如何将整个输入流读到字节数组?
当前回答
如果你碰巧使用谷歌Guava,它将像使用ByteStreams一样简单:
byte[] bytes = ByteStreams.toByteArray(inputStream);
其他回答
这是我的复制粘贴版本:
@SuppressWarnings("empty-statement")
public static byte[] inputStreamToByte(InputStream is) throws IOException {
if (is == null) {
return null;
}
// Define a size if you have an idea of it.
ByteArrayOutputStream r = new ByteArrayOutputStream(2048);
byte[] read = new byte[512]; // Your buffer size.
for (int i; -1 != (i = is.read(read)); r.write(read, 0, i));
is.close();
return r.toByteArray();
}
这对我很有用,
if(inputStream != null){
ByteArrayOutputStream contentStream = readSourceContent(inputStream);
String stringContent = contentStream.toString();
byte[] byteArr = encodeString(stringContent);
}
readSourceContent ()
public static ByteArrayOutputStream readSourceContent(InputStream inputStream) throws IOException {
ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
int nextChar;
try {
while ((nextChar = inputStream.read()) != -1) {
outputStream.write(nextChar);
}
outputStream.flush();
} catch (IOException e) {
throw new IOException("Exception occurred while reading content", e);
}
return outputStream;
}
encodeString()
public static byte[] encodeString(String content) throws UnsupportedEncodingException {
byte[] bytes;
try {
bytes = content.getBytes();
} catch (UnsupportedEncodingException e) {
String msg = ENCODING + " is unsupported encoding type";
log.error(msg,e);
throw new UnsupportedEncodingException(msg, e);
}
return bytes;
}
Java 7及以上版本:
import sun.misc.IOUtils;
...
InputStream in = ...;
byte[] buf = IOUtils.readFully(in, -1, false);
如果有人还在寻找一个没有依赖的解决方案,如果你有一个文件。
DataInputStream
byte[] data = new byte[(int) file.length()];
DataInputStream dis = new DataInputStream(new FileInputStream(file));
dis.readFully(data);
dis.close();
ByteArrayOutputStream
InputStream is = new FileInputStream(file);
ByteArrayOutputStream buffer = new ByteArrayOutputStream();
int nRead;
byte[] data = new byte[(int) file.length()];
while ((nRead = is.read(data, 0, data.length)) != -1) {
buffer.write(data, 0, nRead);
}
RandomAccessFile
RandomAccessFile raf = new RandomAccessFile(file, "r");
byte[] data = new byte[(int) raf.length()];
raf.readFully(data);
Kotlin中的解决方案(当然也可以在Java中工作),其中包括当你知道大小时的两种情况:
fun InputStream.readBytesWithSize(size: Long): ByteArray? {
return when {
size < 0L -> this.readBytes()
size == 0L -> ByteArray(0)
size > Int.MAX_VALUE -> null
else -> {
val sizeInt = size.toInt()
val result = ByteArray(sizeInt)
readBytesIntoByteArray(result, sizeInt)
result
}
}
}
fun InputStream.readBytesIntoByteArray(byteArray: ByteArray,bytesToRead:Int=byteArray.size) {
var offset = 0
while (true) {
val read = this.read(byteArray, offset, bytesToRead - offset)
if (read == -1)
break
offset += read
if (offset >= bytesToRead)
break
}
}
如果您知道大小,那么与其他解决方案相比,它可以节省两倍的内存(在很短的时间内,但仍然可能有用)。这是因为您必须将整个流读到末尾,然后将其转换为字节数组(类似于将数组转换为数组的ArrayList)。
例如,如果你在Android上,你有一些Uri要处理,你可以尝试用这个来获取大小:
fun getStreamLengthFromUri(context: Context, uri: Uri): Long {
context.contentResolver.query(uri, arrayOf(MediaStore.MediaColumns.SIZE), null, null, null)?.use {
if (!it.moveToNext())
return@use
val fileSize = it.getLong(it.getColumnIndex(MediaStore.MediaColumns.SIZE))
if (fileSize > 0)
return fileSize
}
//if you wish, you can also get the file-path from the uri here, and then try to get its size, using this: https://stackoverflow.com/a/61835665/878126
FileUtilEx.getFilePathFromUri(context, uri, false)?.use {
val file = it.file
val fileSize = file.length()
if (fileSize > 0)
return fileSize
}
context.contentResolver.openInputStream(uri)?.use { inputStream ->
if (inputStream is FileInputStream)
return inputStream.channel.size()
else {
var bytesCount = 0L
while (true) {
val available = inputStream.available()
if (available == 0)
break
val skip = inputStream.skip(available.toLong())
if (skip < 0)
break
bytesCount += skip
}
if (bytesCount > 0L)
return bytesCount
}
}
return -1L
}
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