I want to 'read' a .MD file inside the public folder (it's need to be in public folder because the user can change these MD files after the build).
File Structure: https://prnt.sc/pg72yq
I think in loading this MD file asynchronously with nodejs or PHP, but i wish load without that, maybe with AJAX, but i actually don't what exactly to do.
/*
Reads the page
*/
read(page){
let file = require(process.env.PUBLIC_URL + `./pages/${page}/index.md`);
// ... Code
}
Related
In my code I am prompting the user to load a json file.
I am then attempting to copy this file into an sqlite database.
Once I have the data I am then able to manipulate it as needed - but I need to get it there in the first place.
So step 1 is to get the data in.
I have progressed as far as prompting the user to navigate to the file they want - but when I try and read the file I get this error ..
ERROR: resources must reside in the root directory thus must start with a '/' character in Codename One! Invalid resource: file:///tmp/temp3257201851214246357..json
So I think that I need to copy this file to the root directory
I cannot find a link that shows me how to do this.
Here is my code so far ...
case "Import Script":
try
{
JSONParser json = new JSONParser();
if (FileChooser.isAvailable()) {
FileChooser.showOpenDialog(".json", e2-> {
String file = (String)e2.getSource();
if (file == null) {
home.add("No file was selected");
home.revalidate();
} else {
home.add("Please wait - busy importing");
home.revalidate();
String extension = null;
if (file.lastIndexOf(".") > 0) {
extension = file.substring(file.lastIndexOf(".")+1);
}
if ("json".equals(extension)) {
FileSystemStorage fs = FileSystemStorage.getInstance();
try {
InputStream fis = fs.openInputStream(file);
try(Reader r = new InputStreamReader(Display.getInstance().getResourceAsStream(getClass(), file), "UTF-8"))
{
Map<String, Object> data = json.parseJSON(r);
Result result = Result.fromContent(data);
...... I progress from here
The error is occurring on this line ...
try(Reader r = new InputStreamReader(Display.getInstance().getResourceAsStream(getClass(), file), "UTF-8"))
If I hard code a filename and manually place it in the /src folder it works ... like this ...
try(Reader r = new InputStreamReader(Display.getInstance().getResourceAsStream(getClass(), '/test.json'), "UTF-8"))
But that defeats the purpose of them selecting a file
Any help would be appreciated
Thanks
I suggest watching this video.
It explains the different ways data is stored. One of the core sources of confusion is the 3 different ways to store files:
Resources
File System
Storage
getResourceAsStream returns a read only path that's physically embedded in the jar. It's flat so all paths to getResourceAsStream must start with / and must have only one of those. I would suggest avoiding more than one . as well although this should work in theory.
The sqlite database must be stored in file system which is encapsulated as FileSystemStorage and that's really the OS native file system. But you can't store it anywhere you want you need to give the DB name to the system and it notifies you where the file is stored and that's whats explained in the code above.
I'm trying to upload a file and place it in my web/uploads/produits/img directory but the code bellow is not working:
public function getUploadDir()
{
return 'uploads/produits/img';
}
protected function getUploadRootDir()
{
return __DIR__.'/../../../../web/'.$this->getUploadDir();
}
I get the folowing error:
Could not move the file "C:\wamp\tmp\php9265.tmp" to "C:\wamp\www\Projet\src\Arkiglass\ProduitBundle/../../../..\web/uploads/produits/img\." (move_uploaded_file()
[function.move-uploaded-file]: Unable to move 'C:\wamp\tmp\php9265.tmp' to 'C:\wamp\www\Projet\src\Arkiglass\ProduitBundle/../../../..\web/uploads/produits/img\.')
It's seems like it doesn't know the directory __DIR__.'/../../../../web/'...
It doesn't seem your entity lives under the entity directory, but rather directly under the bundles dir. Does it? So you're going up one dir to much. Two options:
Move your entity in the Entity subfolder, adjust namespace and all references. Standard symfony bundle dir layout.
remove one level of ../
I have an application that needs to load an add-on in the form of a dll. The dll needs to take its configuration information from a configuration (app.config) file. I want to dynamically find out the app.config file's name, and the way to do this, as I understand , is AppDomain.CurrentDomain.SetupInformation.ConfigurationFile
However, since it is being hosted INSIDE a parent application, the configuration file that is got from the above piece of code is (parentapplication).exe.config. I am not able to load another appdomain inside the parent application but I'd like to change the configuration file details of the appdomain. How should I be going about this to get the dll's configuration file?
