Develop Reference

file not found exception while trying to read file in netbeans

I am making an simple application to play sounds in Java. I am able to do that when I keep the audio files in D: disk. Here is the code
in = new FileInputStream("D:\\"+selectedSounds[position]+".wav");
//Some code for playing audio
Then I placed the audio files in same package where the Jframe class is present. But when I run it prompts fileNotFound exception. Can some one tell me why this is happening.
in = new FileInputStream(selectedSounds[position]+".wav");
// I have also tried
new FileInputStream("./"+selectedSounds[position]+".wav");
Here is the file path

Your wave file, contained within the "Source Packages" won't be accessible once the program is packaged as a Jar, as the files will be embedded within the Jar itself and no longer accessible as files.
Instead, you should be using Class#getResourceAsStream, for example...
try (InputStream in = getClass().getResourceAsStream("/PlayAudio/" + selectedSounds[position]+".wav")) {
// You now have an InputStream to your resource, have fun
} catch (IOException | NullPointerException exp) {
exp.printStackTrace();
}

working with files in dynamic web project

I am working on a dynamic web project.
On submit button click (present on my form)I want to create a new file and put some data inside.
I have written only these two lines and I am getting failure to create file
try{
File file = new File("C:/database.txt");
file.createNewFile();
}catch(Exception e){
return "error in creating file";
}
If I run the enire code in normal java class everything works fine. Why so?

Your web project is working on application server. The web application can manage files, which are on that server. Other files are not accessible. (of course localhost server is on your computer, but that it's path is not "C:/", so you can't write there). You can find the path of your server running this code (it also create a test file):
String pathWhereYouFindYourFile = new File("").getAbsolutePath();
System.out.println(pathWhereYouFindYourFile);
File f = new File("test.txt");
try {
f.createNewFile();
} catch (IOException e) {
e.printStackTrace();
}

Java: store configuration variables in a file

I'm developing a program with NetBeans 8.0 and JavaFX Scene Builder 2.0 that need store some variables in a file, where admin users can modify it when needed, (like change server IP address, or a number value from a no editable textfield) and if they close and load again the program, the changes made in variables are kept. Like any settings section of a program.
I just try do it with the Properties file, but i have problems to store it in the same folder as .jar file. When the program execute the line new FileOutputStream("configuration.properties"); the file is created at root of the disk. As the folder of the file can be stored anywhere, i not know how indicate the right path.
Creating the properties file in the package of the main project and using getClass().getResourceAsStream("configuration.properties"); i can read it but then i can not write in for change values of variables.
Is there a better method to create a configuration file? Or properties file is the best option for this case?
My other question is whether it is possible to prevent access to the contents of the file or encrypt the content?
PD: I've been testing this part of the code in Linux operating system currently, but the program will be used in Windows 7 when ready.

If you use Maven, you can store your property files in your resources folder, say resources/properties/. When you need to load them, do this:
private Properties createProps(String name)
{
Properties prop = new Properties();
InputStream in = null;
try
{
in = getClass().getResourceAsStream(name);
prop.load(in);
}
catch (IOException ex)
{
System.err.println("failed to load \"" + name + "\": " + ex);
}
finally
{
try
{
if (in != null)
{
in.close();
}
}
catch (IOException ex)
{
System.err.println("failed to close InputStream for \"" + name + "\":\n" + FXUtils.extractStackTrace(ex));
}
}
return prop;
}
Where name is the full path to your properties file within your resources folder. For example, if you store props.properties in resources/properties/, then you would pass in properties/props.properties.
I am not 100% sure if you can carry over this exact procedure to a non-Maven project. You'd need to instruct whatever compiler tool you are using to also include your property files.
As far as your final question goes, in regards to encrypting your properties, I would consider posting that as a separate question (after having done thorough research to try to discover an existing solution that works for you).

At last i found how obtain the absolute path from folder where is .jar file to create properties file in, and read/write it. Here is the code:
File file = new File(System.getProperty("java.class.path"));
File filePath = file.getAbsoluteFile().getParentFile();
String strPath = filePath.toString();
File testFile = new File(strPath+"/configuration.properties");
Tested in Ubuntu 13.04 And Windows 7 and it works.
For encrypt the properties values i found this thread that answer how do it.

