I am trying to developing an AddIn for Visual Studio to get a right click context menu for javascript files and image files. I have managed to add my Addin to the right click of all project items
What I want to achieve is to get the Addin ONLY on javascript files and image files. Something like this (Note:- currently I am getting Addin on ALL file types)
Below is the code I have in the connect
if (connectMode == ext_ConnectMode.ext_cm_UISetup)
{
object[] contextGUIDS = new object[] { };
Commands2 commands = (Commands2)_applicationObject.Commands;
string toolsMenuName = "Tools";
//Place the command on the tools menu.
//Find the MenuBar command bar, which is the top-level command bar holding all the main menu items:
Microsoft.VisualStudio.CommandBars.CommandBar menuBarCommandBar = ((Microsoft.VisualStudio.CommandBars.CommandBars)_applicationObject.CommandBars)["MenuBar"];
//Find the Tools command bar on the MenuBar command bar:
CommandBarControl toolsControl = menuBarCommandBar.Controls[toolsMenuName];
CommandBarPopup toolsPopup = (CommandBarPopup)toolsControl;
Microsoft.VisualStudio.CommandBars.CommandBar itemToolBar = ((Microsoft.VisualStudio.CommandBars.CommandBars)_applicationObject.CommandBars)["Item"];
//This try/catch block can be duplicated if you wish to add multiple commands to be handled by your Add-in,
// just make sure you also update the QueryStatus/Exec method to include the new command names.
try
{
//Add a command to the Commands collection:
Command command = commands.AddNamedCommand2(_addInInstance, "CrmAddin", "CrmAddin", "Executes the command for CrmAddin", true, 59, ref contextGUIDS, (int)vsCommandStatus.vsCommandStatusSupported + (int)vsCommandStatus.vsCommandStatusEnabled, (int)vsCommandStyle.vsCommandStylePictAndText, vsCommandControlType.vsCommandControlTypeButton);
//Add a control for the command to the tools menu:
if ((command != null) && (toolsPopup != null))
{
command.AddControl(toolsPopup.CommandBar, 1);
}
if ((command != null) && (itemToolBar != null))
{
command.AddControl(itemToolBar, 1);
}
}
catch (System.ArgumentException)
{
//If we are here, then the exception is probably because a command with that name
// already exists. If so there is no need to recreate the command and we can
// safely ignore the exception.
}
I tried to filter out the file types in the QueryStatus method like this but it is of no help
if (neededText == vsCommandStatusTextWanted.vsCommandStatusTextWantedNone)
{
if (commandName == "CrmAddin.Connect.CrmAddin")
{
bool supportedFileTypes = true;
foreach (Project project in _applicationObject.Solution.Projects)
{
foreach (ProjectItem projectItem in project.ProjectItems)
{
if (!projectItem.Name.EndsWith(".js"))
{
supportedFileTypes = false;
}
}
}
if (supportedFileTypes)
{
status = (vsCommandStatus)vsCommandStatus.vsCommandStatusSupported | vsCommandStatus.vsCommandStatusEnabled;
}
else
{
status = (vsCommandStatus)vsCommandStatus.vsCommandStatusSupported;
}
return;
}
}
Please help if anyone can point me to the right direction.
Just for info.
Was looking for a solution to similar problem, found this first, then after some searching I saw your question on MSDN (?)
http://social.msdn.microsoft.com/Forums/en-US/vsx/thread/c8f35f82-c694-4a6a-8c4a-a8404a4df11f
Which gave the answer :)
I have a series of QTextEdits and QLineEdits connected to a slot through a QSignalMapper(which emits a textChanged(QWidget*) signal). When the connected slot is called (pasted below), I need to be able to differentiate between the two so I know whether to call the text() or toPlainText() function. What's the easiest way to determine the subclass type of a QWidget?
void MainWindow::changed(QWidget *sender)
{
QTextEdit *temp = qobject_cast<QTextEdit *>(sender);
QString currentText = temp->toPlainText(); // or temp->text() if its
// a QLineEdit...
if(currentText.compare(""))
{
...
}
else
{
...
}
}
I was considering using try-catch but Qt doesn't seem to have very extensive support for Exceptions... Any ideas?
Actually, your solution is already almost there. In fact, qobject_cast will return NULL if it can't perform the cast. So try it on one of the classes, if it's NULL, try it on the other:
QString text;
QTextEdit *textEdit = qobject_cast<QTextEdit*>(sender);
QLineEdit *lineEdit = qobject_cast<QLineEdit*>(sender);
if (textEdit) {
text = textEdit->toPlainText();
} else if (lineEdit) {
text = lineEdit->text();
} else {
// Return an error
}
You can also use sender->metaObject()->className() so you won't make unnecesary casts. Specially if you have a lot of classes to test. The code will be like this:
QString text;
QString senderClass = sender->metaObject()->className();
if (senderClass == "QTextEdit") {
QTextEdit *textEdit = qobject_cast<QTextEdit*>(sender);
text = textEdit->toPlainText();
} else if (senderClass == "QLineEdit") {
QLineEdit *lineEdit = qobject_cast<QLineEdit*>(sender);
text = lineEdit->text();
} else {
// Return an error
}
I know is an old question but I leave this answer just in case it would be useful for somebody...
I want to extract the guard statement from the following method
private void CreateProxy()
{
//extract the following guard statement.
