I am trying to supply a path, and open it with the "native open with" dialog of linux.
This is the native open with dialog seen on ubuntu:
I looked into GioLaunch and am able to launch things into their default application handler, but i cant manage to launch it into this "Open With" dialog.
This is my code for launching into default app:
var launcher = g_desktop_app_info_new_from_filename('blah blah blah');
var error = new TYPES.GError.ptr();
var rez_launch_uris = g_app_info_launch_uris(launcher.address(), null, null, error.address())
Related
Is there a way to open Notification Center on OS X from a Cocoa app?
We tried launching the Notification Center app, but that doesn't work.
Our app is not sandboxed, so we can do whatever is required.
You can open it using ScriptingBridge, but it's not an official API, so it might break someday.
It's a bit odd, but this only works with sandboxed apps. If your app is not sandboxed, you need Accessibility access.
1. Generate Header file
Open Terminal and run the following commands:
cd ~/Desktop/
sdef /System/Library/CoreServices/System\ Events.app | sdp -fh --basename SystemEvents
Copy the file "SystemEvents.h" from your Desktop into your project.
Xcode will show some errors or warnings for a few lines in this file. Just remove these lines.
2. Import ScriptingBridge framework and generated header file.
#import <ScriptingBridge/ScriptingBridge.h>
and
#import "SystemEvents.h"
3. Add code
SystemEventsApplication *systemEventsApp = (SystemEventsApplication *)[[SBApplication alloc] initWithBundleIdentifier:#"com.apple.systemevents"];
SystemEventsApplicationProcess *sysUIServer = [systemEventsApp.applicationProcesses objectWithName:#"SystemUIServer"];
SystemEventsMenuBarItem *item = nil;
for (SystemEventsMenuBar *menuBar in sysUIServer.menuBars) {
item = [menuBar.menuBarItems objectWithName:#"Notification Center"];
if (item != nil && [item.name isEqualToString:#"Notification Center"])
break;
}
[item clickAt:nil];
You can also replace line 4-9 with this code, if you're sure the menu bars won't change in the future:
SystemEventsMenuBarItem *item = [[[sysUIServer.menuBars objectAtLocation:#2] menuBarItems] objectWithName:#"Notification Center"];
4. Add temporary exception for sandbox
Open your projects .entitlements file and add "com.apple.security.temporary-exception.apple-events" as an array. Add "com.apple.systemevents" as String.
Switching to Today view
This requires your app to have Accessibility access, so it won't work in sandboxed apps. You can simply call this script instead of using ScriptingBridge:
tell application "System Events"
click menu bar item "Notification Center" of menu bar 2 of application process "SystemUIServer"
click radio button "Today" of radio group 1 of window "NotificationTableWindow" of application process "NotificationCenter"
end tell
No, unfortunately there is no support for opening the Notification Center. Supposedly because it wouldn't be annoying for the user to have it open without them wanting it to open
I'm making a document-based application where documents are application bundles.
I can open them through "File → Open" menu fine. But when I try to drop them on the Dock icon they get rejected and the icon doesn't highlight.
I've searched this issue and it seems like it should just work given that LSItemContentTypes property is specified. Here is my Info.plist excerpt:
CFBundleDocumentTypes = (
{ CFBundleTypeName = "Example";
CFBundleTypeRole = "Editor";
LSHandlerRank = "Alternate";
LSItemContentTypes = ( "com.apple.application-bundle" );
NSDocumentClass = "Example";
NSExportableTypes = ( "com.apple.application-bundle" );
},
);
(I am developing outside of Xcode, this is a TextMate-formatted plist file.)
Also I tried implementing application:openFile: and application:openFiles: on my app delegate but they never get called.
There are no warnings or other related messages in the log. What am I missing?
Perhaps you need to set LSTypeIsPackage to YES.
That's the only difference I see with your plist stuff and mine that opens packages.
After that, maybe you need to logout/login, restart the app, or whatever to "refresh the system/launch services"?
I want to open all links in the same window instead in new window.
I tried
profile.setPreference("browser.link.open_newwindow", 1)
but the result is:
WARNING: traffic.loop 0 error: Preference browser.link.open_external may not be overridden: frozen value=2, requested value=1
Is there an another way to open the links in the same window ?
