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Android: copy a file to another package as root

I'm facing an issue while writing an app I need for personal use, and I can't find a solution to it.
I need to take a copy of a binary file from an other application, and after some time I need to restore it. The file is taken from /data/data/app.package/files/, so I should not be able to read/write it without root.
I was able to copy the file to an other location using a file explorer, however I'm not able to copy it back with root shell. If I use a file explorer, the file is read again by the app... if I use my code, the original app says that there is data corruption
I've tried several approaches:
cp [origin] [destination] -> file created, but unable to read.
cp -p [origin] [destination] -> now the permissions seems valid, however it still does not work
cat [origin] > [destination] , and after I manually did chown to set the user and group to the correct one (instead of root) -> still no luck
(taken from here: Copy files to another package folder (root, su) )
All of them used this code:
Process suProcess = Runtime.getRuntime().exec("su");
DataOutputStream os = new DataOutputStream(suProcess.getOutputStream());
BufferedReader stdInput = new BufferedReader(new
InputStreamReader(suProcess.getInputStream()));
for(String command: inputCommands){
Log.d("ROOT","Writing: " + command);
os.writeBytes(command+"\n");
}
os.writeBytes("exit\n");
os.flush();
int suProcessRetval = suProcess.waitFor();
/* This part is usually outside the method,
I copied it here to show how I access the output.
String aLine = null;
while ((aLine = stdInput.readLine() )!= null){
Log.d("ROOT", aLine);
}
*/
os.close();
is there a way to do it? or maybe any ideas that could help me investigating on this?
Thank you!

Extract Files from Java project, during run proccess

I am trying to extract few files contained in the java project into a certain path, lets say "c:\temp".
I tried to use this example :
String home = getClass().getProtectionDomain().
getCodeSource().getLocation().toString().
substring(6);
JarFile jar = new JarFile(home);
ZipEntry entry = jar.getEntry("mydb.mdb");
File efile = new File(dest, entry.getName());
InputStream in =
new BufferedInputStream(jar.getInputStream(entry));
OutputStream out =
new BufferedOutputStream(new FileOutputStream(efile));
byte[] buffer = new byte[2048];
for (;;) {
int nBytes = in.read(buffer);
if (nBytes <= 0) break;
out.write(buffer, 0, nBytes);
}
out.flush();
out.close();
in.close();
I think I am doing it wrong and, this code probably looking for a specific jar but not in my project directory. I prefer to figure a way that can retrieve my files from resources package, inside the project folder and extract it to specific folder i choose.
I am using Eclipse, 1.4 J2SE library.

Well, it's hard to guess what's wrong without any code examples.
But as for pair of random guesses I could tell that sometimes you get this kind of error when the file is locked by earlier instance of your program which is still running. Make sure you've got only one running instance of Eclipse.
Also you can try to refresh the project folder by right click --> refresh to sync your file system with Eclipse's internal file system: when it comes to Eclipse, multiple refresh/rebuild someway magically solves project problems :)

Do I need to delete tmp files created by my java application?

I output several temporary files in my application to tmp directories but was wondering if it is best practise that I delete them on close or should I expect the host OS to handle this for me?
I am pretty new to Java, I can handle the delete but want to keep the application as multi-OS and Linux friendly as possible. I have tried to minimise file deletion if I don't need to do it.
This is the method I am using to output the tmp file:
try {
java.io.InputStream iss = getClass().getResourceAsStream("/nullpdf.pdf");
byte[] data = IOUtils.toByteArray(iss);
iss.read(data);
iss.close();
String tempFile = "file";
File temp = File.createTempFile(tempFile, ".pdf");
FileOutputStream fos = new FileOutputStream(temp);
fos.write(data);
fos.flush();
fos.close();
nopathbrain = temp.getAbsolutePath();
System.out.println(tempFile);
System.out.println(nopathbrain);
} catch (IOException ex) {
ex.printStackTrace();
System.out.println("TEMP FILE NOT CREATED - ERROR ");
}

createTempFile() only creates a new file with a unique name, but does not mark it for deletion. Use deleteOnExit() on the created file to achieve that. Then, if the JVM shuts down properly, the temporary files should be deleted.
edit:
Sample for creating a 'true' temporary file in java:
File temp = File.createTempFile("temporary-", ".pdf");
temp.deleteOnExit();
This will create a file in the default temporary folder with a unique random name (temporary-{randomness}.pdf) and delete it when the JVM exits.
This should be sufficient for programs with a short to medium run time (e.g. scripts, simple GUI applications) that do sth. and then exit. If the program runs longer or indefinitely (server application, a monitoring client, ...) and the JVM won't exit, this method may clog the temporary folder with files. In such a situation the temporary files should be deleted by the application, as soon as they are not needed anymore (see delete() or Files helper class in JDK7).
As Java already abstracts away OS specific file system details, both approaches are as portable as Java. To ensure interoperability have a look at the new Path abstraction for file names in Java7.

Java: get absolute path of project

I'm trying to run a exe file in path outside of the current package. My code.java file that runs it is in
%Workspace_path%\Project\src\main\java\com\util\code.java
However the directory of where the exe is
%Workspace_path%\Project\src\main\resources\program.exe
If possible, it seems like the best solution here would be to get the absolute path of the Project then append "src\main\resources\" to it. Is there a good way to do this or is there an alternative solution?
I'm using Eclipse, but it would great if it could be used in other IDEs too. Thanks for any help.

