blocks|key|1762299|text|System.Diagnostics.Debugger.IsAttached|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1762300|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>System.Diagnostics.Debugger.IsAttached
</code></pre>

blocks|key|1547471|text|我不确定在调试模式下运行是否算作real，但一个简单的方法是在检查System.Diagnostics.Debugger.IsAttached的代码中包含一条if语句。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1547472|entityMap^0|X|12|27|2|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

I'm not sure if running in debug mode counts as real, but an easy way is to include an <code>if</code> statement in your code that checkes for <code>System.Diagnostics.Debugger.IsAttached</code>.

blocks|key|1547501|text|检查控件的DesignMode属性：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1547502|if+(!DesignMode)
{
//Do+production+runtime+stuff
}|code-block|syntax|javascript|1547503|请注意，这在您的构造函数中不起作用，因为组件还没有初始化。|1547504|entityMap^0|5|A|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You check the <code>DesignMode</code> property of your control:

<pre><code>if (!DesignMode)
{
//Do production runtime stuff
}
</code></pre>

Note that this won't work in your constructor because the components haven't been initialized yet.

blocks|key|1547565|text|如果您在窗体或控件中，则可以使用DesignMode属性：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1547566|if+(DesignMode)
{
++++++++DesignMode+Only+stuff
}|code-block|syntax|javascript|1547567|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

<pre><code>if (DesignMode)
{
 DesignMode Only stuff
}
</code></pre>

blocks|key|1547578|text|System.ComponentModel.Component.DesignMode+==+true|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1547579|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>System.ComponentModel.Component.DesignMode == true
</code></pre>

blocks|key|1547617|text|最可靠的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1547618|public+bool+isInDesignMode
{
++++get
++++{
++++++++System.Diagnostics.Process+process+=+System.Diagnostics.Process.GetCurrentProcess();
++++++++bool+res+=+process.ProcessName+==+"devenv";
++++++++process.Dispose();
++++++++return+res;
++++}
}|code-block|syntax|javascript|1547619|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The most reliable approach is:

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

blocks|key|1547679|text|我们在UserControls中使用下面的代码，它可以完成这项工作。正如其他成员指出的那样，在使用自定义用户控件的应用程序中，仅使用DesignMode将不起作用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1547680|++++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;
++++}|code-block|syntax|javascript|1547681|基本上，上面的逻辑可以归结为：|1547682|++++public+bool+IsControlDesignerHosted(System.Windows.Forms.Control+ctrl)
++++{
++++++++if+(ctrl+==+null)+return+false;
++++++++if+(ctrl.Site+!=+null+&&+ctrl.Site.DesignMode)+return+true;
++++++++return+IsControlDesignerHosted(ctrl.Parent);
++++}|1547683|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

We use the following code in UserControls and it does the work. Using only DesignMode will not work in your app that uses your custom user controls as pointed out by other members.

<pre><code> 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;
 }
</code></pre>

Basically the logic above boils down to:

<pre><code> public bool IsControlDesignerHosted(System.Windows.Forms.Control ctrl)
 {
 if (ctrl == null) return false;
 if (ctrl.Site != null &amp;&amp; ctrl.Site.DesignMode) return true;
 return IsControlDesignerHosted(ctrl.Parent);
 }
</code></pre>

blocks|key|1762633|text|我发现DesignMode属性有错误，至少在以前的Visual+Studio版本中是这样。因此，我使用以下逻辑创建了自己的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1762634|Process.GetCurrentProcess().ProcessName.ToLower().Trim()+==+"devenv";|code-block|syntax|javascript|1762635|有点像黑客，我知道，但它工作得很好。|1762636|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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

<pre><code>Process.GetCurrentProcess().ProcessName.ToLower().Trim() == "devenv";
</code></pre>

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

blocks|key|1547770|text|最可靠的方法是忽略DesignMode属性，并使用在应用程序启动时设置的您自己的标志。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1547771|类：|1547772|public+static+class+Foo
{
++++public+static+bool+IsApplicationRunning+{+get;+set;+}
}|code-block|syntax|javascript|1547773|Program.cs：|1547774|[STAThread]
static+void+Main()
{
+++++Foo.IsApplicationRunning+=+true;
+++++//+...+code+goes+here+...
}|1547775|那么只要你需要，只要检查旗子就行了。|1547776|if(Foo.IsApplicationRunning)
{
++++//+Do+runtime+stuff
}
else
{
++++//+Do+design+time+stuff
}|1547777|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

The most reliable way to do this is to ignore the DesignMode property and use your own flag that gets set on application startup.

Class:

<pre><code>public static class Foo
{
 public static bool IsApplicationRunning { get; set; }
}
</code></pre>

Program.cs:

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

Then just check the flag whever you need it.