You should modify the firefox profile parameters:
profile = webdriver.FirefoxProfile()
profile.set_preference("browser.link.open_newwindow", 3)
profile.set_preference("browser.link.open_newwindow.restriction", 0)
driver = webdriver.Firefox(firefox_profile=profile)
if this methode does not work, you can set perference using firefox Options:
from selenium.webdriver.firefox.options import Options
opts = Options()
opts.set_preference("browser.link.open_newwindow.restriction", 0)
opts.set_preference("browser.link.open_newwindow", 3)
driver = webdriver.Firefox(firefox_options=opts)
(A) browser.link.open_newwindow - for links in Firefox tabs :
3 : divert new window to a new tab (default)
2 : allow link to open a new window
1 : force new window into same tab
(B) browser.link.open_newwindow.restriction - for links in Firefox tabs
0 : apply the setting under (A) to ALL new windows (even script windows)
2 : apply the setting under (A) to normal windows, but NOT to script windows
with features (default)
1 : override the setting under (A) and always use new windows
I've found a workaround!
JavascriptExecutor js = (JavascriptExecutor) driver;
String script = "document.getElementById('yourFormOrAnchorId').target=''";
js.executeScript(script);
After that you can select your anchor or any of the form elements and click or submit it. The target page will open in the same tab.
This basically changes the current HTML page so that anchors and forms don't force the browser to open new tabs or windows. For testing this might be suboptimal, but it simplifies the writing of tests a lot.
Try this out...
Modify FireFox profile parameters "browser.link.open_newwindow.restriction" and "browser.link.open_newwindow".
FirefoxProfile profile = new FirefoxProfile();
profile.setPreference("browser.link.open_newwindow.restriction", 0);
profile.setPreference("browser.link.open_newwindow", 1);
If you are using Google Chrome then simply install this extension and it will take care of the rest of the task. This extension is also handy to open pop-ups in new tabs which usually opens in new windows. (First you need to download the extension .crx file from given location.)
ChromeOptions options = new ChromeOptions();
options.addExtensions(new File("/path/to/extension.crx"));
DesiredCapabilities capabilities = new DesiredCapabilities();
capabilities.setCapability(ChromeOptions.CAPABILITY, options);
ChromeDriver driver = new ChromeDriver(capabilities);
In the selenium config file:
C:\Python27\Lib\site-packages\selenium\webdriver\firefox\webdriver_prefs.json
change the following line from:
"browser.link.open_newwindow": 2,
to:
"browser.link.open_newwindow": 3,
I test it and it worked
According to Selium docs (https://code.google.com/p/selenium/wiki/FirefoxDriver) the following property webdriver.firefox.profile controls the firefox profile used.
Which is where firefox gets the browser.link.open_newwindow on start up from. To create a new profile for your tests you can follow the instructions here https://support.mozilla.org/en-US/kb/profile-manager-create-and-remove-firefox-profiles detailed configuration of the profile can be done either by editing the profile's pref.js or firing up the profile and editing it via about:config.
hope this of help!
Actually, Selenium is not responsible of the page opens in a new window or in a same window. It is fully depends upon the Browser settings which you used for execution.
For a sake take Firefox browser
If you want to open all the links in a new window. Do these steps
Open Tools
Click Options
Click Tabs menu
Check the box of Open new windows in a new tab instead.
Now click the link which opens a window. It will opens in a new tab of same window.
I have an firefox extension with the name myjavascriptfile.js,As I am new to this addon concepts,just I want to debug this script.So I am using the following statements in this file like
function LOG(text)
{
var consoleService = Components.classes["#mozilla.org/consoleservice;1"].getService(Components.interfaces.nsIConsoleService);
consoleService.logStringMessage(text);
}
observe: function(subject, topic, data)
{
LOG("observe called ");
}
I know this observe is getting called but I dont know where to see my log message.can some one tell me Where it is printing?
Please help.
That text goes to the Error Console. You might need to go to about:config and change devtools.errorconsole.enabled preference to true - the Error Console was removed from the menus by default while ago (strangely enough, I could still see it even without this pref). I think that on OS X you can still open the Error Console via Tools / Web Developer menu, on Windows you have to click the Firefox button and choose Web Developer menu there. Alternatively, Command-Shift-J should do as well.
I am building an internal development tool to manage different processes commonly used in our development environment. The tool shows the list of the monitored processes, indicating their running state and allows to start or stop each process.
I'd like to add the functionality of attaching a debugger to a monitored process from my tool instead of going in Debug -> Attach to process in Visual Studio and finding the process.
My goal is to have something like Debugger.Launch() that would show a list of the available Visual Studio. I can't use Debugger.Launch(), because it launches the debugger on the process that makes the call. I would need something like Debugger.Launch(processId).
How do I achieve this functionality?
A solution could be to implement a command in each monitored process to call Debugger.Launch() when the command is received from the monitoring tool, but I would prefer something that does not require to modify the code of the monitored processes.
