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Java program creates files, but files are not visible

I have a program which replaces text in a file based on command line input. Currently it creates a temporary file, and writes the string with the replaced text in the new temporary file. This program works on a desktop in a computer lab on my campus, but when I try to run it on my personal laptop, the temporary file is created, I can find it by printing its canonical path, and file.exists() returns true, but it does not show up on my desktop.
A search using Windows Explorer yields nothing.
I am running Windows 7 and using TextPad. Does anyone know what might be causing this? I can supply any other necessary information.
Edit: I am running Windows 7 on a Mac Pro 2011, if that makes any difference at all.
Edit: I discovered the problem. I had downloaded Comodo Antivirus software and whenever I created a file it would create it in a VTRoot folder for sandbox purposes. I was able to alter the settings and solved my issue.
import java.io.*;
import java.util.*;
public class ReplaceText{
public static void main(String[] args)throws IOException{
if(args.length != 2){
System.out.println("Incorrect format. Use java ClassName textToReplace filename");
System.exit(1);
}
File source = new File(args[1]);
if(!source.exists()){
System.out.println("Source file " + args[1] + " does not exist.");
System.exit(2);
}
File temp = new File("temp.txt");
try(
Scanner input = new Scanner(source);
PrintWriter output = new PrintWriter(temp);
){
while(input.hasNext()){
String s1 = input.nextLine();
String s2 = s1.replace(args[0], "a");
output.println(s2);
}
}
}
}

If you could not see the temp.txt file it is because it is located in your project directory where your java source code are located.However if you will found it then also it will contain nothing.The reason being you have not closed the output stream to file just place
output.close()
after the while loop.

Try to create file by providing entire file path while creating like
File(URI uri)
This creates a new File instance by converting the given file: URI into an abstract pathname.

I discovered the problem. I had downloaded Comodo Antivirus software and whenever I created a file it would create it in a VTRoot folder for sandbox purposes. I was able to alter the settings and solved my issue.

If you have installed comodo antivirus then you should follow the instruction :
1.Open the comodo,
2.Click on the settings,
3.Then click on the Containment -> click on the Auto-containment,
4.Then at the top, uncheck the enable auto-containment.

FileOutputStream not written to disk on server? Where does the file go?

I've been trying to generate report files with Dynamic Reports, however it just doesn't seem to create files on Server. When I use the same method running locally it generates the file, however when I run it on server, there are not files created. I'm running Tomcat 7 in Eclipse. The file is supposed to be created using FileOutputStream.
Well here's the method that works locally, but not on Tomcat:
StyleBuilder plainStyle = stl.style().setFontName("FreeUniversal");
StyleBuilder boldStyle = stl.style(plainStyle).bold();
StyleBuilder italicStyle = stl.style(plainStyle).italic();
StyleBuilder boldItalicStyle = stl.style(plainStyle).boldItalic();
try {
report().title(
Templates.createTitleComponent("Fonts"),
cmp.text("FreeUniversal font - plain").setStyle(plainStyle),
cmp.text("FreeUniversal font - bold").setStyle(boldStyle),
cmp.text("FreeUniversal font - italic").setStyle(italicStyle),
cmp.text("FreeUniversal font - bolditalic").setStyle(boldItalicStyle))
.toDocx(createFile("docx"))
// .show()
;
} catch (DRException e) {
e.printStackTrace();
}
Oh and the CreateFile(...) method:
private FileOutputStream createFile(String extension) throws FileNotFoundException {
FileOutputStream file;
String filePath = reportsPath + "generated_report." + extension;
filePath = "generated_report." + extension;
System.out.println("FILENAME IS: " + filePath);
file = new FileOutputStream(new File(filePath));
return file;
}
I know the reportsPath here is not active.
So there are no exceptions. By the way, it might be possible that other files don't get created on this server too, because I'm also uploading a file via servlet and it get used, however it doesn't appear anywhere in the path and it doesn't seem to be saved, but I don't need to keep the uploaded files anyway, so it wasn't much of a concern, but now this? I need to be able to find those reports, so the files must get created.
And I'm sure it's not that I can't find the files, I've run the search everywhere, actually in all of my computer on which the server is running, there were no files created with that name...
So, any ideas? It must be a Tomcat configuration issue or something?
Any ideas? Thanks

The output was written to the file you supplied in the constructor. If that was a relative filename, it is relative to the current working directory when you executed the code.
Or else an IOException was thrown and the code didn't execute at all.

