[NOTE: Before you read this, please understand I am a WinForms developer so I can only tell you that I know the code in this article works on WinForms.  I assume it will work for WebForms, but not sure.]

How many times have you been working on a WinForms app writing some logic in the form to handle events, work with objects, etc and when you switch over to ‘Design View’ all you see is a White form with an Exception message?  Now normally this message is the dreaded ‘Object not instance of an object message’.  What you have just encountered is NOT uncommon to everyone one else.  Your WinForm application just tried to access an object before you had an instance of it created (maybe this is valid because it is a global object or is class level object that has not been created yet).

Well, lucky for you Microsoft has created this create property on its component object called DesignMode.  However, if you have ever tried to use it you will get mixed results.  The really good news for you is this, you can write your own DesignMode check with a few lines of code.

Below is the code I use to check for Design Mode       

  public static bool IsInDesignMode()

  {

      bool returnFlag = false;

  #if DEBUG  

      if ( System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime )

      {

            returnFlag = true;

      }

      else if ( Process.GetCurrentProcess().ProcessName.ToUpper().Equals( "DEVENV" ) )

      {

            returnFlag = true;

      }

  #endif

      return returnFlag;

  }

Now if you notice, I wrap my main check in #if DEBUG because I don’t care about this if I am in Release mode. 

There are 2 checks above just to be double safe, In the past I have only used the Process.GetCurrentProcess().ProcessName.ToUpper().Equals( "DEVENV" ) check and that has worked just fine, but I ran across the LicenseManager check and threw that in for good measure.

If anyone else has a better way of doing this, please let me know.