Hello
I am using a XamGrid with many rowlevels.I defined columnLayouts for each sublevel.My Issue is to be able to show (or not) one of the column of the sublevel depending on a property of my current item or the parent viewmodel.Is there any way to realize this?
Thanks in advance
Hello Kasar,
You can bind the Visibility property of a column using a static resource since the column does not derive from FrameworkElement and could not be bound to the DataContext. Here is a forum thread where Stefan has provided a sample on the same: http://www.infragistics.com/community/forums/t/76299.aspx.
Please do not hesitate to let me know if you need further assistance on the same.
Hello Maria,
Thank you for your response. I do really appreciate.
I tried the sample your colleague Stefan provided but it is not exactly what I want. I think I did not explain my self clear enough.
My purpose is to be able to hide a column only on certain row. To make it as clear as possible I attached a picture.
Thank you for the provided screenshot. It helped me better understand what your scenario is.
XamGrid displays the parent and child rows in different layouts. And that being said the columns that are shown within the parent or child layout could be modified. That change would reflect all the records within the specified layout.
You can suggest a product idea for adding the ability to modify the columns collection on row basics in one of our future versions at http://ideas.infragistics.com.
Submitting your idea will allow you to communicate directly with our product management team, track the progress of your idea at any time, see how many votes it got, read comments from other developers in the community, and see if someone from the product team has additional questions for you.
Remember when submitting your idea to explain the context in which a feature would be used and why it is needed as well as anything that would prevent you from accomplishing this today. You can even add screenshots to build a stronger case. Remember that for your suggestion to be successful, you need other members of the community to vote for it. You can also link back to this thread for additional details.
Thank you in advance to submitting your product idea.