How do I convert a java.io.File to a byte[]?
From JDK 7 you can use Files.readAllBytes(Path).
Example:
import java.io.File;
import java.nio.file.Files;
File file;
// ...(file is initialised)...
byte[] fileContent = Files.readAllBytes(file.toPath());
It depends on what best means for you. Productivity wise, don't reinvent the wheel and use Apache Commons. Which is here FileUtils.readFileToByteArray(File input).
Since JDK 7 - one liner:
byte[] array = Files.readAllBytes(Paths.get("/path/to/file"));
No external dependencies needed.
import java.io.RandomAccessFile;
RandomAccessFile f = new RandomAccessFile(fileName, "r");
byte[] b = new byte[(int)f.length()];
f.readFully(b);
Documentation for Java 8: http://docs.oracle.com/javase/8/docs/api/java/io/RandomAccessFile.html
Basically you have to read it in memory. Open the file, allocate the array, and read the contents from the file into the array.
The simplest way is something similar to this:
public byte[] read(File file) throws IOException, FileTooBigException {
if (file.length() > MAX_FILE_SIZE) {
throw new FileTooBigException(file);
}
ByteArrayOutputStream ous = null;
InputStream ios = null;
try {
byte[] buffer = new byte[4096];
ous = new ByteArrayOutputStream();
ios = new FileInputStream(file);
int read = 0;
while ((read = ios.read(buffer)) != -1) {
ous.write(buffer, 0, read);
}
}finally {
try {
if (ous != null)
ous.close();
} catch (IOException e) {
}
try {
if (ios != null)
ios.close();
} catch (IOException e) {
}
}
return ous.toByteArray();
}
This has some unnecessary copying of the file content (actually the data is copied three times: from file to buffer, from buffer to ByteArrayOutputStream, from ByteArrayOutputStream to the actual resulting array).
You also need to make sure you read in memory only files up to a certain size (this is usually application dependent) :-).
You also need to treat the IOException outside the function.
Another way is this:
public byte[] read(File file) throws IOException, FileTooBigException {
if (file.length() > MAX_FILE_SIZE) {
throw new FileTooBigException(file);
}
byte[] buffer = new byte[(int) file.length()];
InputStream ios = null;
try {
ios = new FileInputStream(file);
if (ios.read(buffer) == -1) {
throw new IOException(
"EOF reached while trying to read the whole file");
}
} finally {
try {
if (ios != null)
ios.close();
} catch (IOException e) {
}
}
return buffer;
}
This has no unnecessary copying.
FileTooBigException is a custom application exception.
The MAX_FILE_SIZE constant is an application parameters.
For big files you should probably think a stream processing algorithm or use memory mapping (see java.nio).
As someone said, Apache Commons File Utils might have what you are looking for
public static byte[] readFileToByteArray(File file) throws IOException
Example use (Program.java):
import org.apache.commons.io.FileUtils;
public class Program {
public static void main(String[] args) throws IOException {
File file = new File(args[0]); // assume args[0] is the path to file
byte[] data = FileUtils.readFileToByteArray(file);
...
}
}
If you don't have Java 8, and agree with me that including a massive library to avoid writing a few lines of code is a bad idea:
public static byte[] readBytes(InputStream inputStream) throws IOException {
byte[] b = new byte[1024];
ByteArrayOutputStream os = new ByteArrayOutputStream();
int c;
while ((c = inputStream.read(b)) != -1) {
os.write(b, 0, c);
}
return os.toByteArray();
}
Caller is responsible for closing the stream.
// Returns the contents of the file in a byte array.
public static byte[] getBytesFromFile(File file) throws IOException {
// Get the size of the file
long length = file.length();
// You cannot create an array using a long type.
// It needs to be an int type.
// Before converting to an int type, check
// to ensure that file is not larger than Integer.MAX_VALUE.
if (length > Integer.MAX_VALUE) {
// File is too large
throw new IOException("File is too large!");
}
// Create the byte array to hold the data
byte[] bytes = new byte[(int)length];
// Read in the bytes
int offset = 0;
int numRead = 0;
InputStream is = new FileInputStream(file);
try {
while (offset < bytes.length
&& (numRead=is.read(bytes, offset, bytes.length-offset)) >= 0) {
offset += numRead;
}
} finally {
is.close();
}
// Ensure all the bytes have been read in
if (offset < bytes.length) {
throw new IOException("Could not completely read file "+file.getName());
}
return bytes;
}
You can use the NIO api as well to do it. I could do this with this code as long as the total file size (in bytes) would fit in an int.
