我如何将java.io.File转换为字节[]?
当前回答
你也可以使用NIO api来实现。只要总文件大小(以字节为单位)适合int类型,我就可以用这段代码做到这一点。
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();
}
}
我认为它非常快,因为它使用MappedByteBuffer。
其他回答
让我添加另一个不使用第三方库的解决方案。它重用了Scott提出的异常处理模式(链接)。我把丑陋的部分移动到一个单独的消息(我会隐藏在一些FileUtils类;))
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;
}
你也可以使用NIO api来实现。只要总文件大小(以字节为单位)适合int类型,我就可以用这段代码做到这一点。
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();
}
}
我认为它非常快,因为它使用MappedByteBuffer。
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);
使用与社区wiki答案相同的方法,但更干净,并编译开箱即用(如果你不想导入Apache Commons库,例如在Android上,首选的方法):
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();
}
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);
}
}
}
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