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
I will get input stream from third party library to my application.
I have to write this input stream to a file.
Following is the code snippet I tried:
private void writeDataToFile(Stub stub) {
OutputStream os = null;
InputStream inputStream = null;
try {
inputStream = stub.getStream();
os = new FileOutputStream("test.txt");
int read = 0;
byte[] bytes = new byte[1024];
while ((read = inputStream.read(bytes)) != -1) {
os.write(bytes, 0, read);
}
} catch (Exception e) {
log("Error while fetching data", e);
} finally {
if(inputStream != null) {
try {
inputStream.close();
} catch (IOException e) {
log("Error while closing input stream", e);
}
}
if(os != null) {
try {
os.close();
} catch (IOException e) {
log("Error while closing output stream", e);
}
}
}
}
Is there any better approach to do this ?
Since you are stuck with Java 6, do yourself a favour and use Guava and its Closer:
final Closer closer = Closer.create();
final InputStream in;
final OutputStream out;
final byte[] buf = new byte[32768]; // 32k
int bytesRead;
try {
in = closer.register(createInputStreamHere());
out = closer.register(new FileOutputStream(...));
while ((bytesRead = in.read(buf)) != -1)
out.write(buf, 0, bytesRead);
out.flush();
} finally {
closer.close();
}
Had you used Java 7, the solution would have been as simple as:
final Path destination = Paths.get("pathToYourFile");
try (
final InputStream in = createInputStreamHere();
) {
Files.copy(in, destination);
}
And yourInputStream would have been automatically closed for you as a "bonus"; Files would have handled destination all by itself.
If you're not on Java 7 and can't use fge's solution, you may want to wrap your OutputStream in a BufferedOutputStream
BufferedOutputStream os = new BufferedOutputStream(new FileOutputStream("xx.txt"));
Such buffered output stream will write bytes in blocks to the file, which is more efficient than writing byte per byte.
It can get cleaner with an OutputStreamWriter:
OutputStream outputStream = new FileOutputStream("output.txt");
Writer writer = new OutputStreamWriter(outputStream);
writer.write("data");
writer.close();
Instead of writing a string, you can use a Scanner on your inputStream
Scanner sc = new Scanner(inputStream);
while (sc.HasNext())
//read using scanner methods
I am trying to get file content in bytes in Android application. I have get the file in SD card now want to get the selected file in bytes. I googled but no such success. Please help
Below is the code to get files with extension. Through this i get files and show in spinner. On file selection I want to get file in bytes.
private List<String> getListOfFiles(String path) {
File files = new File(path);
FileFilter filter = new FileFilter() {
private final List<String> exts = Arrays.asList("jpeg", "jpg", "png", "bmp", "gif","mp3");
public boolean accept(File pathname) {
String ext;
String path = pathname.getPath();
ext = path.substring(path.lastIndexOf(".") + 1);
return exts.contains(ext);
}
};
final File [] filesFound = files.listFiles(filter);
List<String> list = new ArrayList<String>();
if (filesFound != null && filesFound.length > 0) {
for (File file : filesFound) {
list.add(file.getName());
}
}
return list;
}
here it's a simple:
File file = new File(path);
int size = (int) file.length();
byte[] bytes = new byte[size];
try {
BufferedInputStream buf = new BufferedInputStream(new FileInputStream(file));
buf.read(bytes, 0, bytes.length);
buf.close();
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Add permission in manifest.xml:
<uses-permission android:name="android.permission.READ_EXTERNAL_STORAGE" />
The easiest solution today is to used Apache common io :
http://commons.apache.org/proper/commons-io/javadocs/api-release/org/apache/commons/io/FileUtils.html#readFileToByteArray(java.io.File)
byte bytes[] = FileUtils.readFileToByteArray(photoFile)
The only drawback is to add this dependency in your build.gradle app :
implementation 'commons-io:commons-io:2.5'
+ 1562 Methods count
Since the accepted BufferedInputStream#read isn't guaranteed to read everything, rather than keeping track of the buffer sizes myself, I used this approach:
byte bytes[] = new byte[(int) file.length()];
BufferedInputStream bis = new BufferedInputStream(new FileInputStream(file));
DataInputStream dis = new DataInputStream(bis);
dis.readFully(bytes);
Blocks until a full read is complete, and doesn't require extra imports.