OK, in the end, I managed to hack something together which works for me. Perhaps this will help;
Using the Assembly.GetExecutingAssembly, from the DLL which has the config file I want to read, I can use the .CodeBase to find where the DLL was before I launched a new AppDomain for it. The *.dll
.config is in that same folder.
Then have to convert the URI (as .CodeBase looks like "file://path/assembly.dll") to get the LocalPath for the ConfigurationManager (which doesn't like Uri formatted strings).
try
{
string assemblyName = Assembly.GetExecutingAssembly().GetName().Name;
string originalAssemblyPath = Path.GetDirectoryName(Assembly.GetExecutingAssembly().CodeBase);
Uri uri = new Uri(String.Format("{0}\\{1}.dll", originalAssemblyPath, assemblyName));
string dllPath = uri.LocalPath;
configuration = ConfigurationManager.OpenExeConfiguration(dllPath);
}
catch { }
Hi I have a wpf application that plays sounds on events such as button click. And the current code I have now plays the sound file, but Ive realized that it has to be in another folder which for example if the user deletes, makes the application pretty much not useable.
I was wondering, How could I get a .wav file without creating a whole new folder in the application Release directory.
Any ideas?
Thanks.
If the .wav files is added to your project at the root then marked as content (Right Click on the .wav file and click properties. Then change the build action to content) then it will be copied directly to the root output folder
You can the reference the wav file in code or XAML as a relative path (simple "Myfile.wav")
I believe this is what you are asking for?
How about putting the .wav-file as en embedded resource in the dll?
Just add the file to the project, right-click on the file and set Build Action to Embedded Resource.
Then you can load the .wav file as a stream and save it to disc like this:
private void WriteEmbeddedResourceToFile(string writePath, string assemblyName)
{
Assembly asm = Assembly.GetExecutingAssembly();
Stream s = asm.GetManifestResourceStream(assemblyName);
if (s != null)
{
var file = new byte[s.Length];
s.Read(file, 0, (int)s.Length);
File.WriteAllBytes(writePath, file);
s.Close();
}
}
The string assemblyName must be set like this: namespace.(if you put the file in a folder, insert folder subpath here).filename.wav
Alternatively, you can drop saving it to disc, and just use the stream directly:
private byte[] GetEmbeddedResource(string assemblyName)
{
Assembly asm = Assembly.GetExecutingAssembly();
Stream s = asm.GetManifestResourceStream(assemblyName);
if (s != null)
{
var file = new byte[s.Length];
s.Read(file, 0, (int)s.Length);
s.Close();
return file;
}
}
Then, when you want to load the file, GetEmbeddedResource will return the byte array.
So I'm trying to dynamically create a folder inside the web pages folder.
I'm making a game database. Everytime a game is added I do this:
public void addGame(Game game) throws DatabaseException {
em.getTransaction().begin();
em.persist(game);
em.getTransaction().commit();
File file = new File("C:\\GameDatabaseTestFolder");
file.mkdir();
}
So everything works here.
The file get's created.
But I want to create the folder like this:
public void addGame(Game game) throws DatabaseException {
em.getTransaction().begin();
em.persist(game);
em.getTransaction().commit();
File file = new File(game.getId()+"/screenshots");
file.mkdir();
}
Or something like that. So it will be created where my jsp files are and it will have the id off the game.
I don't understand where the folder is created by default.
thank you in advance,
David
It's by default relative to the "current working directory", i.e. the directory which is currently open at the moment the Java Runtime Environment has started the server. That may be for example /path/to/tomcat/bin, or /path/to/eclipse/workspace/project, etc, depending on how the server is started.
You should now realize that this condition is not controllable from inside the web application.
You also don't want to store it in the expanded WAR folder (there where your JSPs are), because any changes will get lost whenever you redeploy the WAR (with the very simple reason that those files are not contained in the original WAR).
Rather use an absolute path instead. E.g.
String gameWorkFolder = "/path/to/game/work/folder";
new File(gameWorkFolder, game.getId()+"/screenshots");
You can make it configureable by supplying it as a properties file setting or a VM argument.
See also:
Image Upload and Display in JSP
getResourceAsStream() vs FileInputStream