File f = new File("c:\\wscp.script");
FileInputStream fin = null;
FileChannel ch = null;
try {
fin = new FileInputStream(f);
ch = fin.getChannel();
int size = (int) ch.size();
MappedByteBuffer buf = ch.map(MapMode.READ_ONLY, 0, size);
byte[] bytes = new byte[size];
buf.get(bytes);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} finally {
try {
if (fin != null) {
fin.close();
}
if (ch != null) {
ch.close();
}
} catch (IOException e) {
e.printStackTrace();
}
}
I think its very fast since its using MappedByteBuffer.
Simple way to do it:
File fff = new File("/path/to/file");
FileInputStream fileInputStream = new FileInputStream(fff);
// int byteLength = fff.length();
// In android the result of file.length() is long
long byteLength = fff.length(); // byte count of the file-content
byte[] filecontent = new byte[(int) byteLength];
fileInputStream.read(filecontent, 0, (int) byteLength);
Simplest Way for reading bytes from file
import java.io.*;
class ReadBytesFromFile {
public static void main(String args[]) throws Exception {
// getBytes from anyWhere
// I'm getting byte array from File
File file = null;
FileInputStream fileStream = new FileInputStream(file = new File("ByteArrayInputStreamClass.java"));
// Instantiate array
byte[] arr = new byte[(int) file.length()];
// read All bytes of File stream
fileStream.read(arr, 0, arr.length);
for (int X : arr) {
System.out.print((char) X);
}
}
}
Guava has Files.toByteArray() to offer you. It has several advantages:
It covers the corner case where files report a length of 0 but still have content
It's highly optimized, you get a OutOfMemoryException if trying to read in a big file before even trying to load the file. (Through clever use of file.length())
You don't have to reinvent the wheel.
import java.io.File;
import java.nio.file.Files;
import java.nio.file.Path;
File file = getYourFile();
Path path = file.toPath();
byte[] data = Files.readAllBytes(path);
Using the same approach as the community wiki answer, but cleaner and compiling out of the box (preferred approach if you don't want to import Apache Commons libs, e.g. on Android):
public static byte[] getFileBytes(File file) throws IOException {
ByteArrayOutputStream ous = null;
InputStream ios = null;
try {
byte[] buffer = new byte[4096];
ous = new ByteArrayOutputStream();
ios = new FileInputStream(file);
int read = 0;
while ((read = ios.read(buffer)) != -1)
ous.write(buffer, 0, read);
} finally {
try {
if (ous != null)
ous.close();
} catch (IOException e) {
// swallow, since not that important
}
try {
if (ios != null)
ios.close();
} catch (IOException e) {
// swallow, since not that important
}
}
return ous.toByteArray();
}
This is one of the simplest way
String pathFile = "/path/to/file";
byte[] bytes = Files.readAllBytes(Paths.get(pathFile ));
I belive this is the easiest way:
org.apache.commons.io.FileUtils.readFileToByteArray(file);
ReadFully Reads b.length bytes from this file into the byte array, starting at the current file pointer. This method reads repeatedly from the file until the requested number of bytes are read. This method blocks until the requested number of bytes are read, the end of the stream is detected, or an exception is thrown.
RandomAccessFile f = new RandomAccessFile(fileName, "r");
byte[] b = new byte[(int)f.length()];
f.readFully(b);
If you want to read bytes into a pre-allocated byte buffer, this answer may help.
Your first guess would probably be to use InputStream read(byte[]). However, this method has a flaw that makes it unreasonably hard to use: there is no guarantee that the array will actually be completely filled, even if no EOF is encountered.
Instead, take a look at DataInputStream readFully(byte[]). This is a wrapper for input streams, and does not have the above mentioned issue. Additionally, this method throws when EOF is encountered. Much nicer.
