I am trying to transfer a .mp4 file using WebRTC and it's DataChannel. In order to do that I am breaking the file into chunks like below:
FileInputStream is = new FileInputStream(file);
byte[] chunk = new byte[260000];
int chunkLen = 0;
sentFileByte = new ArrayList<>();
while ((chunkLen = is.read(chunk)) != -1) {
sentFileByte.add(chunk);
}
After that, sending the chunks by index like:
byte[] b = sentFileByte.get(index);
ByteBuffer bb = ByteBuffer.wrap(b);
bb.put(b);
bb.flip();
dataChannel.send(new DataChannel.Buffer(bb, true));
On the receiver end I am receiving the chunks and adding it to an Arraylist
receivedFileByteArr.add(chunkByteArr);
After receiving all the chunks successfully I am trying to convert these in to a file like below:
String path = Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DOWNLOADS).getAbsolutePath() + "/" + fileName;
File file = new File(path);
try {
FileOutputStream fileOutputStream = new FileOutputStream(file);
for (int i = 0; i < receivedFileByteArr.size(); i++) {
fileOutputStream.write(receivedFileByteArr.get(i));
}
fileOutputStream.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
After completing all these steps, file is created successfully. File size is also same. But problem is the file is not playable in any video player. I guess I am making some mistake on FileInputStream and FileOutputStream. I need help to fix this error.
With Java:
I have a byte[] that represents a file.
How do I write this to a file (ie. C:\myfile.pdf)
I know it's done with InputStream, but I can't seem to work it out.
Use Apache Commons IO
FileUtils.writeByteArrayToFile(new File("pathname"), myByteArray)
Or, if you insist on making work for yourself...
try (FileOutputStream fos = new FileOutputStream("pathname")) {
fos.write(myByteArray);
//fos.close(); There is no more need for this line since you had created the instance of "fos" inside the try. And this will automatically close the OutputStream
}
Without any libraries:
try (FileOutputStream stream = new FileOutputStream(path)) {
stream.write(bytes);
}
With Google Guava:
Files.write(bytes, new File(path));
With Apache Commons:
FileUtils.writeByteArrayToFile(new File(path), bytes);
All of these strategies require that you catch an IOException at some point too.
Another solution using java.nio.file:
byte[] bytes = ...;
Path path = Paths.get("C:\\myfile.pdf");
Files.write(path, bytes);
Also since Java 7, one line with java.nio.file.Files:
Files.write(new File(filePath).toPath(), data);
Where data is your byte[] and filePath is a String. You can also add multiple file open options with the StandardOpenOptions class. Add throws or surround with try/catch.
From Java 7 onward you can use the try-with-resources statement to avoid leaking resources and make your code easier to read. More on that here.
To write your byteArray to a file you would do:
try (FileOutputStream fos = new FileOutputStream("fullPathToFile")) {
fos.write(byteArray);
} catch (IOException ioe) {
ioe.printStackTrace();
}
Try an OutputStream or more specifically FileOutputStream
Basic example:
String fileName = "file.test";
BufferedOutputStream bs = null;
try {
FileOutputStream fs = new FileOutputStream(new File(fileName));
bs = new BufferedOutputStream(fs);
bs.write(byte_array);
bs.close();
bs = null;
} catch (Exception e) {
e.printStackTrace()
}
if (bs != null) try { bs.close(); } catch (Exception e) {}
File f = new File(fileName);
byte[] fileContent = msg.getByteSequenceContent();
Path path = Paths.get(f.getAbsolutePath());
try {
Files.write(path, fileContent);
} catch (IOException ex) {
Logger.getLogger(Agent2.class.getName()).log(Level.SEVERE, null, ex);
}
////////////////////////// 1] File to Byte [] ///////////////////
Path path = Paths.get(p);
byte[] data = null;
try {
data = Files.readAllBytes(path);
} catch (IOException ex) {
Logger.getLogger(Agent1.class.getName()).log(Level.SEVERE, null, ex);
}
/////////////////////// 2] Byte [] to File ///////////////////////////
File f = new File(fileName);
byte[] fileContent = msg.getByteSequenceContent();
Path path = Paths.get(f.getAbsolutePath());
try {
Files.write(path, fileContent);
} catch (IOException ex) {
Logger.getLogger(Agent2.class.getName()).log(Level.SEVERE, null, ex);
}
I know it's done with InputStream
Actually, you'd be writing to a file output...
This is a program where we are reading and printing array of bytes offset and length using String Builder and Writing the array of bytes offset length to the new file.
