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Android File Permissions issues on FilesDir()

I need to use a file in my application. If i upload the file to Data/Data/APP/files then it is added with -rw-rw-rw permissions which i can then use in my application. If i programatically write the file to getFilesDir() the exact same directory, i can see the 2 exact files in the same directory, however the programatically saved file has permissions -rw------- i cannot then access the file in my app using getfilesDir().
this is how the file is saved:
public void writeFileOnInternalStorage(Context mcoContext,String sFileName, String sBody){
File file = new File(getApplicationContext().getFilesDir(), "");
if(!file.exists()){
file.mkdir();
}
try{
File gpxfile = new File(file, sFileName);
FileWriter writer = new FileWriter(gpxfile);
writer.append(sBody);
writer.flush();
writer.close();
}catch (Exception e){
e.printStackTrace();
}
}
How can i get the correct permissions to use the file. It may well not be a permissions issue it maybe the way i am saving the file? It is a .graphml extension file.

What do you mean by cannot access your file?
Take a look at this documentation (https://developer.android.com/training/data-storage), you don't need permission to access (read and write) files inside App-Specific Directory.
Edit:
I'm assuming that you already stored the file to the disk.
Please note that to read and write files inside App-Specific Directory doesn't require permission. You can read it using this simple code.
public File readFile(Context context, String fileName) {
File file = new File(context.getFilesDir(), fileName);
// do other stuff, like checking if the file exist, etc.
return file;
}
It doesn't matter what file extension it is, as long as the file exists, you will get it.
Actually there are so many articles that already cover this topic, please take a look to understand this topic better.

Casting multiform part to a file

I have a managed bean. An HTML/JSF page sends a zipped file as a Part. I need to extract the contents from it, but to do that I need it to be File. Is it possible to cast the Part to File like as follows, and then treat it like a normal zip file?
Part xmlFile;
public void myMethod()
{
File zippedFile = (File) getXmlFile();
....
}
If not, how would I go about doing so? I've looked on online but there seems to be very little information.

You can create a temporary file:
File aFile = File.createTempFile(PREFIX, SUFFIX);
And write the contents of your Part payload into that file:
try (InputStream input = part.getInputStream()) {
Files.copy(input, aFile.toPath(), StandardCopyOption.REPLACE_EXISTING);
}
You may also want to make use of File's delete() or deleteOnExit() for cleaning up resources. Bear in mind that deleteOnExit() is only invoked when the JVM exits, which may not be what you want in your case.

How can I access a non-class file inside a jar FROM 'within' the jar

This should be simple but it has cost me hours. Everything I find on this site indicates I am doing it right but the file still cannot be found.
Inside a jar file I have two files 'CDAkeystore.jks' and 'CDAtruststore.jks' at top level.
Yet when I call
securityProps.setProperty("javax.net.ssl.keyStore","CDAkeystore.jks");
I get a system cannot find the file requested error.
The class file calling this method is inside the same jar in the usual package arrangement.
The jar file is as follows:
com ..... (a lot more class files)
org ..... (lots of class files)
META-INF
CDAtruststore.jks
CDAkeystore.jks
How can this be SOOO difficult?!!
---------- Added INfo ------n
Since the object using the path is open source I found the routine they are using to load the file. It is:
InputStream keystoreInputStream = preBufferInputStream(new FileInputStream(keyStoreName));
which according to the documentation of FileInputStream(String name) is
Creates a FileInputStream by opening a connection to an actual file, the file named by the path name 'name' in the file system. So how should this path be expressed?

Use YourClass.class.getResourceAsStream() or this.getClass().getResourceAsStream(). You can also use class loader if you are in multiple class loaders environment.

The answer is, in short, that you can't. At least in this situation. I am stuck with passing a path to a file to a library implementation that I have no control over. So the library method accesses the file on the assumption that the file exists in unzipped form in the OS's file system. It is getting the path from a Property.setProperty(stringKey, stringPath)
So the only solution I found was an ugly hack. I need to take the resource in my jar and copy it to a file on the system. Then I would pass the path to that file in the above setProperty() method. The ugly hack is implemented as follows (if anyone else can come up with a nicer solution I would be happy). It does solve the problem. The library routine is able to find my newly created file.
/* This evil reads a file as a resource inside a jar and dumps the file where ever
* the loader of this jar/application defines as the current directory. This pain is done
* so the SecurityDomian class can load the file; it cannot access the file from
* the jar. This means the 'name' passed in contains the file name inside the jar
* prefixed with "/" so it is not read with an assumed package extension.
*/
private boolean createFileFromResource(String name)
{
// Dont bother if the file already exists
if(new File(name.replace("/", "")).exists())
{
return true;
}
byte[] keystoreFile = new byte[2048];
ByteArrayOutputStream byteArrayOut = new ByteArrayOutputStream(2048);
// Get the resource
InputStream inputStream = this.getClass().getResourceAsStream(name);
try
{
int bytesRead = 0;
while(true)
{
// Read the resource into the buffer keystoreFile in 2048 byte chunks
bytesRead = inputStream.read(keystoreFile);
if(bytesRead < 0)
{
break;
}
// Copy and append the chunks to the ByteArrayOutputStream (this class
// does the auto-extending of the output array as more chunks are
// added so you don't have to do it.
byteArrayOut.write(keystoreFile, 0, bytesRead);
}
inputStream.close();
// Now create a file at the root of where ever the loader happens to think
// the root is. So remove the "/" in front of the file name
FileOutputStream outputStream = new FileOutputStream(name.replace("/", ""));
// Write out the file. Note you will be left with garbage at that location.
byteArrayOut.writeTo(outputStream);
outputStream.flush();
outputStream.close();
}
catch (IOException e)
{
e.printStackTrace();
return false;
}
return true;
}

