InfoPath Dev

Hello, ...is it possible to set the taskpane width programmatically in the onload event? Thanks for any tips. Karsten

Check out this article and I think you might find what you're looking for: http://office.microsoft.com/en-us/assistance/HA011032471033.aspx I hope this helps!

Just to make sure: I've already tried that. i've used the FileSystemObject, but i can not get the file inside the xsn file, because it seems Infopath doesn't extract all files in the same directory at runtime.

Hello, I'm workin with Infopath for about 1 month, and i have to do several formulars for the company i'm working for next time. I'm searching an solution, how I can access files in a XSN archive via an Event. For example: A user clicks to a Buttom, and the script.js copies the file form.xsn\data.txt to the given directory. I need that to save ...

I nees code example to: 1) search file name in the DhsrePoint Library (should it be in the code behinde or will I nedd Web Service to do this?) 2) I will use the SharePoint list adapter to get the firstname and last name that are culomns in the list. How to use the SharePoint list adapter for this?

What are the differents between form in access to form in infophat? what is better to use and in witch case?

Hi all, I was wondering if there was a way of determining whether a given node is a 'Group', or a 'Field', via JScript code. I've tried several ways to get that information, but none have succeeded. Since I can actually see whether it is a Group or Field in the InfoPath GUI, I would figure there is a way of getting that information with code. I ...

Hi Kin, this is a regular on this foras ;-) your view in ShearePoint is as default set to deliver max 100 items. go to the properties of your view and remove the max limit: Modify Settings & Columns > MyView > Item Limit > Number of items to display

I need to extract XSL file from XSN file in C#. I try to unzip XSN using library Java.util.zip. When I use class ZipFile I get the following exception: ''Exception of type java.util.zip.ZipException was thrown.'' without any additional information. If I use class ZipInputStream no entry is available: FileInputStream fis=new ...

Thanks a lot for the idea MFaus. The only problem with your solution is that it adds a ''my:'' at the start. However, after some research, I found a way to get just the name of the field: If you use XDocument.DOM.selectSingleNode(''/my:myFields/my:field1'').baseName it will return field1 only. Thanks again for your help. - Reg.