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

zip folder in windows using command line

I am writing a program that needs to zip a file.
This will run over both linux and windows machines. It works just fine in Linux but I am not able to get anything done in windows.
To send commands I am using the apache-net project. I've also tried using Runtime().exec
but it isn't working.
Can somebody suggest something?
CommandLine cmdLine = new CommandLine("zip");
cmdLine.addArgument("-r");
cmdLine.addArgument("documents.zip");
cmdLine.addArgument("documents");
DefaultExecutor exec = new DefaultExecutor();
ExecuteWatchdog dog = new ExecuteWatchdog(60*1000);
exec.setWorkingDirectory(new File("."));
exec.setWatchdog(dog);
int check =-1;
try {
check = exec.execute(cmdLine);
} catch (ExecuteException e) {
} catch (IOException e) {
}

Java provides its own compression library in java.util.zip.* that supports the .zip format. An example that zips a folder can be found here. Here's a quickie example that works on a single file. The benefit of going with native Java is that it will work on multiple operating systems and is not dependent on having specific binaries installed.
public static void zip(String origFileName) {
try {
String zipName=origFileName + ".zip";
ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(new FileOutputStream(zipName)));
byte[] data = new byte[1000];
BufferedInputStream in = new BufferedInputStream(new FileInputStream(origFileName));
int count;
out.putNextEntry(new ZipEntry(origFileName));
while((count = in.read(data,0,1000)) != -1) {
out.write(data, 0, count);
}
in.close();
out.flush();
out.close();
} catch (Exception ex) {
ex.printStackTrace();
}
}

The same code won't work in Windows. Windows doesn't have a "zip" program the way that Linux does. You will need to see if Windows 7 has a command line zip program (I don't think it does; see here: http://answers.microsoft.com/en-us/windows/forum/windows_vista-files/how-to-compress-a-folder-from-command-prompt/02f93b08-bebc-4c9d-b2bb-907a2184c8d5). You will likely need to do two things
Make sure the user has a suitable 3rd party zip program
Do OS detection to execute the proper command.

You can use inbuilt compact.exe to compress/uncompress in dos
It displays or alters the compression of files on NTFS partitions.
COMPACT [/C | /U] [/S[:dir]] [/A] [/I] [/F] [/Q] [filename [...]]
/C Compresses the specified files. Directories will be marked so that files added afterward will be compressed.
/U Uncompresses the specified files. Directories will be marked so that files added afterward will not be compressed.
/S Performs the specified operation on files in the given directory and all subdirectories. Default "dir" is the current directory.
/A Displays files with the hidden or system attributes. These files are omitted by default.
/I Continues performing the specified operation even after errors have occurred. By default, COMPACT stops when an error is encountered.
/F Forces the compress operation on all specified files, even those that are already compressed. Already-compressed files are skipped by default.
/Q Reports only the most essential information.
filename Specifies a pattern, file, or directory.
Used without parameters, COMPACT displays the compression state of the current directory and any files it contains. You may use multiple filenames and wildcards. You must put spaces between multiple parameters.
Examples
compact
Display all the files in the current directory and their compact status.
compact file.txt
Display the compact status of the file file.txt
compact file.txt /C
Compacts the file.txt file.

Hosting an executable within Android application

I am working on an Android application that depends on an ELF binary:
our Java code interacts with this binary to get things done. This
runtime needs to be started and terminated on Application startup and
application exit / on demand.
Questions:
I am assuming that we will be able to execute this binary using the
Runtime.exec() API. Is there any constraints as to where I
need to be putting my library in the folder structure? How would the system runtime locate this executable? Is there some sort of class path setting?
Since the application has dependencies on this Runtime, I was
thinking of wrapping it around a service so that it can be started or
stopped as required. What is the best way to handle such executables
in Android project?
What are other alternatives, assuming that I do not have source code for this executable?
Please advice.
Thanks.

