My app's local storage stored some MS office files, how can open and show them from my app?
If your app just downloaded a file and it went into the storage folder, you can use Launcher class to launch an app from the phone which can open the downloaded file for you.
Class StorageFolder will help you get your StorageFile.
Considering you already have an object of kind StorageFile, you can use the following line to launch a suitable app automatically from your phone ::
await Launcher.LaunchFileAsync(storageFile);
This will open the app associated with the file type.
I hope i got your question clearly and answered in the right direction.
Thanks.
Hopefully, the code is similar to selecting a photo.
If it is, then consider the following snippet:
FileOpenPicker openPicker = new FileOpenPicker();
openPicker.ViewMode = PickerViewMode.Thumbnail;
openPicker.SuggestedStartLocation = PickerLocationId.DocumentsLibrary;
openPicker.FileTypeFilter.Add(".doc");
openPicker.FileTypeFilter.Add(".docx");
// Launch file open picker and caller app is suspended and may be terminated if required
openPicker.PickSingleFileAndContinue();
I'm using sqldb to connect to Firebird from within my DLL. This fails because it cannot find fbclient.dll which is actually present in the same directory as my DLL. GetCurrentDir returns the path to the Windows system folder. Performing a SetCurrentDir with the path of the DLL successfully changes the current directory, but still it won't work. What can I do to get sqldb to use fbclient.dll at a location of my choosing?
From the wiki page seems there is no way to explicitly specify the directory from where the Firebird client library could be loaded. So as a workaround you may use the SetDllDirectory function which will add a directory provided to its only parameter to the search path used to locate DLL libraries for the application. A subsequent call to LoadLibrary function used to load the Firebird's client library will go through the search list and find it in the location you added by the SetDllDirectory function call.
I have a scenario where a user who is not on the domain is trying to open a file that is on the network. Trying to determine if the path exists using the Dir() function. Here is what my code looks like...
If Len(Dir("\\xx\xxxxx\Shared\Virtual Machine\_Testing\Update\", vbDirectory)) > 0 Then Return True
I get the Run-Time error Bad file name or number (error number 52).
Yep, dir() on a bad/inaccessible unc causes a runtime error, unlike the behaviour for a local file.
You can either wrap it in an error handler or use the GetFileAttributes API and look for the directory attribute flag (the built in getattr() won't work for this).
For the time-being, if the user doesn't mind mapping a network drive, he could open the file with your code the way it is.
Try below steps.
Restart your machine
After restart, access/open the share path through windows explorer
Provide the network credentials and select the option "remember my
credentials"
Now run/debug your application. it should work!
Does anyone know of a "Global" Application data for Adobe Air?
File.applicationStorageDirectory stores within the users area?
But we need the application to store information for the whole computer.
Something such as Windows: C:\ProgramData or AllUsers but there doesn't seem to be an official way to use these areas. (Trying to keep this as Standard as possible)
But there doesn't seem to be a File.allUsersApplicationStorage or File.globalApplicationData anything.
Now I know I could do something such as below for Windows Vista/7
var _path:String = File.applicationDirectory.nativePath;
// _np: C:\Program Files (x86)\ProgName\
_path = _path.substr(0, _np.indexOf(File.separator) + 1);
// _np: C:\
var _file = new File(_path).resolvePath("ProgramData/ProgName");
but I do not know what to do for WinXP, MacOS or Linux.
Any help would be welcome.
Couldn't you just use File.applicationDirectory? (Or rather, a subdirectory of it.) That would store your information in the application directory, and would be accessible from all users.
It might not be ideal, but it works.
I've recently searched how I could get the application's directory in Java. I've finally found the answer but I've needed surprisingly long because searching for such a generic term isn't easy. I think it would be a good idea to compile a list of how to achieve this in multiple languages.
Feel free to up/downvote if you (don't) like the idea and please contribute if you like it.
Clarification:
There's a fine distinction between the directory that contains the executable file and the current working directory (given by pwd under Unix). I was originally interested in the former but feel free to post methods for determining the latter as well (clarifying which one you mean).
In Java the calls
System.getProperty("user.dir")
and
new java.io.File(".").getAbsolutePath();
return the current working directory.
The call to
getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
returns the path to the JAR file containing the current class, or the CLASSPATH element (path) that yielded the current class if you're running directly from the filesystem.
Example:
Your application is located at
C:\MyJar.jar
Open the shell (cmd.exe) and cd to C:\test\subdirectory.
Start the application using the command java -jar C:\MyJar.jar.
The first two calls return 'C:\test\subdirectory'; the third call returns 'C:\MyJar.jar'.
When running from a filesystem rather than a JAR file, the result will be the path to the root of the generated class files, for instance
c:\eclipse\workspaces\YourProject\bin\
The path does not include the package directories for the generated class files.
A complete example to get the application directory without .jar file name, or the corresponding path to the class files if running directly from the filesystem (e.g. when debugging):
String applicationDir = getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
if (applicationDir.endsWith(".jar"))
{
applicationDir = new File(applicationDir).getParent();
}
// else we already have the correct answer
In .NET (C#, VB, …), you can query the current Assembly instance for its Location. However, this has the executable's file name appended. The following code sanitizes the path (using System.IO and using System.Reflection):
Directory.GetParent(Assembly.GetExecutingAssembly().Location)
Alternatively, you can use the information provided by AppDomain to search for referenced assemblies:
System.AppDomain.CurrentDomain.BaseDirectory
VB allows another shortcut via the My namespace:
My.Application.Info.DirectoryPath
In Windows, use the WinAPI function GetModuleFileName(). Pass in NULL for the module handle to get the path for the current module.
