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Write multiple Zip files to an OutputStream using IOUtils.copy method

This question feels very difficult to explain to me but ill do my best.
Currently I have a method that returns an InputStream with a Zip file that i have to add to a main zip file. Problem is when I write something into the OutputStream it overwrites previous written data. I tried using ZipOutputStream with ZipEntries but this recompresses the file and does weird things, so its not a solution. Things that I'm required to use and are not negotiable are:
Retrieving the file with the method that returns an InputStream
Using IOUtils.copy() method to download the file (this may be optional if u have another solution that allows me to download the file through a browser)
This is the code so far:
OutputStream os = null;
InputStream is = null;
try {
os = response.getOutputStream();
for (int i = 0; i < splited.length; i += 6) {
String[] file= //an array with the data to retrieve the file
is = FileManager.downloadFile(args);
int read;
byte[] buffer = new byte[1024];
while (0 < (read = is.read(buffer))) {
os.write(buffer, 0, read);
}
}
} catch (Exception ex) {
//Exception captures
}
response.setHeader("Content-Disposition", "attachment; filename=FileName");
response.setContentType("application/zip");
IOUtils.copy(is, os);
os.close();
is.close();
return forward;

You can use ZipOutputStream wrapping the response's OutputStream but instead of close call finish, and do not call close on the ResponseOutputStream.
You must start with the HTTP headers.
response.setHeader("Content-Disposition", "attachment; filename=FileName");
response.setContentType("application/zip");
try {
ZipOutputStream os = new ZipOutputStream(response.getOutputStream());
for (int i = 0; i < splited.length - 5; i += 6) {
String[] file= //an array with the data to retrieve the file
try (InputStream is = FileManager.downloadFile(args)) {
os.putNextEntry(new ZipEntry(filePath));
is.TransferTo(os);
os.closeEntry();
}
}
os.finish();
} catch (Exception ex) {
//Exception captures
}
return forward;
Since java 9 transferTo copies Input/OutputStreams.
One can also copy a Path with Files.copy(Path, OutputStream) where Path is an URI based generalisation of File, so also URLs might immediately be copied.
Here try-with-resources ensures that every is is closed.

Java write exe file

is it possible to write/create an exe file in Java?
I can successfully read it but writing the exact same data that has been read to a new file seems to create some trouble because Windows tell's me it's not supported for my pc anymore.
This is the code I'm using to read the file where path is a String given with the actual path (it's in the .jar itself that's why I'm using ResourceAsStream()):
try {
InputStream inputStream = FileIO.class.getResourceAsStream(path);
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream));
ArrayList<String> _final = new ArrayList<String>();
String line;
while ((line = reader.readLine()) != null) {
_final.add(line);
}
inputStream.close();
return _final.toArray(new String[_final.size()]);
}catch(Exception e) {
return null;
}
This is the code I'm using to write the file:
public static void writeFileArray(String path, String[] data) {
String filename = path;
try{
FileWriter fileWriter = new FileWriter(filename);
BufferedWriter bufferedWriter = new BufferedWriter(fileWriter);
for(String d : data) {
bufferedWriter.write(d + "\n");
}
bufferedWriter.close();
}
catch(IOException ex){
System.out.println("FileIO failed to write file, IO exception");
}
}
So it doesn't give me any error's or something and the file size of the original .exe and the 'transferred' .exe stays the same, but it doesn't work anymore. Am I just doing it wrong? Did I forget something? Can u even do this with Java?
Btw I'm not that experienced with reading/writing files..
Thanks for considering my request.