`Enter code here
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
//*This is a program where we are reading and printing array of bytes offset and length using StringBuilder and Writing the array of bytes offset length to the new file*//
public class ReadandWriteAByte {
public void readandWriteBytesToFile(){
File file = new File("count.char"); //(abcdefghijk)
File bfile = new File("bytefile.txt");//(New File)
byte[] b;
FileInputStream fis = null;
FileOutputStream fos = null;
try{
fis = new FileInputStream (file);
fos = new FileOutputStream (bfile);
b = new byte [1024];
int i;
StringBuilder sb = new StringBuilder();
while ((i = fis.read(b))!=-1){
sb.append(new String(b,5,5));
fos.write(b, 2, 5);
}
System.out.println(sb.toString());
}catch (IOException e) {
e.printStackTrace();
}finally {
try {
if(fis != null);
fis.close(); //This helps to close the stream
}catch (IOException e){
e.printStackTrace();
}
}
}
public static void main (String args[]){
ReadandWriteAByte rb = new ReadandWriteAByte();
rb.readandWriteBytesToFile();
}
}
O/P in console : fghij
O/P in new file :cdefg
You can try Cactoos:
new LengthOf(new TeeInput(array, new File("a.txt"))).value();
More details: http://www.yegor256.com/2017/06/22/object-oriented-input-output-in-cactoos.html
This question already has answers here:
Convert InputStream to byte array in Java
(34 answers)
Closed 4 years ago.
I am trying to convert an InputStream into a byte array to write it in a file, to generate a PDF.
I have a File type with the url of a PDF, and with that, i have the inputStream of that.
File fichero_pdf = new File("C:/Users/agp2/Desktop/PDF_TRIAXE.pdf");
InputStream stream4 = new FileInputStream(fichero_pdf);
Until here everything is perfect, the problem appears when i try to transform this InputStream to a byte[] and write it in a new File.
I have these two methods:
to convert the Stream to a byte[]:
private static byte[] getArrayFromInputStream(InputStream is) {
BufferedReader br = null;
StringBuilder sb = new StringBuilder();
String line;
try {
br = new BufferedReader(new InputStreamReader(is));
while ((line = br.readLine()) != null) {
sb.append(line+"\n");
}
} catch (IOException e) {
e.printStackTrace();
} finally {
if (br != null) {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
return sb.toString().getBytes();
}
To write the byte[] in the new file:
...
File file=new File(dto.getTitulo());
InputStream stream=dto.getContenido();
byte[] array=getStringFromInputStream(stream);
OutputStream salida=new FileOutputStream(file);
salida.write(array);
salida.close();
stream.close();
helper.addAttachment(file.getName(), file);
}
mailSender.send(message);
...
The Email is sent at perfection, but when i can't open the .pdf.
Also, i compare the code of the new pdf with the code of the first, and is a little bit different.
I need to create a valid pdf file from an inputStream.
You have 2 problems:
You are trying to read bytes as strings, but you don't have to do it. In your case you should use byte streams(FileInputStream, BufferedInputStream), not char streams(InputStreamReader, BufferedReader).
You loose data when you convert String to bytes here:
return sb.toString().getBytes();
I would like to suggest you to use java 7 try-with-resources instead of try-catch-finally.
You can read the whole file to a byte array using ByteArrayOutputStream.
Sample code does the following:
getArrayFromInputStream() - reads all file bytes to byte array
writeContent() - writes content to a new file, in my example pdf_sample2.pdf
Example:
public class ReadAllBytes {
// as example - write to resources folder
private static String DIR = "src\\main\\resources\\";
public static void main(String[] args) throws IOException {
try {
byte[] fileAsBytes = getArrayFromInputStream(new FileInputStream(new File(DIR + "pdf-sample.pdf")));
writeContent(fileAsBytes, DIR + "pdf_sample2.pdf");
} catch (Exception e){
e.printStackTrace();
}
}
private static byte[] getArrayFromInputStream(InputStream inputStream) throws IOException {
byte[] bytes;
byte[] buffer = new byte[1024];
try(BufferedInputStream is = new BufferedInputStream(inputStream)){
ByteArrayOutputStream bos = new ByteArrayOutputStream();
int length;
while ((length = is.read(buffer)) > -1 ) {
bos.write(buffer, 0, length);
}
bos.flush();
bytes = bos.toByteArray();
}
return bytes;
}
private static void writeContent(byte[] content, String fileToWriteTo) throws IOException {
File file = new File(fileToWriteTo);
try(BufferedOutputStream salida = new BufferedOutputStream(new FileOutputStream(file))){
salida.write(content);
salida.flush();
}
}
}
I've created a Huffman coding algorithm, and then I wrote binary code in String and put it in binary file using FileOutputStream and DataOutputStream.
But now I cant understand how to read it? I need to get 1 and 0 sequence from binary file.
