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在Java中将文件转换为byte []

如何将Java.io.File转换为byte[]

658
Ben Noland

这取决于你最好的方法。生产力明智,不要重新发明轮子并使用Apache Commons。这是 IOUtils.toByteArray(InputStream input)

432
svachon

JDK 7 你可以使用 Files.readAllBytes(Path)

例:

import Java.io.File;
import Java.nio.file.Files;

File file;
// ...(file is initialised)...
byte[] fileContent = Files.readAllBytes(file.toPath());
1161
Alan
import Java.io.RandomAccessFile;
RandomAccessFile f = new RandomAccessFile(fileName, "r");
byte[] b = new byte[(int)f.length()];
f.readFully(b);

Java 8的文档: http://docs.Oracle.com/javase/8/docs/api/Java/io/RandomAccessFile.html

159
Dmitry Mitskevich

从JDK 7开始 - 一个班轮:

byte[] array = Files.readAllBytes(Paths.get("/path/to/file"));

不需要外部依赖。

144
Paulius Matulionis

基本上你必须在内存中阅读它。打开文件,分配数组,并将文件中的内容读入数组。

最简单的方法是这样的:

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();
}

这有一些不必要的文件内容复制(实际上数据被复制三次:从文件到buffer,从bufferByteArrayOutputStream,从ByteArrayOutputStream复制到实际的结果数组)。

您还需要确保在内存中只读取一定大小的文件(这通常取决于应用程序):-)。

您还需要在函数外部处理IOException

另一种方式是这样的:

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;
}

这没有不必要的复制。

FileTooBigException是自定义应用程序异常。 MAX_FILE_SIZE常量是一个应用程序参数。

对于大文件,您应该考虑使用流处理算法或使用内存映射(请参阅Java.nio)。

76
Mihai Toader

正如有人所说, Apache Commons File Utils 可能有你想要的东西

public static byte[] readFileToByteArray(File file) throws IOException

使用示例(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);
        ...
    }
}
68
Tom

您也可以使用NIO api来完成它。只要总文件大小(以字节为单位)适合int,我就可以使用此代码执行此操作。

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();
    }
}

我认为它使用MappedByteBuffer非常快。

22
Amit

如果您没有Java 8,并且同意我的意见,包括一个大型库以避免编写几行代码是一个坏主意:

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();
}

呼叫者负责关闭流。

19
Jeffrey Blattman
// 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;
    }
19
Cuga

简单的方法:

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);
16
Sudip Bhandari

从文件中读取字节的最简单方法

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);
        }
    }
}
13
Muhammad Sadiq

番石榴有 Files.toByteArray() 为你提供。它有几个优点:

  1. 它涵盖了文件报告长度为0但仍有内容的角落情况
  2. 它是高度优化的,如果在尝试加载文件之前尝试读取大文件,则会出现OutOfMemoryException。 (通过巧妙地使用file.length())
  3. 你不必重新发明轮子。
11
jontejj
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);
10
BlondCode

使用与社区维基答案相同的方法,但更清晰和开箱即用的编译(如果您不想导入Apache Commons库,首选方法,例如在Android上):

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();
}
9
manmal

ReadFully 从当前文件指针开始,将此文件中的b.length个字节读入字节数组。此方法从文件中重复读取,直到读取所请求的字节数。此方法将一直阻塞,直到读取所请求的字节数,检测到流的末尾或抛出异常。

RandomAccessFile f = new RandomAccessFile(fileName, "r");
byte[] b = new byte[(int)f.length()];
f.readFully(b);
5
Tarun M

如果要将字节读入预先分配的字节缓冲区,则此答案可能会有所帮助。

你的第一个猜测可能是使用 InputStream read(byte[]) 。但是,这种方法存在一个缺陷,使其难以使用:即使没有遇到EOF,也无法保证数组实际上会被完全填充。

相反,看看 DataInputStream readFully(byte[]) 。这是输入流的包装器,没有上述问题。此外,遇到EOF时抛出此方法。好多了。

3
Laurens Holst

让我在不使用第三方库的情况下添加另一种解决方案。它重用了由 Scottlink )提出的异常处理模式。我把丑陋的部分移动到一个单独的消息中(我会隐藏在一些FileUtils类中;))

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;
}
3
Andreas_D
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);
        }
    }
}
2
mazatwork

另一种从文件中读取字节的方法

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();
        }
    }
1
Muhammad Aamir Ali

试试这个 :

    try
    {
        String path="";

        InputStream inputStream=new FileInputStream(path);

        byte[] data=IOUtils.readFully(inputStream,-1,false);

    }
    catch (Exception e)
    {

    }

别忘了“ 导入Sun.misc.IOUtils;

0
Maifee Ul Asad

下面的方法不仅将Java.io.File转换为byte [],我还发现它是读取文件的最快方式,当测试许多不同的 Java文件读取方法时 相互对立:

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);
    }
  }
}
0
gomisha