Host selected = this.comboBox1.SelectedItem as Host;
if (selected == null)
{
return;
}
this.SearchProxy = ServiceProxy.ProxyFactory.CreateSearchProxy(GetSelectedIP().ToString());
this.StreamProxy = ServiceProxy.ProxyFactory.CreatePlayerProxy(GetSelectedIP().ToString());
}
//extracted guard method
public bool IsHostSelected()
{
Host selected = this.comboBox1.SelectedItem as Host;
if (selected == null)
{
return false;
}
return true;
}
see? now i have to add return value for the extracted method, is this kinda ugly?
any better solution to avoid adding the return value for the extracted method?
I don't see the big deal. First, I would rewrite it as:
static bool SelectedItemIsHost(ComboBox box) {
return box.SelectedItem is Host;
}
Note the rename, the ComboBox as a parameter, and the body change.
Now, this makes your code read more clearly:
void CreateProxy() {
if(SelectedItemIsHost(this.comboBox1)) {
this.SearchProxy = ServiceProxy.ProxyFactory.CreateSearchProxy(GetSelectedIP().ToString());
this.StreamProxy = ServiceProxy.ProxyFactory.CreatePlayerProxy(GetSelectedIP().ToString());
}
}
So now it reads "if the selected item is a Host then do stuff."
Now, this goes way beyond your question, but this looks like a big coupling of UI logic and domain logic. You might want to reconsider a decoupling there.
any better solution to avoid adding the return value for the extracted method?
Yes:
//extracted guard method
public bool IsHostSelected()
{
Host selected = this.comboBox1.SelectedItem as Host;
return selected != null;
}
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?
if (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime)
{
// Design time logic
}
To find out if you're in "design mode":
Windows Forms components (and controls) have a DesignMode property.
Windows Presentation Foundation controls should use the IsInDesignMode attached property.
The Control.DesignMode property is probably what you're looking for. It tells you if the control's parent is open in the designer.
In most cases it works great, but there are instances where it doesn't work as expected. First, it doesn't work in the controls constructor. Second, DesignMode is false for "grandchild" controls. For example, DesignMode on controls hosted in a UserControl will return false when the UserControl is hosted in a parent.
There is a pretty easy workaround. It goes something like this:
public bool HostedDesignMode
{
get
{
Control parent = Parent;
while (parent!=null)
{
if(parent.DesignMode) return true;
parent = parent.Parent;
}
return DesignMode;
}
}
I haven't tested that code, but it should work.
The most reliable approach is:
public bool isInDesignMode
{
get
{
System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess();
bool res = process.ProcessName == "devenv";
process.Dispose();
return res;
}
}
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:
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 whever you need it.
if(Foo.IsApplicationRunning)
{
// Do runtime stuff
}
else
{
// Do design time stuff
}
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 an answer to a different thread, which I will repeat here in case the link is ever broken:
protected static bool IsInDesigner
{
get { return (Assembly.GetEntryAssembly() == null); }
}
The devenv approach stopped working in VS2012 as the designer now has its own process. Here is the solution I am currently using (the 'devenv' part is left there for legacy, but without VS2010 I am not able to test that though).
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;
}
}
/// <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);
}
System.Diagnostics.Debugger.IsAttached
It's hack-ish, but if you're using VB.NET and when you're running from within Visual Studio My.Application.Deployment.CurrentDeployment will be Nothing, because you haven't deployed it yet. I'm not sure how to check the equivalent value in C#.
using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess())
{
bool inDesigner = process.ProcessName.ToLower().Trim() == "devenv";
return inDesigner;
}
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:
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;
}
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.
You check the DesignMode property of your control:
if (!DesignMode)
{
//Do production runtime stuff
}
Note that this won't work in your constructor because the components haven't been initialized yet.
When running a project, its name is appended with ".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;
}
}
It works for me.
I'm not sure if running in debug mode counts as real, but an easy way is to include an if statement in your code that checkes for System.Diagnostics.Debugger.IsAttached.
If you created a property that you don't need at all at design time, 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*/ } }
It stopped my Visual Studio crashing every time I made a change to the form with NotImplementedException() and tried to save. Instead, Visual Studio knows that I don't want to serialize this property, so it can skip it. It only displays some weird string in the properties box of the form, but it seems to be safe to ignore.
Please note that this change does not take effect until you rebuild.
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.
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;
}
Basically the logic above boils down to:
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);
}
If you are in a form or control you can use the DesignMode property:
if (DesignMode)
{
DesignMode Only stuff
}
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:
Process.GetCurrentProcess().ProcessName.ToLower().Trim() == "devenv";
Kind of a hack, I know, but it works well.
System.ComponentModel.Component.DesignMode == true
To solve the problem, you can also code as below:
private bool IsUnderDevelopment
{
get
{
System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess();
if (process.ProcessName.EndsWith(".vshost")) return true;
else return false;
}
}
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 }
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 which returns the DesignMode's Property value via 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 ProcessName mess, and while reflection should not be used lightly, in this case it did all the difference! ;)
EDIT:
You can also make the second function an extension method like this:
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();
}
For WPF (hopefully this is useful for those WPF people stumbling upon this question):
if (System.ComponentModel.DesignerProperties.GetIsInDesignMode(new DependencyObject()))
{
}
GetIsInDesignMode requires a DependencyObject. If you don't have one, just create one.
/// <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");
}
}
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.
string testString1 = "\\bin\\";
//string testString = "\\bin\\Debug\\";
//string testString = "\\bin\\Release\\";
if (AppDomain.CurrentDomain.BaseDirectory.Contains(testString))
{
//Your code here
}