1) No, there should be no constrains, besides those that access system files and thus require root. The best place would be straight to /data/data/[your_package_name] to avoid polluting elsewhere.
2) A very thorough discussion about compiling against native libraries can be found here: http://www.aton.com/android-native-libraries-for-java-applications/ . Another option is a cross-compiler for arm (here is the one used to compile the kernel, it's free: http://www.codesourcery.com/sgpp/lite/arm ). If you plan to maintain a service that executes your cammand, be warned that services can be stopped and restarted by android at any moment.
3) Now, if you don't have the source code, I hope that your file is at least compiled as an arm executable. If not, I don't see how you could even run it.
You will execute the file by running the following commands in your java class:
String myExec = "/data/data/APPNAME/FILENAME";
Process process = Runtime.getRuntime().exec(myExec);
DataOutputStream os = new DataOutputStream(process.getOutputStream());
DataInputStream osRes = new DataInputStream(process.getInputStream());
I know nothing about your executable, so you may or may not need to actually get the inputStream and outputStream.
I am assuming that running adb to push the binary file is out of the question, so
I was looking for a neat way to package it. I found a great post about including an executable in your app. Check it out here:
http://gimite.net/en/index.php?Run%20native%20executable%20in%20Android%20App
The important part is this one (emphasis mine):
From Android Java app, using assets folder
Include the binary in the assets folder.
Use getAssets().open(FILENAME) to get an InputStream.
Write it to /data/data/APPNAME (e.g. /data/data/net.gimite.nativeexe), where your application has access to write files and make it executable.
Run /system/bin/chmod 744 /data/data/APPNAME/FILENAME using the code above.
Run your executable using the code above.
The post uses the assets folder, insted of the raw folder that android suggests for static files:
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).
To access the data folder, you can follow the instructions here:
http://developer.android.com/guide/topics/data/data-storage.html#filesInternal
Also, there's the File#setExecutable(boolean); method that should works instead of the shell command.
So, putting everything together, I would try:
InputStream ins = context.getResources().openRawResource (R.raw.FILENAME)
byte[] buffer = new byte[ins.available()];
ins.read(buffer);
ins.close();
FileOutputStream fos = context.openFileOutput(FILENAME, Context.MODE_PRIVATE);
fos.write(buffer);
fos.close();
File file = context.getFileStreamPath (FILENAME);
file.setExecutable(true);
Of course, all this should be done only once after installation. You can have a quick check inside onCreate() or whatever that checks for the presence of the file and executes all this commands if the file is not there.
Let me know if it works. Good luck!

Here is a complete guide for how to package and run the executable. I based it on what I found here and other links, as well as my own trial and error.
1.) In your SDK project, put the executable file in your /assets folder
2.) Programmatically get the String of that files directory (/data/data/your_app_name/files) like this
String appFileDirectory = getFilesDir().getPath();
String executableFilePath = appFileDirectory + "/executable_file";
3.) In your app's project Java code: copy the executable file from /assets folder into your app's "files" subfolder (usually /data/data/your_app_name/files) with a function like this:
private void copyAssets(String filename) {
AssetManager assetManager = getAssets();
InputStream in = null;
OutputStream out = null;
Log.d(TAG, "Attempting to copy this file: " + filename); // + " to: " + assetCopyDestination);
try {
in = assetManager.open(filename);
Log.d(TAG, "outDir: " + appFileDirectory);
File outFile = new File(appFileDirectory, filename);
out = new FileOutputStream(outFile);
copyFile(in, out);
in.close();
in = null;
out.flush();
out.close();
out = null;
} catch(IOException e) {
Log.e(TAG, "Failed to copy asset file: " + filename, e);
}
Log.d(TAG, "Copy success: " + filename);
}
4.) Change the file permissions on executable_file to actually make it executable. Do it with Java calls:
File execFile = new File(executableFilePath);
execFile.setExecutable(true);
5.) Execute the file like this:
Process process = Runtime.getRuntime().exec(executableFilePath);
Note that any files referred to here (such as input and output files) must have their full path Strings constructed. This is because this is a separate spawned process and it has no concept of what the "pwd" is.
If you want to read the command's stdout you can do this, but so far it's only working for me for system commands (like "ls"), not the executable file:
BufferedReader reader = new BufferedReader(
new InputStreamReader(process.getInputStream()));
int read;
char[] buffer = new char[4096];
StringBuffer output = new StringBuffer();
while ((read = reader.read(buffer)) > 0) {
output.append(buffer, 0, read);
}
reader.close();
process.waitFor();
Log.d(TAG, "output: " + output.toString());

For executing binary file starting from Android 10 it's only possible from read-only folder. It means that you should pack binary with your app. Android doc
Put android:extractNativeLibs="true" into AndroidManifest;
Put your binary to src/main/resources/lib/* directory, where * – stands for architecture of CPU, for instance armeabi-v7a;
Use code like this for executing:
private fun exec(command: String, params: String): String {
try {
val process = ProcessBuilder()
.directory(File(filesDir.parentFile!!, "lib"))
.command(command, params)
.redirectErrorStream(true)
.start()
val reader = BufferedReader(
InputStreamReader(process.inputStream)
)
val text = reader.readText()
reader.close()
process.waitFor()
return text
} catch (e: Exception) {
return e.message ?: "IOException"
}
}
Here is discussion with answer from android team on reddit.

I've done something like this using the NDK. My strategy was to recompile the program using the NDK and write some wrapper JNI code that called into the program's main function.
I'm not sure what the lifecycle of NDK code is like. Even services that are intended to be long-running can be started and stopped by the system when convenient. You would probably have to shutdown your NDK thread and restart it when necessary.