There is no method like .readString() in DataInputStream
try{
FileChooser fileChooser = new FileChooser();
fileChooser.getExtensionFilters().add(new FileChooser.ExtensionFilter("Binary", "*.bin"));
FileOutputStream fileOutputStream = new FileOutputStream(fileChooser.showSaveDialog(window));
DataOutputStream outputStream = new DataOutputStream(fileOutputStream);
outputStream.writeChars(Main.string_ready_encode);
}catch (IOException e){
e.printStackTrace();
}
Main.string_ready_encode contains ready sequence
The problem with your writing code is that you have specified no file format. We now can only read the file if we know how many bytes it has. If you do know that, you can read it by doing the following:
try (DataInputStream stream = new DataInputStream(new FileInputStream(f))) {
byte[] bytes = new byte[NUMBER_OF_BYTES];
stream.read(bytes);
String content = new String(bytes);
System.out.println(content);
} catch (IOException e) {
e.printStackTrace();
}
But I would actually advise you to rewrite you file with some known file format, like so:
try (Writer writer = new OutputStreamWriter(new FileOutputStream(f), Charsets.UTF_8)) {
writer.write(Main.stringReadyEncode, 0, Main.stringReadyEncode.length());
} catch (IOException x) {
e.printStackTrace();
}
And read it like you would read any other file:
try (BufferedReader r = Files.newBufferedReader(f.toPath(), Charsets.UTF_8)) {
String line;
while((line = r.readLine()) != null) {
// do whatever you want with line
System.out.println(line);
}
} catch (IOException e) {
e.printStackTrace();
}
Just make sure to replace Charsets.UTF_8 with whatever encoding you used while writing to the file.
I've been trying to tackle this problem for a day or two and can't seem to figure out precisely how to add text files to a zip file, I was able to figure out how to add these text files to a 7zip file which was insanely easy, but a zip file seems to me much more complicated for some reason. I want to return a zip file for user reasons btw.
Here's what I have now:
(I know the code isn't too clean at the moment, I plan to tackle that after getting the bare functionality down).
private ZipOutputStream addThreadDumpsToZipFile(File file, List<Datapoint<ThreadDump>> allThreadDumps, List<Datapoint<String>> allThreadDumpTextFiles) {
ZipOutputStream threadDumpsZipFile = null;
try {
//creat new zip file which accepts input stream
//TODO missing step: create text files containing each thread dump then add to zip
threadDumpsZipFile = new ZipFile(new FileOutputStream(file));
FileInputStream fileInputStream = null;
try {
//add data to each thread dump entry
for(int i=0; i<allThreadDumpTextFiles.size();i++) {
//create file for each thread dump
File threadDumpFile = new File("thread_dump_"+i+".txt");
FileUtils.writeStringToFile(threadDumpFile,allThreadDumpTextFiles.get(i).toString());
//add entry/file to zip file (creates block to add input to)
ZipEntry threadDumpEntry = new ZipEntry("thread_dump_"+i); //might need to add extension here?
threadDumpsZipFile.putNextEntry(threadDumpEntry);
//add the content to this entry
fileInputStream = new FileInputStream(threadDumpFile);
byte[] byteBuffer = new byte[(int) threadDumpFile.length()]; //see if this sufficiently returns length of data
int bytesRead = -1;
while ((bytesRead = fileInputStream.read(byteBuffer)) != -1) {
threadDumpsZipFile.write(byteBuffer, 0, bytesRead);
}
}
threadDumpsZipFile.flush();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
fileInputStream.close();
} catch(Exception e) {
}
}
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return threadDumpsZipFile;
}
As you can sort of guess, I have a set of Thread Dumps that I want to add to my zip file and return to the user.
Let me know if you guys need any more info!
PS: There might be some bugs in this question, I just realized with some breakpoints that the threadDumpFile.length() won't really work.
Look forward to your replies!
Thanks,
Arsa
Here's a crack at it. I think you'll want to keep the file extensions when you make your ZipEntry objects. See if you can implement the below createTextFiles() function; the rest of this works -- I stubbed that method to return a single "test.txt" file with some dummy data to verify.
void zip()
{
try {
FileOutputStream fos = new FileOutputStream("yourZipFile.zip");
ZipOutputStream zos = new ZipOutputStream(fos);
File[] textFiles = createTextFiles(); // should be an easy step
for (int i = 0; i < files.length; i++) {
addToZipFile(file[i].getName(), zos);
}
zos.close();
fos.close();
} catch (Exception e) {
e.printStackTrace();
}
}
void addToZipFile(String fileName, ZipOutputStream zos) throws Exception {
File file = new File(fileName);
FileInputStream fis = new FileInputStream(file);
ZipEntry zipEntry = new ZipEntry(fileName);
zos.putNextEntry(zipEntry);
byte[] bytes = new byte[1024];
int length;
while ((length = fis.read(bytes)) >= 0) {
zos.write(bytes, 0, length);
}
zos.closeEntry();
fis.close();
}