How to determine if .NET code is executed by Visual Studio Designer

Question

I get some errors that occurred in my code when I open a Windows Forms form in Visual Studio Designer. I would like to include in my code and perform a different initialization if the form is opened by the designer than if it were really executed.

How can I determine at runtime if the code is executed as part of the constructor that opens the form?

Answer 1

To find out if you are in "development mode":

• Windows Forms components (and controls) have the DesignMode property.
• Windows Presentation Foundation controls must use the IsInDesignMode attached property.

Answer 2

if (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime) { // Design time logic } 

Answer 3

The Control.DesignMode property is probably what you are looking for. It reports whether the parent control is open in the designer.

In most cases, it works fine, but there are times when it does not work as expected. Firstly, it does not work in the designer of controls. Secondly, DesignMode is false for grandchild elements. For example, DesignMode on controls hosted in UserControl will return false if UserControl is hosted in the parent.

There is a fairly simple solution. This happens something like this:

 public bool HostedDesignMode { get { Control parent = Parent; while (parent!=null) { if(parent.DesignMode) return true; parent = parent.Parent; } return DesignMode; } } 

I have not tested this code, but it should work.

Answer 4

The most reliable way:

 public bool isInDesignMode { get { System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess(); bool res = process.ProcessName == "devenv"; process.Dispose(); return res; } } 

Answer 5

The most reliable way to do this is to ignore the DesignMode property and use the custom flag that is set when the application starts.

Grade:

 public static class Foo { public static bool IsApplicationRunning { get; set; } } 

Program.cs:

 [STAThread] static void Main() { Foo.IsApplicationRunning = true; // ... code goes here ... } 

Then just check the flag you need.

 if(Foo.IsApplicationRunning) { // Do runtime stuff } else { // Do design time stuff } 

Answer 6

The devenv approach stops working in VS2012, as the designer now has his own process. Here is the solution I am currently using (the "devenv" part is left for inheritance, but without VS2010 I cannot verify this).

 private static readonly string[] _designerProcessNames = new[] { "xdesproc", "devenv" }; private static bool? _runningFromVisualStudioDesigner = null; public static bool RunningFromVisualStudioDesigner { get { if (!_runningFromVisualStudioDesigner.HasValue) { using (System.Diagnostics.Process currentProcess = System.Diagnostics.Process.GetCurrentProcess()) { _runningFromVisualStudioDesigner = _designerProcessNames.Contains(currentProcess.ProcessName.ToLower().Trim()); } } return _runningFromVisualStudioDesigner.Value; } } 

Answer 7

I had the same problem in Visual Studio Express 2013. I tried many of the suggested solutions, but the one that worked for me was a response to another thread , which I will repeat here if the connection is ever broken:

 protected static bool IsInDesigner { get { return (Assembly.GetEntryAssembly() == null); } } 

Answer 8

 using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess()) { bool inDesigner = process.ProcessName.ToLower().Trim() == "devenv"; return inDesigner; } 

I tried the above code (using statement added) and this may not work for me in some cases. Testing in the usercontrol constructor is placed directly on the form with the constructor loading at startup. But will work in other places.

What worked for me in all places:

 private bool isDesignMode() { bool bProcCheck = false; using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess()) { bProcCheck = process.ProcessName.ToLower().Trim() == "devenv"; } bool bModeCheck = (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime); return bProcCheck || DesignMode || bModeCheck; } 

It may be a bit overkill, but it works, so for me it is good enough.

The success in the example noted above is bModeCheck, so probably DesignMode is redundant.

Answer 9

 /// <summary> /// Are we in design mode? /// </summary> /// <returns>True if in design mode</returns> private bool IsDesignMode() { // Ugly hack, but it works in every version return 0 == String.CompareOrdinal( "devenv.exe", 0, Application.ExecutablePath, Application.ExecutablePath.Length - 10, 10); } 

Answer 10

 System.Diagnostics.Debugger.IsAttached 

Answer 11

I'm not sure if debugging is considered real, but an easy way is to include an if in your code that checks System.Diagnostics.Debugger.IsAttached .

Answer 12

This is hack-ish, but if you use VB.NET and when you work from Visual Studio My.Application.Deployment.CurrentDeployment will be Nothing because you have not deployed it yet. I am not sure how to check the equivalent value in C #.