I'm going to guess that you're using a Reader when you should be using a raw input stream. Use BufferedInputStream instead of BufferedReader.
BufferedInputStream in = new BufferedInputStream( inputStream );
The problem is that Reader interprets the binary as your local character set instead of the data you want.
Edit: if you need a bigger hint start with this. I just noticed you're using a BufferedWriter too, that won't work either.
try {
InputStream inputStream = FileIO.class.getResourceAsStream(path);
BufferedInputStream in = new BufferedInputStream( inputStream );
ByteArrayOutputStream bos = new ByteArrayOutputStream();
byte[] bytes = new byte[ 1024 ];
for( int length; ( length = ins.read( bytes ) ) != -1; )
bos.write( bytes, 0, length );
}
inputStream.close();
return bos;

When you are using Java 7 or newer, you should copy a resource to a file using
public static void copyResourceToFile(String resourcePath, String filePath) {
try(InputStream inputStream = FileIO.class.getResourceAsStream(resourcePath)) {
Files.copy(inputStream, Paths.get(filePath));
}
catch(IOException ex){
System.out.println("Copying failed. "+ex.getMessage());
}
}
This construct ensures correct closing of the resources even in the exceptional case and the JRE method ensures correct and efficient copying of the data.
It accepts additional options, e.g. to specify that the target file should be overwritten in case it already exists, you would use
public static void copyResourceToFile(String resourcePath, String filePath) {
try(InputStream inputStream = FileIO.class.getResourceAsStream(resourcePath)) {
Files.copy(inputStream, Paths.get(filePath), StandardCopyOption.REPLACE_EXISTING);
}
catch(IOException ex){
System.out.println("Copying failed. "+ex.getMessage());
}
}

You are using InputStreams for strings, .exe files are bytes!
Try using a ByteArrayInputStream and ByteArrayOutputStream.
Edit: completing with markspace's answer:
new BufferedInputStream(new ByteArrayInputStream( ... ) )

How to read audio file

In an Android project I am reading an audio file into an InputStream and consequently write it to another location on the device. When reading, in.read(buffer) as shown in the snippet returns -1. After running the app, I locate the file using the phone's file manager app to play. But it doesn't play because the file is empty, i.e. size is 0 bytes. What is the right way to read and write audio files in using InputStreams and OutputStreams?
try {
DocumentFile newFile = pickedDir.createFile("audio/mp3", "New File");
OutputStream out = getContentResolver().openOutputStream(newFile.getUri());
InputStream in = new FileInputStream("/storage/emulated/0/beat.mp3");
// To check whether the source file exists
File testFile = File("/storage/emulated/0/beat.mp3");
Log.d("App", "exists: " + testFile.exists() + " len: " + testFile.length());//exists: true len: 0
byte[] buffer = new byte[1024];
int read;
while ((read = in.read(buffer)) != -1) {
out.write(buffer, 0, read);
}
in.close();
out.flush();
out.close();
} catch (Exception e) {
Log.d("Exception ", e.getMessage());
}

to be pedantic, use:
FileInputStream in = new FileInputStream("/storage/emulated/0/beat.mp3");
And just make sure the path is correct and the file exists. Other than that, from the info you gave, I cannot think of anything else

OutputStream out = getContentResolver().openOutputStream(newFile.getUri());
InputStream in = new FileInputStream("/storage/emulated/0/beat.mp3");
File newFile = File("/storage/emulated/0/beat.mp3");
Seem your code have problem here, newFile.getUri() is called before it init?

Download a text file from the web - Android/ Java

I'm trying to get save a text file from the internet into a folder in my res directory (res/files) so I can then read and interpret it. My android manifest has set the appropiate permissions but when I test it in the simulator it fails.
<uses-permission android:name="android.permission.INTERNET"></uses-permission>
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE"></uses-permission>
Here's the method to get the file:
public void getTextFile(){
String path ="http://hullmc.org.uk/cjvize/test.txt";
URL u = null;
try {
u = new URL(path);
BufferedReader in = new BufferedReader(new InputStreamReader(u.openStream()));
int i = 0;
String replicated = "";
do{
String str = in.readLine();
replicated = replicated + "/n" + str;
i++;
}while(i<85);
in.close();
}
catch(Exception e){
welcome.setText("Failed");
}
}
Can anyone suggest why this is not working? Many thanks!

