What is the purpose of DesignModeEnabled

Question

0.00/5 (No votes)

See more:

Hello,

I have Window 8.1 Application code. It is using MVVM design to manage files in ViewModel and model.

But at each ViewModel class it is using below if block. i am not able to understand what is the purpose of using this block.

C#

if (Windows.ApplicationModel.DesignMode.DesignModeEnabled)
{
 
}

I also gone through with http://msdn.microsoft.com/en-us/library/windows/apps/windows.applicationmodel.designmode.designmodeenabled.aspx[^]. How we can use this in our program is not described here.

Can someone help me to understand this ?

Thanks

Posted 9-Mar-14 23:37pm

girishmeena

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Accepted Answer · 2014-03-10T00:03:00

Solution 1

The description you link to is pretty clear:

True if the process is running in design mode; otherwise false.
Remarks

Use the DesignModeEnabled property when your custom types require special logic when running in a visual designer.

I.e. it allows your code to determine if it is running in the Visual Studio designer or on a live form.
If it's on a designer, you might want to fill it with dummy data, or put a border round it, or anything else to make the designers job easier, but that won't affect the "Live" form.

Posted 10-Mar-14 0:03am

OriginalGriff

Comments

girishmeena 10-Mar-14 7:10am

Thanks, That's make sense to me. but what i need to do so that it will run under DesignModeEnabled = true.
Right now i am debugging the code in VS 2013 in my local machine but it always return DesignModeEnabled = false.

OriginalGriff 10-Mar-14 7:20am

That's because it's under the debugger - which means it's running code. That's not in the designer any more, it's in a live form!

girishmeena 10-Mar-14 11:39am

Ok. when it will go to DesignModeEnabled = true ?

OriginalGriff 10-Mar-14 11:52am

When it's running in the designer!
I.e. when you drop an instance of your control onto the Visual Studio Form in Design view.
See here: http://msdn.microsoft.com/en-us/library/5ytx0z24(v=vs.90).aspx
It describes how to design-time debug a control.

girishmeena 11-Mar-14 0:25am

Thanks You very much Griff. And Thanks again for giving many answers in a single question.

OriginalGriff 11-Mar-14 3:21am

You're welcome!