Answer 13

You check the DesignMode property of your control:

 if (!DesignMode) { //Do production runtime stuff } 

Note that this will not work in your constructor, because the components are not yet initialized.

Answer 14

 System.ComponentModel.Component.DesignMode == true 

Answer 15

We use the following code in UserControls and it does the job. Using only DesignMode will not work in your application that uses your user controls as indicated by other members.

  public bool IsDesignerHosted { get { return IsControlDesignerHosted(this); } } public bool IsControlDesignerHosted(System.Windows.Forms.Control ctrl) { if (ctrl != null) { if (ctrl.Site != null) { if (ctrl.Site.DesignMode == true) return true; else { if (IsControlDesignerHosted(ctrl.Parent)) return true; else return false; } } else { if (IsControlDesignerHosted(ctrl.Parent)) return true; else return false; } } else return false; } 

Answer 16

When the project starts, its name is added using ".vshost".

So, I use this:

  public bool IsInDesignMode { get { Process p = Process.GetCurrentProcess(); bool result = false; if (p.ProcessName.ToLower().Trim().IndexOf("vshost") != -1) result = true; p.Dispose(); return result; } } 

This works for me.

Answer 17

If you created a property that you do not need at all during development, you can use the DesignerSerializationVisibility attribute and set it to Hidden. For example:

 protected virtual DataGridView GetGrid() { throw new NotImplementedException("frmBase.GetGrid()"); } [DesignerSerializationVisibility(DesignerSerializationVisibility.Hidden)] public int ColumnCount { get { return GetGrid().Columns.Count; } set { /*Some code*/ } } 

This stopped my work in Visual Studio every time I made changes to the form using NotImplementedException() and tried to save. Instead, Visual Studio knows that I do not want to serialize this property, so it may skip it. It only displays some weird line in the form properties window, but it looks like it can be ignored.

Please note that this change does not take effect until recovery.

Answer 18

If you are in a form or control, you can use the DesignMode property:

 if (DesignMode) { DesignMode Only stuff } 

Answer 19

I found the DesignMode property to be erroneous, at least in previous versions of Visual Studio. Therefore, I made my own using the following logic:

 Process.GetCurrentProcess().ProcessName.ToLower().Trim() == "devenv"; 

Kind of a hack, I know, but it works well.

Answer 20

To solve this problem, you can also enter the code as shown below:

 private bool IsUnderDevelopment { get { System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess(); if (process.ProcessName.EndsWith(".vshost")) return true; else return false; } } 

Answer 21

Here's another one:

  //Caters only to thing done while only in design mode if (App.Current.MainWindow == null){ // in design mode } //Avoids design mode problems if (App.Current.MainWindow != null) { //applicaiton is running } 

Answer 22

After testing most of the answers here, unfortunately, nothing worked for me (VS2015). So I added a little twist to JohnV's answer, which didn't work out of the box, since DesignMode is a protected property in the Control class.

First, I made an extension method that returns the value of the DesignMode property through Reflection:

 public static Boolean GetDesignMode(this Control control) { BindingFlags bindFlags = BindingFlags.Instance | BindingFlags.NonPublic | BindingFlags.Static; PropertyInfo prop = control.GetType().GetProperty("DesignMode", bindFlags); return (Boolean)prop.GetValue(control, null); } 

and then I made a function like JohnV:

 public bool HostedDesignMode { get { Control parent = Parent; while (parent != null) { if (parent.GetDesignMode()) return true; parent = parent.Parent; } return DesignMode; } } 

This is the only method that worked for me, avoiding all the mess of ProcessName, and while reflection should not be used lightly, in this case it all mattered !;)

EDIT:

You can also make the second function an extension method as follows:

 public static Boolean IsInDesignMode(this Control control) { Control parent = control.Parent; while (parent != null) { if (parent.GetDesignMode()) { return true; } parent = parent.Parent; } return control.GetDesignMode(); } 

Answer 23

  /// <summary> /// Whether or not we are being run from the Visual Studio IDE /// </summary> public bool InIDE { get { return Process.GetCurrentProcess().ProcessName.ToLower().Trim().EndsWith("vshost"); } } 

Answer 24

Here's a flexible way that you can adapt to where you compile, as well as whether you care about which mode you are in.

 string testString1 = "\\bin\\"; //string testString = "\\bin\\Debug\\"; //string testString = "\\bin\\Release\\"; if (AppDomain.CurrentDomain.BaseDirectory.Contains(testString)) { //Your code here }