<pre><code>if(Foo.IsApplicationRunning)
{
 // Do runtime stuff
}
else
{
 // Do design time stuff
}
</code></pre>

blocks|key|1547790|text|if+(System.ComponentModel.LicenseManager.UsageMode+==+System.ComponentModel.LicenseUsageMode.Designtime)
{
++//+Design+time+logic
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1547791|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-cs prettyprint-override"><code>if (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime)
{
 // Design time logic
}
</code></pre>

blocks|key|1547820|text|在运行工程时，其名称会附加".vshost“。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1547821|所以，我使用这个：|1547822|++++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;
++++++++}
++++}|code-block|syntax|javascript|1547823|这对我很管用。|1547824|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

When running a project, its name is appended with ".vshost".

So, I use this:

<pre><code> 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;
 }
 }
</code></pre>

It works for me.

blocks|key|1547857|text|using+(System.Diagnostics.Process+process+=+System.Diagnostics.Process.GetCurrentProcess())
{
++++bool+inDesigner+=+process.ProcessName.ToLower().Trim()+==+"devenv";
++++return+inDesigner;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1547858|我尝试了上面的代码(添加了一条using语句)，但在某些情况下可能会失败。在用户控件的构造函数中测试，该控件直接放置在窗体中，设计器在启动时加载。但可以在其他地方工作。|unstyled|1547859|在所有地方，对我起作用的是：|1547860|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+%7C%7C+DesignMode+%7C%7C+bModeCheck;
}|1547861|也许有点夸张，但它是有效的，所以对我来说已经足够了。|1547862|上面提到的示例中的成功是bModeCheck，所以DesignMode可能是剩余的。|1547863|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$B|C]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|W|8|@]|9|@]|A|$]]]|P|$]]

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

I tried the above code (added a using statement) and this would fail on some occasions for me. Testing in the constructor of a usercontrol placed directly in a form with the designer loading at startup. But would work in other places.

What worked for me, in all locations is:

<pre><code>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;
}
</code></pre>

Maybe a bit overkill, but it works, so is good enough for me.

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

blocks|key|1762787|text|为了解决这个问题，您还可以编写如下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1762788|private+bool+IsUnderDevelopment
{
++++get
++++{
++++++++System.Diagnostics.Process+process+=+System.Diagnostics.Process.GetCurrentProcess();
++++++++if+(process.ProcessName.EndsWith(".vshost"))+return+true;
++++++++else+return+false;
++++}

}|code-block|syntax|javascript|1762789|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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

}
</code></pre>

blocks|key|1762818|text|++++///+<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");
++++++++}
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1762819|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

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

blocks|key|1762823|text|///+<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);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1762824|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>/// &lt;summary&gt;
/// Are we in design mode?
/// &lt;/summary&gt;
/// &lt;returns&gt;True if in design mode&lt;/returns&gt;
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);
}
</code></pre>

blocks|key|1762911|text|我在Visual+Studio+Express+2013中也遇到了同样的问题。我尝试了这里建议的许多解决方案，但对我有效的是an+answer+to+a+different+thread，我将在这里重复一遍，以防链接中断：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1762912|protected+static+bool+IsInDesigner
{
++++get+{+return+(Assembly.GetEntryAssembly()+==+null);+}
}|code-block|syntax|javascript|1762913|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/21054933/1287392^0|1Q|V|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

I had the same problem in Visual Studio Express 2013. I tried many of the solutions suggested here but the one that worked for me was <a href="https://stackoverflow.com/a/21054933/1287392">an answer to a different thread</a>, which I will repeat here in case the link is ever broken:

<pre><code>protected static bool IsInDesigner
{
 get { return (Assembly.GetEntryAssembly() == null); }
}
</code></pre>

blocks|key|1762941|text|这是另一个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1762942|++++++++//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+}|code-block|syntax|javascript|1762943|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's another one:

<pre><code> //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 }
</code></pre>

blocks|key|1548040|text|这里有一种灵活的方式，它可以适应您编译的位置，以及您是否关心您所处的模式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1548041|string+testString1+=+"\\bin\\";
//string+testString+=+"\\bin\\Debug\\";
//string+testString+=+"\\bin\\Release\\";

if+(AppDomain.CurrentDomain.BaseDirectory.Contains(testString))
{
++++//Your+code+here
}|code-block|syntax|javascript|1548042|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's a flexible way that is adaptable to where you compile from as well as whether or not you care which mode you're in.

<pre><code>string testString1 = "\\bin\\";
//string testString = "\\bin\\Debug\\";
//string testString = "\\bin\\Release\\";

if (AppDomain.CurrentDomain.BaseDirectory.Contains(testString))
{
 //Your code here
}
</code></pre>

I am getting some errors thrown in my code when I open a Windows Forms form in Visual Studio's designer. I would like to branch in my code and perform a different initialization if the form is being opened by designer than if it is being run for real. 

How can I determine at run-time if the code is being executed as part of designer opening the form?

How to tell if .NET code is being run by Visual Studio designer

Visual Studio

Windows

当我在Visual Studio的设计器中打开Windows窗体窗体时，我的代码中抛出了一些错误。如果窗体是由设计器打开的，我想在我的代码中进行分支并执行不同的初始化。如何在运行时确定代码是否正在作为打开窗体的设计器的一部分执行？

问如何判断Visual Studio designer是否正在运行.NET代码
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