This is working fine for me :
Use of class variable for View and Activity allow to keep code centralaized and shared, passing view as parameter, updated in constructor :)
1) Code to store the file locally
View newReport;
Activity reportActivity;
private void downloadFile(String fileUrl, String fileName) {
try{
InputStream is = (InputStream) new URL(fileUrl).getContent();
FileOutputStream output = reportActivity.openFileOutput(fileName, newReport.getContext().MODE_PRIVATE);
byte data[] = new byte[1024];
int count;
while ((count = is.read(data)) != -1)
output.write(data, 0, count);
output.flush();
output.close();
is.close();
} catch (IOException e) {
e.printStackTrace();
}
}
It saves the file on the internal storage.
Then to save a file from URL, just call:
downloadFile(myFileUrl, mySaveToFileName);
And to list your local files available:
String[] fileList = newReport.getContext().fileList();
for (String s : fileList){
System.out.println("File found : "+s);
}
Note: you do not require to save it locally to read it. If you prefer just to read it (to extract some info), let me know.
2) Code to "read and save to database", this should resolve:
// After InputStream declaration:
BufferedReader in = new BufferedReader(new InputStreamReader(is));
String inputLine;
while ((inputLine = in.readLine()) != null)
{
//TODO Update database row concatenating inputLine to existing text value.
}
in.close();
in=null;
is.close();

you can't save into the resource folder of your app. you can't even store files into the assets folder.
there aren't even such folders when you install the app - they are all zipped into the APK . the res folder is a special one too, because each file there also creates a constant in the "R.java" file, so that it would be easier to reach and use. you can't reach such a thing when it's dynamic...
what you can do is to choose the right folder for you (read here), and download the file into there, using something like this :
InputStream input = new BufferedInputStream(url.openStream());
OutputStream output = new FileOutputStream(fullFilePath);
byte data[] = new byte[1024];
int count;
while ((count = input.read(data)) != -1)
output.write(data, 0, count);
//todo close streams and handle exceptions
if you use Apache commons library, you could minimize the code to just one line:
IOUtils.copy(new BufferedInputStream(url.openStream()), new FileOutputStream(fullFilePath));

How to get Java File absolute path from InputStream?

I'm on Java 6 and I have a method that scans the runtime classpath for a file called config.xml. If found, I would like to read the contents of the file into a string:
InputStream istream = this.getClass().getClassLoader().getResourceAsStream("config.xml");
if(istream != null) {
System.out.println("Found config.xml!");
StringBuffer fileData = new StringBuffer(1000);
BufferedReader reader;
try {
reader = new BufferedReader(new FileReader(fileName));
char[] buf = new char[1024];
int numRead = 0;
while((numRead=reader.read(buf)) != -1) {
String readData = String.valueOf(buf, 0, numRead);
fileData.append(readData);
buf = new char[1024];
reader.close();
}
} catch (FileNotFoundException fnfExc) {
throw new RuntimeException("FileNotFoundException: " + fnfExc.getMessage());
} catch (IOException ioExc) {
throw new RuntimeException("IOException: " + ioExc.getMessage());
}
}
When I run this code, I get the following console output:
Found config.xml!
Exception in thread "main" java.lang.RuntimeException: FileNotFoundException: config.xml (No such file or directory)
at com.me.myapp.Configurator.readConfigFileFromClasspath(Configurator.java:556)
at com.me.myapp.Configurator.<init>(Configurator.java:34)
...rest of stack trace omitted for brevity
So the classpath scan for config.xml is successful, but then the reader can't seem to find the file. Why??? My only theory is that when config.xml is found on the classpath, it doesn't contain an absolute path to the location of the file on the file system, and perhaps that's what the reader code is looking for.

You use a resource from a classloader.
Instead of doing:
InputStream istream = this.getClass().getClassLoader().getResourceAsStream("config.xml");
do:
URL url = getClass().getResource("config.xml");
That URL will have the path (use .toURI().getPath()). To open the matching input stream afterwards, use .openStream().
You know at least that the resource exists: if it doesn't, .getResource{,AsStream}() both return null (instead of throwing an IOException, which is doubtful imho)

From your given example, it is not clear what fileName refers to. You should just use the stream you got from getResourceAsStream()to read you file, something along
reader = new BufferedReader(new InputStreamReader(istream));
And you should avoid to repeatedly allocating buf new for every read cycle, once is enough.