The de facto approach to solving this is to bundle the EXE as a classpath resource. It seems you have arranged for this already.
When working with classpath resources, a mature program should not assume that the resource is in the filesystem. The resources could be packaged in a JAR file, or even in a WAR file. The only thing you can trust at that point is the standard methods for accessing resources in Java, as hinted below.
The way to solve your problem, then, is to access the resource contents using the de facto standard of invoking Class.getResourceAsStream (or ClassLoader.getResourceAsStream), save the contents to a temporary file, and execute from that file. This will guarantee your program works correctly regardless of its packaging.
In other words:
Invoke getClass().getResourceAsStream("/program.exe"). From static methods, you can't call getClass, so use the name of your current class instead, as in MyClass.class.getResourceAsStream. This returns an InputStream.
Create a temporary file, preferably using File.createTempFile. This returns a File object identifying the newly created file.
Open an OutputStream to this temp file.
Use the two streams to copy the data from the resource into the temp file. You can use IOUtils.copy if you're into Apache Commons tools. Don't forget to close the two streams when done with this step.
Execute the program thus stored in the temporary file.
Clean up.
In other words (code snippet added later):
private void executeProgramFromClasspath() throws IOException {
// Open resource stream.
InputStream input = getClass().getResourceAsStream("/program.exe");
if (input == null) {
throw new IllegalStateException("Missing classpath resource.");
}
// Transfer.
OutputStream output = null;
try {
// Create temporary file. May throw IOException.
File temporaryFile = File.createTempFile(getClass().getName(), "");
output = new FileOutputStream(temporaryFile);
output = new BufferedOutputStream(output);
IOUtils.copy(input, output);
} finally {
// Close streams.
IOUtils.closeQuietly(input);
IOUtils.closeQuietly(output);
}
// Execute.
try {
String path = temporaryFile.getAbsolutePath();
ProcessBuilder processBuilder = new ProcessBuilder(path);
Process process = processBuilder.start();
process.waitFor();
} catch (InterruptedException e) {
// Optional catch. Keeps the method signature uncluttered.
throw new IOException(e);
} finally {
// Clean up
if (!temporaryFile.delete()) {
// Log this issue, or throw an error.
}
}
}

Well,in your context,the project root is happen to be the current path
.
,that is where the java.exe start to execute,so a easy way is:
String exePath="src\\main\\resources\\program.exe";
File exeFile=new File(".",exePath);
System.out.println(exeFile.getAbusolutePath());
...
I tested this code on Eclipse,It's ok. I think is should work on different ide.
Good Luck!

writing temporary files in Tomcat

I need to create temporary directory but I'm always getting access denied when I try to create a file into the temporary directory.
java.io.FileNotFoundException: C:\tmpDir7504230706415790917 (Access Denied)
here's my code:
public static File createTempDir() throws IOException {
File temp = File.createTempFile("tmpDir", "", new File("C:/"));
temp.delete();
temp.mkdir();
return temp;
}
public File createFile(InputStream inputStream, File tmpDir ) {
File file = null;
if (tmpDir.isDirectory()) {
try {
file = new File(tmpDir.getAbsolutePath());
// write the inputStream to a FileOutputStream
OutputStream out = new FileOutputStream(file);
int read = 0;
byte[] bytes = new byte[1024];
while ((read = inputStream.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
inputStream.close();
out.flush();
out.close();
System.out.println("New file created!");
} catch (IOException e) {
System.out.println(e.getMessage());
}
}
return file;
}
I'm working on a web application and I'm using tomcat. Is there a way to create temporary file on tomcat server memory? I know that's bizarre, but I don't know ... maybe it's possible.

You could use Tomcat's temp folder.
If you use
<%=System.getProperty("java.io.tmpdir")%>
in a JSP you can get path to it.

This line in your code says create a file whose name starts with text "tmpDir" in the directory "C:\". That is not what you want.
File temp = File.createTempFile("tmpDir","",new File("C:/"));
The operating system is properly disallowing that because C:\ is a protected directory. Use the following instead:
File temp = File.createTempFile("tmp",null);
This will let Java determine the appropriate temporary directory. Your file will have the simple prefix "tmp" followed by some random text. You can change "tmp" to anything meaningful for your app, in case you need to manually clean out these temp files and you want to be able to quickly identify them.

You usually cannot write onto C:\ directly due to the default permission setting. I sometime have permission issue for doing so. However, you can write your temporary file in your user folder. Usually, this is C:\Documents and Settings\UserName\ on XP or C:\Users\UserName\ on vista and Windows 7. A tool called SystemUtils from Apache Lang can be very useful if you want to get the home directory depending on OS platform.
For example:
SystemUtils.getUserDir();
SystemUtils.getUserHome();
Update
Also, you create a temp file object but you call mkdir to make it into a directory and try to write your file to that directory object. You can only write a file into a directory but not on the directory itself. To solve this problem, either don't call temp.mkdir(); or change this file=new File(tmpDir.getAbsolutePath()); to file=new File(tmpDir, "sometempname");

On Linux with tomcat7 installation:
So if you are running web application this is the temp directory Tomcat uses for the creation of temporary files.
TOMCAT_HOME/temp
In my case TOMCAT_HOME => /usr/share/tomcat7
If you are running Java program without tomcat, by default it uses /tmp directory.
Not sure if it affects but i ran this command too.
chmod 777 on TOMCAT_HOME/temp