I am working on a Windows C++ App, where I get a URI similar to file:///C:/test 1/file.foo. Now I want to e.g. open that URI with ifstream.
Is there any C/C++ API available on Windows to convert such a path?
My Google Foo seems to be weak today.
There are multiple file URI "versions" so you should not parse it yourself, some of the APIs are broken as well.
If you just want a Windows style path, call PathCreateFromUrl.
If you don't want to convert the path then you must use CreateURLMonikerEx or SHParseDisplayName (with a bind context) but then you end up with a Windows IStream instead.
There is PathCreateFromUrl() :
https://msdn.microsoft.com/en-us/library/bb773581(v=vs.85).aspx
Related
In the application that I am developing (using C++ and Qt), I am using QApplication::applicationDirPath() to access some resources, with respect of the application's path.
As an example, since I want to open a HTML manual from the application, I act this way:
void MainWindow::on_actionHelp_triggered()
{
QString link = QApplication::applicationDirPath() + "/Guide/guide.html";
bool r = QDesktopServices::openUrl(QUrl::fromLocalFile(link));
}
This snippet works if the project's structure presents the path "ProjectName/bin/Release/Guide/guide.html" (since the .exe file is in "ProjectName/bin/Release/AppName.exe").
But what can I do to refer to a higher-directory-level resource? As an example, I wish my HTML file to be in "ProjectName/data/Guide/guide.html". But this way, it seems not possible to compose the path in the way I'm acting.
EDIT: After #olive's comment, I wish to clarify a thing:
"Why am I not using '../'?"
Because it won't work from Visual Studio, where I am massively launch the application to test it. From VS, in fact, I shall use "../data/Guide/guide.html", when "from the outside", I'd have to do "../../data/Guide/guide.html".
That's why (I think) QApplication::applicationDirPath() exists. However, I am not an expert, so don't blame me and correct any eventual mistake of mine, please!
Just use ... QApplication::applicationDirPath() + "/../../data/Guide/guide.html" is perfectly valid path!
Of course there is another problem. When the application is installed, the relative path will probably be different again. You either need to configure the paths in visual studio so that the relative path works both during development and after deployment, or you need to detect the layout.
I am trying Share an Image in Windows 8 Metro C++ Application using Share Charm. To do so, I need to load image to StorageFile^ first. I assume it should looks like:
create_task(imageFile->GetFileFromPathAsync("Textures/title.png")).then([this](StorageFile^ storageFile)
{
imageFile = storageFile;
});
where imageFile is defined in header file
Windows::Storage::StorageFile^ imageFile;
This actual code would throw this exeption
An invalid parameter was passed to a function that considers invalid parameters fatal.
This seems to be very trivial, but there is a very little documentation about Sharing in Metro, and the only Microsoft example shows how to do sharing using FilePicker.
Would be very grateful if someone knows how to do it properly.
If "Textures" is coming from your application package, you should use StorageFile::GetFileFromApplicationUriAsync instead:
Uri^ uri = ref new Uri("ms-appx:///Assets/Logo.png");
create_task(StorageFile::GetFileFromApplicationUriAsync(uri)).then([](task<StorageFile^> t)
{
auto storageFile = t.get();
auto f = storageFile->FileType;
});
You can also use a task-based continuation (as I show above) in order to inspect the exception information more closely. In your case, the inner exception is: The specified path (Assets/Logo.png) contains one or more invalid characters.
This is due to the forward-slash, if you change it to a backslash you'll see: The specified path (Assets\Logo.png) is not an absolute path, and relative paths are not permitted.
If you want to use GetFileFromPathAsync I would recommend using
Windows::ApplicationModel::Package::Current->InstalledLocation
To figure out where your application is installed and building up your path from there.
Im' trying to get the module handle of a module in C++ on windows.
My problem is that I have only the base address and File handle of the module but GetModuleHandle receives only its name as a parameter.