Not only does the following way convert a java.io.File to a byte[], I also found it to be the fastest way to read in a file, when testing many different Java file reading methods against each other:
java.nio.file.Files.readAllBytes()
import java.io.File;
import java.io.IOException;
import java.nio.file.Files;
public class ReadFile_Files_ReadAllBytes {
public static void main(String [] pArgs) throws IOException {
String fileName = "c:\\temp\\sample-10KB.txt";
File file = new File(fileName);
byte [] fileBytes = Files.readAllBytes(file.toPath());
char singleChar;
for(byte b : fileBytes) {
singleChar = (char) b;
System.out.print(singleChar);
}
}
}
//The file that you wanna convert into byte[]
File file=new File("/storage/0CE2-EA3D/DCIM/Camera/VID_20190822_205931.mp4");
FileInputStream fileInputStream=new FileInputStream(file);
byte[] data=new byte[(int) file.length()];
BufferedInputStream bufferedInputStream=new BufferedInputStream(fileInputStream);
bufferedInputStream.read(data,0,data.length);
//Now the bytes of the file are contain in the "byte[] data"
Let me add another solution without using third-party libraries. It re-uses an exception handling pattern that was proposed by Scott (link). And I moved the ugly part into a separate message (I would hide in some FileUtils class ;) )
public void someMethod() {
final byte[] buffer = read(new File("test.txt"));
}
private byte[] read(final File file) {
if (file.isDirectory())
throw new RuntimeException("Unsupported operation, file "
+ file.getAbsolutePath() + " is a directory");
if (file.length() > Integer.MAX_VALUE)
throw new RuntimeException("Unsupported operation, file "
+ file.getAbsolutePath() + " is too big");
Throwable pending = null;
FileInputStream in = null;
final byte buffer[] = new byte[(int) file.length()];
try {
in = new FileInputStream(file);
in.read(buffer);
} catch (Exception e) {
pending = new RuntimeException("Exception occured on reading file "
+ file.getAbsolutePath(), e);
} finally {
if (in != null) {
try {
in.close();
} catch (Exception e) {
if (pending == null) {
pending = new RuntimeException(
"Exception occured on closing file"
+ file.getAbsolutePath(), e);
}
}
}
if (pending != null) {
throw new RuntimeException(pending);
}
}
return buffer;
}
public static byte[] readBytes(InputStream inputStream) throws IOException {
byte[] buffer = new byte[32 * 1024];
int bufferSize = 0;
for (;;) {
int read = inputStream.read(buffer, bufferSize, buffer.length - bufferSize);
if (read == -1) {
return Arrays.copyOf(buffer, bufferSize);
}
bufferSize += read;
if (bufferSize == buffer.length) {
buffer = Arrays.copyOf(buffer, bufferSize * 2);
}
}
}
Another Way for reading bytes from file
Reader reader = null;
try {
reader = new FileReader(file);
char buf[] = new char[8192];
int len;
StringBuilder s = new StringBuilder();
while ((len = reader.read(buf)) >= 0) {
s.append(buf, 0, len);
byte[] byteArray = s.toString().getBytes();
}
} catch(FileNotFoundException ex) {
} catch(IOException e) {
}
finally {
if (reader != null) {
reader.close();
}
}
Try this :
import sun.misc.IOUtils;
import java.io.IOException;
try {
String path="";
InputStream inputStream=new FileInputStream(path);
byte[] data=IOUtils.readFully(inputStream,-1,false);
}
catch (IOException e) {
System.out.println(e);
}
Can be done as simple as this (Kotlin version)
val byteArray = File(path).inputStream().readBytes()
EDIT:
I've read docs of readBytes method. It says:
Reads this stream completely into a byte array.
Note: It is the caller's responsibility to close this stream.
So to be able to close the stream, while keeping everything clean, use the following code:
val byteArray = File(path).inputStream().use { it.readBytes() }
Thanks to #user2768856 for pointing this out.
try this if you have target version less than 26 API
private static byte[] readFileToBytes(String filePath) {
File file = new File(filePath);
byte[] bytes = new byte[(int) file.length()];
// funny, if can use Java 7, please uses Files.readAllBytes(path)
try(FileInputStream fis = new FileInputStream(file)){
fis.read(bytes);
return bytes;
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return null;
}
In JDK8
Stream<String> lines = Files.lines(path);
String data = lines.collect(Collectors.joining("\n"));
lines.close();
I have this code snippet in my application and I am quite sure that I have
closed all the streams.