Here is a solution that guarantees entire file will be read, that requires no libraries and is efficient:
byte[] fullyReadFileToBytes(File f) throws IOException {
int size = (int) f.length();
byte bytes[] = new byte[size];
byte tmpBuff[] = new byte[size];
FileInputStream fis= new FileInputStream(f);;
try {
int read = fis.read(bytes, 0, size);
if (read < size) {
int remain = size - read;
while (remain > 0) {
read = fis.read(tmpBuff, 0, remain);
System.arraycopy(tmpBuff, 0, bytes, size - remain, read);
remain -= read;
}
}
} catch (IOException e){
throw e;
} finally {
fis.close();
}
return bytes;
}
NOTE: it assumes file size is less than MAX_INT bytes, you can add handling for that if you want.
If you want to use a the openFileInput method from a Context for this, you can use the following code.
This will create a BufferArrayOutputStream and append each byte as it's read from the file to it.
/**
* <p>
* Creates a InputStream for a file using the specified Context
* and returns the Bytes read from the file.
* </p>
*
* #param context The context to use.
* #param file The file to read from.
* #return The array of bytes read from the file, or null if no file was found.
*/
public static byte[] read(Context context, String file) throws IOException {
byte[] ret = null;
if (context != null) {
try {
InputStream inputStream = context.openFileInput(file);
ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
int nextByte = inputStream.read();
while (nextByte != -1) {
outputStream.write(nextByte);
nextByte = inputStream.read();
}
ret = outputStream.toByteArray();
} catch (FileNotFoundException ignored) { }
}
return ret;
}
In Kotlin you can simply use:
File(path).readBytes()
You can also do it this way:
byte[] getBytes (File file)
{
FileInputStream input = null;
if (file.exists()) try
{
input = new FileInputStream (file);
int len = (int) file.length();
byte[] data = new byte[len];
int count, total = 0;
while ((count = input.read (data, total, len - total)) > 0) total += count;
return data;
}
catch (Exception ex)
{
ex.printStackTrace();
}
finally
{
if (input != null) try
{
input.close();
}
catch (Exception ex)
{
ex.printStackTrace();
}
}
return null;
}
A simple InputStream will do
byte[] fileToBytes(File file){
byte[] bytes = new byte[0];
try(FileInputStream inputStream = new FileInputStream(file)) {
bytes = new byte[inputStream.available()];
//noinspection ResultOfMethodCallIgnored
inputStream.read(bytes);
} catch (IOException e) {
e.printStackTrace();
}
return bytes;
}
Following is the working solution to read the entire file in chunks and its efficient solution to read the large files using a scanner class.
try {
FileInputStream fiStream = new FileInputStream(inputFile_name);
Scanner sc = null;
try {
sc = new Scanner(fiStream);
while (sc.hasNextLine()) {
String line = sc.nextLine();
byte[] buf = line.getBytes();
}
} finally {
if (fiStream != null) {
fiStream.close();
}
if (sc != null) {
sc.close();
}
}
}catch (Exception e){
Log.e(TAG, "Exception: " + e.toString());
}
To read a file in bytes, often used to read binary files, such as pictures, sounds, images, etc.
Use the method below.
public static byte[] readFileByBytes(File file) {
byte[] tempBuf = new byte[100];
int byteRead;
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
try {
BufferedInputStream bufferedInputStream = new BufferedInputStream(new FileInputStream(file));
while ((byteRead = bufferedInputStream.read(tempBuf)) != -1) {
byteArrayOutputStream.write(tempBuf, 0, byteRead);
}
bufferedInputStream.close();
return byteArrayOutputStream.toByteArray();
} catch (Exception e) {
e.printStackTrace();
return null;
}
}