Is there a way in c++ on windows to obtain a module handle without knwoing the module's name?
Have you tried using GetModuleHandleEx (GET_MODULE_HANDLE_EX_FLAG_FROM_ADDRESS | GET_MODULE_HANDLE_EX_FLAG_UNCHANGED_REFCOUNT, (LPCTSTR)lpBaseAddress, &module)?
Use GetModuleHandleEx if you use WindowsXP or later. For previous Windows versions like Windows2000 this function is not available, but there is alternate solution, check comments at the bottom of this page: http://msdn.microsoft.com/en-us/library/ms683200(v=VS.85).aspx
I'm new to MFC, once I create my first app, in myApp::InitInstance() . I have
SetRegistryKey(_T("Local AppWizard-Generated Applications"));
Can I delete this and save settings to my own ini construct ?
Edit: After further testing, the solution below does not work if your app class is derived from CWinAppEx ! It does work if your app is directly derived from CWinApp.
To store values in an .ini file instead of the registry:
Omit the call to SetRegistryKey.
In your app class, set m_pszProfileName to the full path of your .ini file. The filename string must be allocated using malloc, because the framework will call free on it when your app shuts down. First free the existing value, then assign your new string:
free((void*)m_pszProfileName);
m_pszProfileName = ::_tcsdup(_T("C:\\somedir\\myini.ini"));
Call CWinApp::GetProfileInt, CWinApp::WriteProfileInt and similar functions as usual.
I strongly recommend using a path under APPDATA for storing your .ini file.
Yes you can. CWinApp::SetProfileXXX() does this for you, actually - but I wouldn't use these methods anymore in 2010, they were OK when ppl moved from .ini to the registry.
I am not sure if this is possible as a .ini file has only strings for your program. You can create an operating system script (.bat for windows, .sh for unix etc) and call it using system() call.
Use win32 APIs WriteProfileString (write to INI file) and GetProfileString (read from INI file)
For more help
ms-help://MS.MSDNQTR.v90.en/sysinfo/base/writeprofilestring.htm
I wrote some code to read a file in my Java Servlet class. (I'm using Netbeans on Windows along with Tomcat server). However, my servlet cannot find the file!
After much digging, I found that I had to place the file I wanted to read in Tomcat/bin folder. That's very surprising. How can I get the path to my Webapps/ folder? Let's assume my website project is called "Web1".
Essentially what I'm doing is I'm trying to read my .xsl file for converting my DOM Document to be an HTML. At first I tried placing this .xsl file everywhere (at the same level as my index.jsp, in the same directory as my servlet class file, etc...but didnt work at all)
Also, when I finished transform(), my HTML file also goes into the Tomcat/bin folder~!
Can you use javax.servlet.ServletContext.getRealPath(String path)?
Returns a String containing the real path for a given virtual path. For example, the path "/index.html" returns the absolute file path on the server's filesystem would be served by a request for "http://host/contextPath/index.html", where contextPath is the context path of this ServletContext..
The real path returned will be in a form appropriate to the computer and operating system on which the servlet container is running, including the proper path separators. This method returns null if the servlet container cannot translate the virtual path to a real path for any reason (such as when the content is being made available from a .war archive).
Where are you consuming that XSL? If from your Java code place the file into src/java/resources so it will end up in the top of your classpath when the WAR is assembled /WEB-INF/classes/foo.xsl. Then you can use Class#getResource("foo.xsl") or even better if you are using DOM4J or equivalent there are ways of loading the file.
Now if it is you JavaScript that performs the transformation on the client that's a different story
Something like this might be more convenient for you:
java.net.URL url = ClassLoader.getSystemResource(file_name);
try {
InputStream is = url.openStream());
//Read the file and do stuff
} catch(IOException e) {
System.err.println("Error: could not load the file");
}
This will allow you to get an InputStream for a file within the classpath (in your case, something in the webapps folder). As for writing results, I'm not sure.