here's my problem.
I have a txt file in the /raw folder and I want to write into it.
From the Android guide i saw this:
Tip: If you want to save a static file in your application at compile time, save the file in your project res/raw/ directory. You can open it with openRawResource(), passing the R.raw. resource ID. This method returns an InputStream that you can use to read the file (but you cannot write to the original file).
So in my code I did this:
FileOutputStream myFile = openRawResource(R.raw.max_easy, Context.MODE_PRIVATE);
But gives me error in openRawResource()... How do i solve this?
Thank you for the help!
I have a txt file in the /raw folder and I want to write into it
That is not possible at runtime. Resources and assets are read-only at runtime.
But gives me error in openRawResource()... How do i solve this?
openRawResource() is for reading in the resource, and it gives you an InputStream. You are welcome to write your data to an ordinary file, such as on internal storage.
Related
I'm trying to read a text file located in src/main/resources/test/file.txt. I'm trying to get the path of the file using String path = getClass().getResource("/text/file.txt").getFile(); but when I try to read it I get a FileNotFoundException. I tried putting many different paths, all of which failed. How can I go about doing this?
The idea of putting something into the src/main/resources tree is that it will be copied into the JAR file that you build from your project. It will then be available to your application via the Class methods getResource(String) and getResourceAsStream(String) methods.
When you are running in your application in the development environment, it is certainly possible to use FileInputStream etcetera to access the resource. But this won't work in production. In production, the resources will then be inside your app's JAR file. FileInputStream cannot open a JAR file and its contents by name.
When you do this:
getClass().getResource("/text/file.txt");
you get a URL for the resource, which will look something like this:
jar:file:/path/to/your.jar!/text/file.txt"
It is not possible to turn that into a pathname the FileInputStream will understand. Whatever you try will give you a FileNotFoundException ... or something that is not the resource you want to read.
So what to do?
You have a few options, depending on your application's requirements.
You can use getResourceAsStream and use the resulting input stream directly.
You can copy the contents of getResourceAsStream to a temporary file, and then use the pathname of the temporary file.
You can create an application specific directory (e.g. in the user's home directory) and extract the file you need from the JAR into the directory. You might do this the first time the application runs.
You could open the JAR file as a JarFile and use that API to open an InputStream for the resource. But this assumes that that the resources are in a JAR ... and on some platforms (e.g. Windows) you may encounter problems with file locking. (And it would be a bad idea to attempt to update the resource in the JAR.)
Try giving complete path of the file from the disk.
C:\Users\MyUser\Desktop\file name with extension
I'm trying to read from a file in Android Studio, in a small Java app. So I'm trying this:
File test = new File("C:\\testing\\testFile.dat");
if (test.exists()) {
System.out.println("test exists");
}
else {
System.out.println("test doesn't exist");
}
The file definitely exists, but it keeps on reporting that the file doesn't exist. I was able to work around this with another file by using the AssetManager and reading it through a stream, but the method I'm calling now requires a File's absolute path, but it's point blank refusing to find the file.
Am I doing something dumb, or misunderstanding something?
UPDATE
Ok, thanks for the input, I've now solved the problem. First I had to upload the file I wanted into the virtual device's storage, then I was able to get the path to it.
File test = new File(this.getFilesDir().getAbsolutePath(), "testFile.dat");
but the method I'm calling now requires a File's absolute path
Assets are files on your development machine. They are not files on the device.
Ideally, you switch to some library that supports InputStream or similar options, rather than requiring a filesystem path. If that is not an option, you can always get the InputStream from AssetManager and use that to make a copy of the data in some file that you control (e.g., in getCacheDir()). You can then pass the path to that file to this method.
You can place this file in your assets/ folder inside the android project and access using the following code.
val inputSteam = assets.open("testFile.dat")
or place it inside the res/raw folder and access it like below.
val inputStream = resources.openRawResource(R.raw.testFile)
We can't access a file on a development machine like this and won't be available on an android device so it will break so it's better if we move this somewhere inside the project and access it as above.
I am trying to make a dictionary in android.
But i am facing problem to get input from text file.
I have put the text file in res/raw directory.
and write the code:
InputStream inputStream=getResources().openRawResource(R.raw.Dictionary);
but having exception my IDE indicates a problem under raw word.
How can I fix this ?
Place the file in the assets folder and open it like this
getResources().getAssets().open("dictionary.txt");
I have a Java application in Eclipse that references .XML files as templates for other functionality. Usually I package the .JAR file without these files, because placing them within the same folder as the .JAR file seems to work fine with this reference:
File myFile = new File("templates/templateA.xsd");
I now require that these templates be placed within the same .JAR file as this application. I can include them with no problems, but these references no longer seem to work.
Is there a correct way of referencing the .XML file from within the same .JAR that the application is running from?
You need to know how to load the files from class path.
one of the ways is as follows
class XMLLoader {
public String loadXML(String fileName){
InputStream inputStream = this.getClass().getClassLoader().getResourceAsStream(fileName);
// do the loading of the file from the given input stream.
}
}
you know that the "templates" folder should be inside of your jar.
If you just need to read this file, you might not need a java.io.File but just an InputStream that you can get via
this.getClass().getResourceAsStream("templates/templateA.xsd")
If you really need a java.io.File... I do not know... The last time a really needed a File, I just copied the InputStream to a temporary file but this is ugly.
I have a file which I am using as a temporary file until the user selects to save this file, How do I go about making a copy of this file but at the same time re-naming it. Example below:
File 1:
/sdcard/Folder/file1.wav
File 2:
Copy of file 1, renamed to file2.wav in /sdcard/Folder/file2.wav
I understand using Environment.getExternalStorageDirectory().getAbsolutePath() gets me the path to the external storage so i'm not asking how I get to the sdcard or whatever. just to make a copy of a file and re-name it.
Try Apache Commons IO FileUtils.moveFile(File, File).