s and >'s to <'s and >'s so I can see your XML. But I would say if you have the actual XmlNode object representing the node that has the error, you could get its parent. With the parent node, search through its child nodes to find the current node you're holding. You should be able to get an index. Add 1 to it and get the position.

Logifusion[^]

Re: Creating XPath from given node [modified]

mhadamji7-Jul-06 6:35

mhadamji

7-Jul-06 6:35

hope you can see the xml now

I will try what you have suggested... just to let you know i am using XmlValidatingReader to derive the IXmlLineInfo which has the lineinfo

I will try and see if I could use XPathDocument or XMLDataDocument but not sure if i can get the node using the lineinfo

will try and post back..

all have nice weekend

cheers

Mustafa
http://www.hostukdomain.com[^]

-- modified at 12:35 Friday 7th July, 2006

FOP it [modified]

Support1236-Jul-06 22:48

Support123

6-Jul-06 22:48

Hi All,

I'm trying to get my flow right with the xml and xsl files...

is it possible to do the following...
my xml file looks like this...
<ClientLookUp>
<Title>
<Heading>WhatEver</Heading>
<Date>TodayOrWhatEver</Date>
</Title>
<TableEntries key="0">
<ClientName>exampleJohn</ClientName>
<ClientSurname>exampleSmith</ClientSurname>
<ClientAge>example21</ClientAge>
<ClientID>example8501014341988</ClientID>
<ClientType>WhateEver</ClientType>
<ClientBday>example01January</ClientBday>
<ClientTelH>example903-0201</ClientTelH>
<ClientFaxH>example903-0202</ClientFaxH>
<ClientEmailH>exampleJohn@Home.co.za</ClientEmailH>
<ClientTelW>example903-0203</ClientTelW>
<ClientFaxW>example903-0204</ClientFaxW>
<ClientEmailW>exampleJohn@Work.co.za</ClientEmailW>
<ClientCell>example903-0205</ClientCell>
</TableEntries>
<TableEntries key="1">
<ClientName>exampleSam</ClientName>
<ClientSurname>exampleLooker</ClientSurname>
<ClientAge>example22</ClientAge>
<ClientID>example8501014341982</ClientID>
<ClientType>WhateEver</ClientType>
<ClientBday>example01January</ClientBday>
<ClientTelH>example903-0301</ClientTelH>
<ClientFaxH>example903-0302</ClientFaxH>
<ClientEmailH>exampleSam@Home.co.za</ClientEmailH>
<ClientTelW>example903-0303</ClientTelW>
<ClientFaxW>example903-0304</ClientFaxW>
<ClientEmailW>exampleSam@Work.co.za</ClientEmailW>
<ClientCell>example903-0305</ClientCell>
</TableEntries>
</ClientLookUp>

is it possible to, Depending on whitch of these tags are created, that they could follow up on each other whith spaces between them?

He who laughs last...

-- modified at 4:48 Friday 7th July, 2006

Re: FOP it

Support1236-Jul-06 22:57

Support123

6-Jul-06 22:57

Like for instance....

if only the ClientName and the ClientTelH was selected in my program and i created the ClientName and the ClientTelH tags only...

i want it to display like this...
exampleJohn 903-0201
exampleSam 903-0301

better yet
can i create a tabel and the cell only as tags are created? to be able to align the output?

for instance...
exampleJohn 903-0201
exampleSam 903-0301

any help whould be greatly appreciated.
Thank you in advance...

He who laughs last...

Re: FOP it [modified]

Anh_Tuan7-Jul-06 1:58

Anh_Tuan

7-Jul-06 1:58

Use

<xsl:for-each select="//ClientLookUp/TableEntries">

to iterate thru each entry and

use

<xsl:if test="">

to find out if the node exist then dump the data as you want

Jup

-- modified at 11:03 Friday 7th July, 2006

Re: FOP it

Dustin Metzgar7-Jul-06 3:28

Dustin Metzgar

7-Jul-06 3:28

Can you modify your post so that you change all the <'s to <'s and so forth? Nobody can see what's in your post.

Logifusion[^]

Re: FOP it

Dustin Metzgar7-Jul-06 5:31

Dustin Metzgar

7-Jul-06 5:31

Thanks!

Logifusion[^]

Replace node value with multiple line text???

JimFeng5-Jul-06 3:14

JimFeng

5-Jul-06 3:14

I need replace node value with multiple line text in a Word XML Document with ASP.net 2.0 application. I tried to use InnerText method in XMLNode object but it failed:

oNode.InnerText = “Address line 1 “ & chr(13) & “ Address Line 2 “ & chr(13) & “ City, State: 12345”
Output on the document: “Address line 1 Address Line 2 City, State: 12345”. I also tried the following:
oNode.InnerText = “<w:p><w:t>Address line 1 <w:p><w:t> Address Line 2 <w:p><w:t> City, State: 12345”
Output on the document: “<w:p><w:t>Address line 1 <w:p><w:t> Address Line 2 <w:p><w:t> City, State: 12345”
Any help is appreciated.