Java: File output help

Fixed: Instead of calling isFile() I used exists() and it seems to be working fine. If possible could someone explain why this change worked?
I'm attempting to write out to an excel file but am having a problem when trying to create that file if the name already exists.
Basically I am taking a file that is uploaded to a server, reading it, and then outputting a report file in a new location with the same filename. I tried to do this by simply checking if the file already existed and then adding a number onto the filename. My code works if the file doesn't exist or if it exists without a number (e.g. filename.xls). If a file exists with the name "filename1.xls" the server just seems to hang when trying to write the file. What can do to fix this?
Here is my code:
String destination = "c:/apache-tomcat-7.0.8/webapps/reports/" + fileName.substring( fileName.lastIndexOf("\\")+1, fileName.lastIndexOf(".")) + ".xls";
int filenum = 1;
while (new File(destination).isFile()) {
destination = "c:/apache-tomcat-7.0.8/webapps/reports/" + fileName.substring( fileName.lastIndexOf("\\")+1, fileName.lastIndexOf(".")) + filenum + ".xls";
filenum++;
}
WritableWorkbook workbook = Workbook.createWorkbook(new File(destination));

That will happen if some process is still keeping the file open. E.g. you've created a FileInputStream on the file to read it, but are never calling close() on it after reading.
Unrelated to the problem, the expanded WAR folder is not the best place to use as a permanent storage. All those files in the expanded WAR folder will get lost whenever you redeploy the WAR. Also hardcoding a servletcontainer-specific path in the code makes it totally unportable.
If your actual intent is to return the Excel file on a per-request basis to the client using a servlet, then you should be using
WritableWorkbook workBook = Workbook.createWorkbook(response.getOutputStream());
// ...
This way it writes to the response immediately without the need for an intermediate file.

Use the File.createTempFile(prefix, suffix, directory) API:
String localName = new File(fileName).getName();
String nameNoExt = localName.substring(0, fileName.lastIndexOf("."));
String extension = localName.substring(fileName.lastIndexOf(".")); // need to include the .
File directory = new File("c:/apache-tomcat-7.0.8/webapps/reports/");
File destFile = File.createTempFile(nameNoExt, extension, directory)

zip a folder structure using java

I am trying to zip the following file structure on my machine,
parent/
parent/test1
parent/test1/image1.jpeg
parent/test2
The problem here is i cant zip the above file structure using java. I have google and found following code sample but it only zip the files only inside a given folder.
File inFolder=new File("out");
File outFolder=new File("Out.zip");
ZipOutputStream out = new ZipOutputStream(new
BufferedOutputStream(new FileOutputStream(outFolder)));
BufferedInputStream in = null;
byte[] data = new byte[1000];
String files[] = inFolder.list();
for (int i=0; i<files.length; i++)
{
in = new BufferedInputStream(new FileInputStream
(inFolder.getPath() + "/" + files[i]), 1000);
out.putNextEntry(new ZipEntry(files[i]));
int count;
while((count = in.read(data,0,1000)) != -1)
{
out.write(data, 0, count);
}
out.closeEntry();
}
out.flush();
out.close();
In the above code the out is a folder and we need to have some files..also folder cannot be empty if so it throws a exception java.util.zip.ZipException or cant contain any sub folders even files inside it (eg:out\newfolder\image.jpeg) if so it throws a java.io.FileNotFoundException: out\newfolder (Access is denied).
In my case im costructig the above file structure by quering the database sometime empty folders along the folder structure can be have.
Can some one please tell me a solution?
Thank You.

What is probably happening is that you're trying to treat every entry as a FileInputStream. However, for a directory, this is not true. Since the path is not to a file, when you try to read it, a FileNotFoundException is thrown. For directories, you still want to create the ZipEntry, but instead of trying to read in any data, just skip it and move on to the next path.

write two methods. The first one takes dirpath, makes a zip stream and calls another method which copies files to the zip stream and calls itself recursively for directories as below:
open an entry in the zip stream for the given directory
list files and dirs in the given directory, loop through them
if an entry is a file, open an entry, copy file content to the entry, close it
if an entry is a directory, call this method. Pass the zip stream
close the entry.
The first method closes the zip stream.