But, surprisingly, I keep getting:
A resource was acquired at attached stack trace but never released. See java.io.Closeable for information on avoiding resource leaks.
java.lang.Throwable: Explicit termination method 'close' not called
Any pointers would be very useful.
if (fd != null) {
InputStream fileStream = new FileInputStream(fd.getFileDescriptor());
ByteArrayOutputStream bos = new ByteArrayOutputStream();
byte[] buf = new byte[1024];
try {
for (int readNum; (readNum = fileStream.read(buf)) != -1;) {
bos.write(buf, 0, readNum);
}
content = bos.toByteArray();
} catch (IOException ex) {
ex.printStackTrace();
} finally {
try {
if (fileStream != null) {
fileStream.close();
}
if (bos != null) {
bos.close();
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
try moving the instantiation of the streams into the try
InputStream fileStream = null;
ByteArrayOutputStream bos = null;
byte[] buf = new byte[1024];
try {
fileStream = new FileInputStream(fd.getFileDescriptor());
bos = new ByteArrayOutputStream();
Try using try with resources. This eliminates the need to close resources in a finally block. https://docs.oracle.com/javase/tutorial/essential/exceptions/tryResourceClose.html
if (fd != null) {
byte[] buf = new byte[1024];
try (InputStream fileStream = new FileInputStream(fd.getFileDescriptor());
ByteArrayOutputStream bos = new ByteArrayOutputStream()) {
for (int readNum; (readNum = fileStream.read(buf)) != -1;) {
bos.write(buf, 0, readNum);
}
content = bos.toByteArray();
} catch (IOException ex) {
ex.printStackTrace();
}
}
Using try with resources would resolve the issue. https://docs.oracle.com/javase/tutorial/essential/exceptions/tryResourceClose.html. You can have a look at AutoCloseable interface here which has been introduced in Java 7 http://docs.oracle.com/javase/7/docs/api/java/lang/AutoCloseable.html
What I see, is that bos won't be closed if an Exception occurres while closing the fileStream.
And as stated before: Use the try-with-resources Statement:
https://docs.oracle.com/javase/tutorial/essential/exceptions/tryResourceClose.html
In my Java program, I would like to display the progress of moving a file. I use the following snippet of code to copy files, which allows me to track the bytes copied and shows it in a progress bar. I was wondering if the code code be adapted to move files rather than just copy them?
BufferedInputStream bis = new BufferedInputStream(new FileInputStream(sourceFile));
BufferedOutputStream bos = new BufferedOutputStream(new FileOutputStream(targetFile));
int theByte;
while((theByte = bis.read()) != -1)
{
bos.write(theByte);
}
bis.close();
bos.close();
Okay, so a "move" operation is a copy with a "delete" at the end, for example...
BufferedInputStream bis = null;
BufferedOutputStream bos = null;
try {
bis = new BufferedInputStream(new FileInputStream(sourceFile));
bos = new BufferedOutputStream(new FileOutputStream(targetFile));
int theByte;
while((theByte = bis.read()) != -1)
{
bos.write(theByte);
}
bos.close();
bis.close();
// You may want to verify that the file's are the same (ie the file size for example)
if (!sourceFile.delete()) {
throw new IOException("Failed to remove source file " + sourceFile);
}
} catch (IOException exp) {
exp.printStackTrace();
} finally {
try {
bis.close();
} catch (Exception exp) {
}
try {
bos.close();
} catch (Exception exp) {
}
}
I am trying to send a file from client to server. Below is the code i have tried. But at times, there is a packet loss during the transfer. I am not sure where i am wrong.
SERVER SIDE CODE:
public static void ReadAndWrite(byte[] aByte, Socket clientSocket,
InputStream inputStream, String fileOutput)
throws FileNotFoundException, IOException {
int bytesRead;
FileOutputStream fileOutputStream = null;
BufferedOutputStream bufferedOutputStream = null;
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
try
{
fileOutputStream = new FileOutputStream( fileOutput );
bufferedOutputStream = new BufferedOutputStream(fileOutputStream);
bytesRead = inputStream.read(aByte, 0, aByte.length);
System.out.println("The length is "+bytesRead);
int count = 0;
do {
count++;
byteArrayOutputStream.write(aByte);
bytesRead = inputStream.read(aByte);
} while (bytesRead != -1);
System.out.println("The count is "+count);
System.out.println("The length is "+byteArrayOutputStream.size());
bufferedOutputStream.write(byteArrayOutputStream.toByteArray());
bufferedOutputStream.flush();
bufferedOutputStream.close();
clientSocket.close();
}
catch(Exception ex)
{
Logger.writeLog(ex,Listen.class.getName(), LogType.EXCEPTION);
throw ex;
}
CLIENT SIDE CODE:
public void readByteArrayAndWriteToClientSocket(
Socket connectionSocket, BufferedOutputStream outToClient, String fileToSend ) throws Exception
{
try{
if (outToClient != null)
{
File myFile = new File(fileToSend);
System.out.println(myFile.length());
byte[] byteArray = new byte[(int) myFile.length()];
FileInputStream fileInputStream = null;
try {
fileInputStream = new FileInputStream(myFile);
} catch (IOException ex) {
Logger.writeLog(ex, FileUtility.class.getName(), LogType.EXCEPTION);
throw ex;
}
BufferedInputStream bufferedInputStream = new BufferedInputStream(fileInputStream);
try {
bufferedInputStream.read(byteArray, 0, byteArray.length);
outToClient.write(byteArray, 0, byteArray.length);
outToClient.flush();
outToClient.close();
connectionSocket.close();
return;
} catch (IOException ex) {
Logger.writeLog(ex, FileUtility.class.getName(), LogType.EXCEPTION);
throw ex;
}
}
}catch (Exception e) {
Logger.writeLog(e, getClass().getName(), LogType.EXCEPTION);
throw e;
}
}
There is no 'packet loss', just bugs in your code.
