How to have code in the constructor that will NOT be executed at design time by Visual Studio?

Question

I have a method call in the constructor of my user control that does something that won't work at design time (connecting to a database), and Visual Studio just bailed out when I tried to add that control to the GUI designer.
Sure, I can factor out that code to a separate method, but I don't like the idea that every time I use that object I need to remember to execute a certain method which is essential to that object's function (that's what the constructor is for!).

Is there something like a preprocessor symbol that I can mark my code with so that Visual Studio won't try to execute that code at design time?

Is that a good design? IMO, one should not do too much (connecting to database, in your case) in the constructor. I would love to see what experts have to say. — shahkalpesh, Nov 21 '09 at 06:19
Please add a tag to indicate what technology you are talking about here. Is it windows forms? WPF? AspNet WebForms? Something else? — julealgon, Sep 05 '22 at 12:48

Matt Davis · Accepted Answer · 2009-11-21T06:38:19.520

28

As others have stated, you can use the DesignMode property of the Component class. However, you will not be able to do this in the constructor of your control. The DesignMode property is always false in the constructor and methods called by the constructor. To get around this, re-factor your code to connect to the database in the OnLoad() callback. The DesignMode property is valid at that point. See here for the reasoning (look for the DesignMode section of the article).

I just ran across this blog entry that describes how to use the System.ComponentModel.LicenseManager.UsageMode property for doing the same thing. The blog describes an additional shortcoming of the DesignMode property when dealing with nested controls. Apparently, the UsageMode property doesn't have the same shortcomings and is available for use in the constructor. I cannot personally vouch for it, but might be worthwhile looking into.

edited Nov 21 '09 at 06:38

answered Nov 21 '09 at 06:27

Matt Davis

45,297
16
93
124

The blog post you linked works perfectly for me! (at least so far) – polyglot Nov 21 '09 at 06:46
That's good to know. I may leverage that technique the next time I need a custom control. – Matt Davis Nov 21 '09 at 06:48
11

Blog is now dead but basically it's just: `if (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime)` – Gareth Feb 06 '14 at 16:16

Michael Petrotta · Answer 2 · 2009-11-21T07:07:57.247

16

In Windows Forms?

if (!DesignMode)
{
    // code that shouldn't be executed at design time
}

As others have mentioned, this won't work in the constructor. It's often used in the Form.Load event.

edited Nov 21 '09 at 07:07

answered Nov 21 '09 at 06:20

Michael Petrotta

59,888
27
145
179

3

Brilliant, exactly the same time X-) – Adriaan Stander Nov 21 '09 at 06:20
Quick and right to the point! I was expecting some preprocessor symbol since that won't affect the program after compilation, but the impact of this small check is of course minimal ;) – polyglot Nov 21 '09 at 06:26
2

please be careful ! DesingMode property has some issues that should be considered.You can find an article here: http://weblogs.asp.net/fmarguerie/archive/2005/03/23/395658.aspx – Beatles1692 Nov 21 '09 at 06:41
Yes I realize that (by the hard way) soon enough. The answer I marked as answer works though. – polyglot Nov 21 '09 at 06:45
Sorry about that; I could've sworn I've used it in constructors. – Michael Petrotta Nov 21 '09 at 07:08
This approach worked perfectly for my workflow. I used `if (!IsInDesignMode) {...}` for WPF. – B.K. Dec 09 '14 at 07:52

score 1 · Answer 3 · answered Nov 21 '09 at 06:20

1

Have a look at this

Component.DesignMode Property

answered Nov 21 '09 at 06:20

Adriaan Stander

162,879
31
289
284

B.K. · Answer 4 · 2014-12-09T08:16:46.577

0

I liked Michael Petrotta's approach for Windows Forms. If anyone wants to apply the same technique to WPF, simply use IsInDesignMode.

Example:

public SomeViewModel()
{
    if (!IsInDesignMode)
    {
        DoWork();
    }
}

edited Dec 09 '14 at 08:16

answered Dec 09 '14 at 08:00

B.K.

9,982
10
73
105

score 0 · Answer 5 · edited Mar 29 '17 at 07:28

public SomeViewModel()
{
    if (!IsInDesignMode)
    {
        DoWork();
    }
}

This code above if you are working on the actual UI that you are trying to work on. In a situation that you have something like this on a control, when you switch back to the designer for that control it's ok and no design time error. Now if you added that control that contains the code above to some other Form or another control via dragging it from the toolbox, it will show some design time errors.

score 0 · Answer 6 · answered Sep 05 '22 at 12:39

This is the only code that worked for me, using WPF UserControl on a Windows Form.

private bool? inDesignMode;
public bool IsDesignMode
{
  get
  {
    if (inDesignMode == null)
    {
      var prop = System.ComponentModel.DesignerProperties.IsInDesignModeProperty;

      inDesignMode = (bool)System.ComponentModel.DependencyPropertyDescriptor
          .FromProperty(prop, typeof(FrameworkElement))
          .Metadata.DefaultValue;

      if (!inDesignMode.GetValueOrDefault(false) && System.Diagnostics.Process.GetCurrentProcess().ProcessName.StartsWith("devenv", StringComparison.Ordinal))
        inDesignMode = true;
    }

    return inDesignMode.GetValueOrDefault(false);
  }
}

How to have code in the constructor that will NOT be executed at design time by Visual Studio?

6 Answers6

Linked