Re: Replace node value with multiple line text???

Dustin Metzgar7-Jul-06 3:30

Dustin Metzgar

7-Jul-06 3:30

Maybe try these characters for a newline: 


Logifusion[^]

Creating a typed dataset of MS Project 2003 XML files

anderslundsgard5-Jul-06 1:35

anderslundsgard

5-Jul-06 1:35

Where can I find the typed dataset for MS Project XML files?
Iv’e tried to create my own but when trying to read an MS Project XML-file like this:

MSProjectDataSet dataset = new MSProjectDataSet();<br />
dataset.ReadXML("savedXMLfileFromProject.xml");

I get different exceptions.

All tips in this topic are welcomed!!

_____________________________

...and justice for all

APe

if statements...

Support1233-Jul-06 22:12

Support123

3-Jul-06 22:12

Hi all,

How do you use the if statements in a stylesheet?

The thing i want to do is to check if a tag in the xml file exists and if it does'nt, then combine 2 cells.

i think it should be something like this...

<xsl:if (does the "test=" part here mean i have to insert the "Check if the tag exists?" statement)>

He who laughs last...

Re: if statements... [modified]

Elina Blank4-Jul-06 5:23

Elina Blank

4-Jul-06 5:23

<code><xsl:if test="*[self::elementName]"> or

<xsl:if test="*[name() = 'elementName']"></code>

Sincerely,
Elina

Life is great!!!
Enjoy every moment of it! Blush | :O

-- modified at 11:24 Tuesday 4th July, 2006

Re: if statements...

Support1234-Jul-06 19:22

Support123

4-Jul-06 19:22

Thank you

He who laughs last...

XML with xsd

anil@kalkitech.in3-Jul-06 21:06

anil@kalkitech.in

3-Jul-06 21:06

hi
I am using MS XML 4.0 and VC++.

When I tried to load the XML file with XSD referecvv , my XPAth queries are returning NULL .

What could be reason?

Please help me.

thanks in advance

XML search

dzthomas3-Jul-06 6:03

dzthomas

3-Jul-06 6:03

Hi,

I'm looking to search through child nodes, and then capture the value of the node and also that of the parent node. For example, the user enters details into a field, which then gets checked against the child node, if matched then it gets the value of that node and also the parent value. Which then will be populated into a treeView

I'm trying to use XMLPathNavigator and it doesn't appear to be working. Any ideas? Or other ways to attempt this. I'm using c#.net 2005 to do this.

It sounds extremely easy, but just can't get it to work.
Cheers... Dan

Cheers.. Dan

Re: XML search

Igor Sukhov3-Jul-06 18:45

Igor Sukhov

3-Jul-06 18:45

Perform the search in XPathDocument or XmlDocument by using SelectSingleNode method. When element is found it is easy to navigate back to the parent node (ParentNode property).

Best regards,

-----------
Igor Sukhovhttp://sukhov.net

Problem using expressions

sishya2-Jul-06 20:06

sishya

2-Jul-06 20:06

Hi,
in xslt i have written an template where in i have created text box(txtbx2) and binded a value as follows:

<asp:DataGrid id="DataGrid1" style="Z-INDEX: 101; LEFT: 465px; POSITION: absolute; TOP: 151px" runat="server">
<Columns>
<asp:TemplateColumn HeaderText="col1">
<ItemTemplate>
<asp:TextBox ID="txtbx2" Runat="server" text='<%# DataBinder.Eval(Container,"DataItem.BookNum") %&gt'></asp:TextBox>
</ItemTemplate>
</asp:TemplateColumn>
</Columns>
</asp:DataGrid>
</xsl:template>
</xsl:stylesheet>

but on running it i get an exception that scripts cant be run.Pls help me out

Making Elements have custom fomatting

relaeir1-Jul-06 11:52

relaeir

1-Jul-06 11:52

I am entering some of my data straight into an XML doc and like the nodes to display like the top form. But calling XmlDocument.Save(filepath) changes them to the 2nd form. Is there any way to keep the formating or write out the elements? Or do you know of an Xml Editor to help me?

<KICategory>
        <Name>TIME</Name>
        <Type>GENERAL</Type>
        <KnownItems Add="true">
          <BaseKI><Name>FULL_DATE</Name></BaseKI>
          <BaseKI><Name>MONTH_DAY</Name></BaseKI>
          <BaseKI><Name>MONTH_YEAR</Name></BaseKI>
          <BaseKI><Name>YEAR</Name></BaseKI>
        </KnownItems>
      </KICategory>


<KICategory>
        <Name>TIME</Name>
        <Type>GENERAL</Type>
        <KnownItems Add="true">
          <BaseKI>
            <Name>FULL_DATE</Name>
          </BaseKI>
          <BaseKI>
            <Name>MONTH_DAY</Name>
          </BaseKI>
          <BaseKI>
            <Name>MONTH_YEAR</Name>
          </BaseKI>
          <BaseKI>
            <Name>YEAR</Name>
          </BaseKI>
        </KnownItems>
      </KICategory>

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