Python
path = os.path.dirname(__file__)
That gets the path of the current module.
Objective-C Cocoa (Mac OS X, I don't know for iPhone specificities):
NSString * applicationPath = [[NSBundle mainBundle] bundlePath];
In Java, there are two ways to find the application's path. One is to employ System.getProperty:
System.getProperty("user.dir");
Another possibility is the use of java.io.File:
new java.io.File("").getAbsolutePath();
Yet another possibilty uses reflection:
getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
In VB6, you can get the application path using the App.Path property.
Note that this will not have a trailing \ EXCEPT when the application is in the root of the drive.
In the IDE:
?App.Path
C:\Program Files\Microsoft Visual Studio\VB98
In .Net you can use
System.IO.Directory.GetCurrentDirectory
to get the current working directory of the application, and in VB.NET specifically you can use
My.Application.Info.DirectoryPath
to get the directory of the exe.
Delphi
In Windows applications:
Unit Forms;
path := ExtractFilePath(Application.ExeName);
In console applications:
Independent of language, the first command line parameter is the fully qualified executable name:
Unit System;
path := ExtractFilePath(ParamStr(0));
Libc
In *nix type environment (also Cygwin in Windows):
#include <unistd.h>
char *getcwd(char *buf, size_t size);
char *getwd(char *buf); //deprecated
char *get_current_dir_name(void);
See man page
Unix
In unix one can find the path to the executable that was started using the environment variables. It is not necessarily an absolute path, so you would need to combine the current working directory (in the shell: pwd) and/or PATH variable with the value of the 0'th element of the environment.
The value is limited in unix though, as the executable can for example be called through a symbolic link, and only the initial link is used for the environment variable. In general applications on unix are not very robust if they use this for any interesting thing (such as loading resources). On unix, it is common to use hard-coded locations for things, for example a configuration file in /etc where the resource locations are specified.
In bash, the 'pwd' command returns the current working directory.
In PHP :
<?php
echo __DIR__; //same as dirname(__FILE__). will return the directory of the running script
echo $_SERVER["DOCUMENT_ROOT"]; // will return the document root directory under which the current script is executing, as defined in the server's configuration file.
echo getcwd(); //will return the current working directory (it may differ from the current script location).
?>
in Android its
getApplicationInfo().dataDir;
to get SD card, I use
Environment.getExternalStorageDirectory();
Environment.getExternalStoragePublicDirectory(String type);
where the latter is used to store a specific type of file (Audio / Movies etc). You have constants for these strings in Environment class.
Basically, for anything to with app use ApplicationInfo class and for anything to do with data in SD card / External Directory using Environment class.
Docs :
ApplicationInfo ,
Environment
In Tcl
Path of current script:
set path [info script]
Tcl shell path:
set path [info nameofexecutable]
If you need the directory of any of these, do:
set dir [file dirname $path]
Get current (working) directory:
set dir [pwd]
Java:
On all systems (Windows, Linux, Mac OS X) works for me only this:
public static File getApplicationDir()
{
URL url = ClassLoader.getSystemClassLoader().getResource(".");
File applicationDir = null;
try {
applicationDir = new File(url.toURI());
} catch(URISyntaxException e) {
applicationDir = new File(url.getPath());
}
return applicationDir;
}
in Ruby, the following snippet returns the path of the current source file:
path = File.dirname(__FILE__)
In CFML there are two functions for accessing the path of a script:
getBaseTemplatePath()
getCurrentTemplatePath()
Calling getBaseTemplatePath returns the path of the 'base' script - i.e. the one that was requested by the web server.
Calling getCurrentTemplatePath returns the path of the current script - i.e. the one that is currently executing.
Both paths are absolute and contain the full directory+filename of the script.
To determine just the directory, use the function getDirectoryFromPath( ... ) on the results.
So, to determine the directory location of an application, you could do:
<cfset Application.Paths.Root = getDirectoryFromPath( getCurrentTemplatePath() ) />
Inside of the onApplicationStart event for your Application.cfc
To determine the path where the app server running your CFML engine is at, you can access shell commands with cfexecute, so (bearing in mind above discussions on pwd/etc) you can do:
Unix:
<cfexecute name="pwd"/>
for Windows, create a pwd.bat containing text #cd, then:
<cfexecute name="C:\docume~1\myuser\pwd.bat"/>
(Use the variable attribute of cfexecute to store the value instead of outputting to screen.)
In cmd (the Microsoft command line shell)
You can get the name of the script with %* (may be relative to pwd)
This gets directory of script:
set oldpwd=%cd%
cd %0\..
set app_dir=%pwd%
cd %oldpwd%
If you find any bugs, which you will. Then please fix or comment.
I released https://github.com/gpakosz/whereami which solves the problem in C and gives you:
the path to the current executable
the path to the current module (differs from path to executable when calling from a shared library).
It uses GetModuleFileNameW on Windows, parses /proc/self/maps on Linux and Android and uses _NSGetExecutablePath or dladdr on Mac and iOS.
Note to answer "20 above regarding Mac OSX only: If a JAR executable is transformed to an "app" via the OSX JAR BUNDLER, then the getClass().getProtectionDomain().getCodeSource().getLocation(); will NOT return the current directory of the app, but will add the internal directory structure of the app to the response. This internal structure of an app is /theCurrentFolderWhereTheAppReside/Contents/Resources/Java/yourfile
Perhaps this is a little bug in Java. Anyway, one must use method one or two to get the correct answer, and both will deliver the correct answer even if the app is started e.g. via a shortcut located in a different folder or on the desktop.
carl
SoundPimp.com