Java 7 zipfs: File is also a folder

I've been using Java 7's ZipFS support.
https://gist.github.com/stain/5591420
shows the behaviour, which I find a bit odd. Basically you can create a ZIP file system, make a file with a given name, and then also make a folder with the same name.
The reason for this seems to be that internally the folder gets "/" appended to its name - however this new name is not returned, therefore you end up in a strange situation where Files.isDirectory() returns false immediately after a successful Files.createDirectory().
try (FileSystem fs = tempZipFS()) {
Path folder = fs.getPath("folder");
Files.createFile(folder);
assertTrue(Files.isRegularFile(folder));
assertFalse(Files.isDirectory(folder));
// try {
Files.createDirectory(folder);
// } catch (FileAlreadyExistsException ex) {
// Is not thrown!
// }
// but a second createDirectory() fails correctly
try {
Files.createDirectory(folder);
} catch (FileAlreadyExistsException ex) {
}
// Look, it's both a file and folder!
Path child = folder.resolve("child");
Files.createFile(child);
// Can this be tested?
assertTrue(Files.isRegularFile(folder));
// Yes, if you include the final /
assertTrue(Files.isDirectory(fs.getPath("folder/")));
// But not the parent
// assertTrue(Files.isDirectory(child.getParent()));
// Or the original Path
// assertTrue(Files.isDirectory(folder));
}
So as long as you have the "/" as the suffix, you can even work with both, and that's how they are listed if you do a directory listing of the root.
Now the ZIP format itself allows this as it only deals with entries in a ZIP file (even allowing multiple entries with the same name), however normal use of a "FileSystem" would normally not allow multiple entries with the same name ; as can be seen when I try to create the folder twice.
The produced ZIP file can be browsed correctly with infozip, 7Zip and Windows 8; but trying to unzip will obviously fail because the native file system don't support such duality.
So is this a feature, bug or something in between?

Understanding Simple XML Parser - New File Output - Java

I am trying to learn how to use the Simple XML Framework as detailed in this thread : Best practices for parsing XML.
I am using the following code :
public class SimpleXMLParserActivity extends Activity {
/** Called when the activity is first created. */
#Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
//setContentView(R.layout.main);
Serializer serializer = new Persister();
Example example = new Example("Example message", 123);
File result = new File("example.xml");
try {
Log.d("Start", "Starting Serializer");
serializer.write(example, result);
} catch (Exception e) {
// TODO Auto-generated catch block
Log.d("Self", "Error");
e.printStackTrace();
}
}
}
I am having a problem understanding the line
File result = new File("example.xml");
1) Does this line create a new file in my app called example.xml ? If so where is this file located.
2) Or does this line look for an existing file called example.xml and then add to it ? If so where should the example.xml file be placed in my app bundle so that it can be found. I do notice at the moment I am getting an error message :
java.io.FileNotFoundException: /example.xml (Read-only file system)
Thank you.

File result = new File("example.xml")
This line will just store the filename "example.xml" in a new File object. There is no check if that file actually exists and it does not try to create it either.
A file without specifying an absolute path (starting with / like new File("/sdcard/example.xml")) is considered to be in the current working directory which I guess is / for Android apps (-> /example.xml (Read-only file system))
I guess serializer.write(example, result); tries to create the actual file for your but fails since you can't write to '/'.
You have to specify a path for that file. There are several places you can store files, e.g.
Context#getFilesDir() will give you a place in your app's home directory (/data/data/your.package/files/) where only you can read / write - without extra permission.
Environment#getExternalStorageDirectory() will give you the general primary storage thing (might be /sdcard/ - but that's very different for devices). To write here you'll need the WRITE_EXTERNAL_STORAGE permission.
there are more places available in Environment that are more specialized. E.g. for media files, downloads, caching, etc.
there is also Context#getExternalFilesDir() for app private (big) files you want to store on the external storage (something like /sdcard/Android/data/your.package/)
to fix your code you could do
File result = new File(Environment.getExternalStorageDirectory(), "example.xml");
Edit: either use the provided mechanisms to get an existing directory (preferred but you are limited to the folders you are supposed to use):
// via File - /data/data/your.package/app_assets/example.xml
File outputFile = new File(getDir("assets", Context.MODE_PRIVATE), "example.xml");
serializer.write(outputFile, result);
// via FileOutputStream - /data/data/your.package/files/example.xml
FileOutputStream outputStream = openFileOutput("example.xml", Context.MODE_PRIVATE);
serializer.write(outputStream, result);
or you may need to create the directories yourself (hackish way to get your app dir but it should work):
File outputFile = new File(new File(getFilesDir().getParentFile(), "assets"), "example.xml");
outputFile.mkdirs();
serializer.write(outputFile, result);
Try to avoid specifying full paths like "/data/data/com.simpletest.test/assets/example.xml" since they might be different on other devices / Android versions. Even the / is not guaranteed to be /. It's safer to use File.separatorChar instead if you have to.

2 solutions to do it cleanly :
use openFileOutput to write a private file in the application private directory (which could be located in the internal memory or the external storage if the app was moved there). See here for a snippet
or use the File constructor to create the File anywhere your app has write access. This is if you want to store the file on the SDCard for example. Instantiating a file doesn't create it on the file system, unless you start writiung to it (with FileOutputStream for example)
I'd recommend approach 1, it's better for users because these files get erased when your app is uninstalled. If the file is large, then using the External Storage is probably better.

What I read on the Android pages, I see it creates a file with that name:
File constructor
I think it writes it to the /data/data/packagname directory
edit: the 'packagename' was not shown in the tekst above. I put it between brackets. :s

Try saving to /sdcard/example.xml.