Side question:
When using Debugger.Launch(), instances of Visual Studio that already have a debugger attached are not listed. Visual Studio is not limited to one attached debugger, you can attach on multiple process when using Debug → Attach to process.
How do I bypass this limitation when using Debugger.Launch() or an alternative?
A coworker ended up with a solution using DTE, and I posted the code on PasteBin.
The methods of interest are AttachVisualStudioToProcess and TryGetVsInstance
Source Code
public static void AttachVisualStudioToProcess(Process visualStudioProcess, Process applicationProcess)
{
_DTE visualStudioInstance;
if (TryGetVsInstance(visualStudioProcess.Id, out visualStudioInstance))
{
//Find the process you want the Visual Studio instance to attach to...
DTEProcess processToAttachTo = visualStudioInstance.Debugger.LocalProcesses.Cast<DTEProcess>().FirstOrDefault(process => process.ProcessID == applicationProcess.Id);
// Attach to the process.
if (processToAttachTo != null)
{
processToAttachTo.Attach();
ShowWindow((int)visualStudioProcess.MainWindowHandle, 3);
SetForegroundWindow(visualStudioProcess.MainWindowHandle);
}
else
{
throw new InvalidOperationException("Visual Studio process cannot find specified application '" + applicationProcess.Id + "'");
}
}
}
private static bool TryGetVsInstance(int processId, out _DTE instance)
{
IntPtr numFetched = IntPtr.Zero;
IRunningObjectTable runningObjectTable;
IEnumMoniker monikerEnumerator;
IMoniker[] monikers = new IMoniker[1];
GetRunningObjectTable(0, out runningObjectTable);
runningObjectTable.EnumRunning(out monikerEnumerator);
monikerEnumerator.Reset();
while (monikerEnumerator.Next(1, monikers, numFetched) == 0)
{
IBindCtx ctx;
CreateBindCtx(0, out ctx);
string runningObjectName;
monikers[0].GetDisplayName(ctx, null, out runningObjectName);
object runningObjectVal;
runningObjectTable.GetObject(monikers[0], out runningObjectVal);
if (runningObjectVal is _DTE && runningObjectName.StartsWith("!VisualStudio"))
{
int currentProcessId = int.Parse(runningObjectName.Split(':')[1]);
if (currentProcessId == processId)
{
instance = (_DTE)runningObjectVal;
return true;
}
}
}
instance = null;
return false;
}
WinDbg does the chain debugging for native code by default. If you want to launch another instance of Visual Studio, check Launch the Debugger Automatically on MSDN:
To automate the existing debugger, use Marshal.GetActiveObject to get the current EnvDTE.Debugger then let it attach to the process you just created.
Sometimes, you may need to debug the startup code for an application that is launched by another process. Examples include services and custom setup actions. In these scenarios, you can have the debugger launch and automatically attach when your application starts.
To setup an application to launch the debugger automatically
Start the Registry Editor (regedit).
In the Registry Editor, open the HKEY_LOCAL_MACHINE folder.
Navigate to HKEY_LOCAL_MACHINE\Software\Microsoft\Windows NT\currentversion\image file execution options.
In the Image File Execution Options folder, locate the name of the application you want to debug, such as myapp.exe. If you cannot find the application you want to debug:
a. Right-click the Image File Execution Options folder, and on the shortcut menu, click New Key.
b. Right-click the new key, and on the shortcut menu, click Rename.
c. Edit the key name to the name of your application; myapp.exe, in this example.
Right-click the myapp.exe folder, and on the shortcut menu, click New String Value.
Right-click the new string value, and on the shortcut menu, click Rename.
Change the name to debugger.
Right-click the new string value, and on the shortcut menu, click Modify.
The Edit String dialog box appears.
In the Value data box, type vsjitdebugger.exe.
Click OK.
From the Registry menu, click Exit.
The directory containing vsjitdebugger.exe must be in your system path. To add it to the system path, follow these steps:
a. Open the Control Panel in Classic view, and double-click System.
b. Click Advanced System Settings.
c. In System Properties, click the Advanced tab.
d. On the Advanced tab, click Environment Variables.
e. In the Environment Variables dialog box, under System variables, select Path, then click the Edit button.
f. In the Edit System Variable dialog box, add the directory to the Variable value box. Use a semicolon to separate it from other entries in the list.
g. Click OK to close the Edit System Variable dialog box.
h. Click OK to close the Environment Variables dialog box.
i. Click OK to close the System Properties dialog box.
Now, use any method to start your application. Visual Studio will start and load the application.
Here is some information about how you can programmatically attach the debugger to multiple processes:
Attach to locally running processes
Attach to remotely running processes