The canonical way to copy a stream in Java is as follows:
while ((count = in.read(buffer)) > 0)
{
out.write(buffer, 0, count);
}
If you know the number of bytes in advance and the sender must keep the connection open after the transfer, it becomes:
while (total < expected && (count = in.read(buffer, 0, expected-total > buffer.length ? buffer.length : (int)(expected-total))) > 0)
{
out.write(buffer, 0, count);
total += count;
}
Forget all the ByteArrayInput/OutputStreams and the extra copies. Just read from the file and send to the socket, or read from the socket and write to the file.
The sockets read method will return when its has obtained all the bytes you asked for, OR, when it stops receiving data from the network.
As transmission is often interrupted in any real network you need to keep issuing read calls until you have the number of bytes you want.
You need code something like this:
char [] buffer = new char[1024];
int expect = 1000;
int sofar = 0;
int chars_read;
try
{
while((chars_read = from_server.read(buffer[sofar])) != -1)
{
sofar = sofar + chars_read;
if (sofar >= expected) break;
}
}
catch(IOException e)
{
to_user.println(e);
}
i need to open a file of 12 Megabytes, but actually i'm doing it creating a buffer of 12834566-byte, because the size of the file is 12MB and i am developing this app for Android mobile systems.
Then, i supose i have to read with blocks of 1024 Kbytes instead of one block of 12 Mbytes, with a for, but i don't know how to do it, i need a little help with it.
This is my actual code:
File f = new File(getCacheDir()+"/berlin.mp3");
if (!f.exists()) try {
InputStream is = getAssets().open("berlin.mp3");
int size = is.available();
byte[] buffer = new byte[size];
is.read(buffer);
is.close();
FileOutputStream fos = new FileOutputStream(f);
fos.write(buffer);
fos.close();
} catch (Exception e) { throw new RuntimeException(e); }
Please, can someone tell me what i have to changue in this code to read blocks of 1024 Kbytes instead of one block of 12 Mbytes?
THanks!
Try copying 1 KB at a time.
File f = new File(getCacheDir()+"/berlin.mp3");
if (!f.exists()) try {
byte[] buffer = new byte[1024];
InputStream is = getAssets().open("berlin.mp3");
FileOutputStream fos = new FileOutputStream(f);
int len;
while((len = is.read(buffer)) > 0)
fos.write(buffer, 0, len);
} catch (Exception e) {
throw new RuntimeException(e);
} finally {
IOUtils.close(is); // utility to close the stream properly.
IOUtils.close(fos);
}
Does Android support symbolic or hand links like UNIX? If it does, this would be faster/more efficient.
File f = new File(getCacheDir()+"/berlin.mp3");
InputStream is = null;
FileOutputStream fos = null;
if (!f.exists()) try {
is = getAssets().open("berlin.mp3");
fos = new FileOutputStream(f);
byte[] buffer = new byte[1024];
while (is.read(buffer) > 0) {
fos.write(buffer);
}
} catch (Exception e) {
throw new RuntimeException(e);
} finally {
// proper stream closing
if (is != null) {
try { is.close(); } catch (Exception ignored) {} finally {
if (fos != null) {
try { fos.close(); } catch (Exception ignored2) {}
}
}
}
}
import org.apache.commons.fileupload.util.Streams;
InputStream in = getAssets().open("berlin.mp3");
OutputStream out = new FileOutputStream(f);
Streams.copy(in, out, true);