Java Android下载二进制文件问题
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Android download binary file problems
提问by Isaac Waller
I am having problems downloading a binary file (video) in my app from the internet. In Quicktime, If I download it directly it works fine but through my app somehow it get's messed up (even though they look exactly the same in a text editor). Here is a example:
我在从互联网下载我的应用程序中的二进制文件（视频）时遇到问题。在 Quicktime 中，如果我直接下载它，它工作正常，但通过我的应用程序不知何故它会变得一团糟（即使它们在文本编辑器中看起来完全相同）。下面是一个例子：
URL u = new URL("http://www.path.to/a.mp4?video");
HttpURLConnection c = (HttpURLConnection) u.openConnection();
c.setRequestMethod("GET");
c.setDoOutput(true);
c.connect();
FileOutputStream f = new FileOutputStream(new File(root,"Video.mp4"));
InputStream in = c.getInputStream();
byte[] buffer = new byte[1024];
int len1 = 0;
while ( (len1 = in.read(buffer)) > 0 ) {
f.write(buffer);
}
f.close();
采纳答案by Ry4an Brase
I don't know if it's the only problem, but you've got a classic Java glitch in there: You're not counting on the fact that read() is alwaysallowed to return fewer bytes than you ask for. Thus, your read could get less than 1024 bytes but your write always writes out exactly 1024 bytes possibly including bytes from the previous loop iteration.
我不知道这是否是唯一的问题，但你有一个经典的 Java 故障：你没有指望 read()总是允许返回比你要求的更少的字节。因此，您的读取可能会少于 1024 字节，但您的写入总是准确地写出 1024 字节，可能包括来自前一次循环迭代的字节。
Correct with:
更正：
while ( (len1 = in.read(buffer)) > 0 ) {
f.write(buffer,0, len1);
}
Perhaps the higher latency networking or smaller packet sizes of 3G on Android are exacerbating the effect?
也许 Android 上 3G 的更高延迟网络或更小数据包大小正在加剧这种影响？
回答by Clint
One problem is your reading of the buffer. If every read of the input stream is not an exact multiple of 1024 you will copy bad data. Use:
一个问题是您对缓冲区的读取。如果对输入流的每次读取都不是 1024 的精确倍数，您将复制错误数据。用：
byte[] buffer = new byte[1024];
int len1 = 0;
while ( (len1 = in.read(buffer)) != -1 ) {
f.write(buffer,0, len1);
}
回答by Ry4an Brase
public class download extends Activity {
private static String fileName = "file.3gp";
private static final String MY_URL = "Your download url goes here";
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
try {
URL url = new URL(MY_URL);
HttpURLConnection c = (HttpURLConnection) url.openConnection();
c.setRequestMethod("GET");
c.setDoOutput(true);
c.connect();
String PATH = Environment.getExternalStorageDirectory()
+ "/download/";
Log.d("Abhan", "PATH: " + PATH);
File file = new File(PATH);
if(!file.exists()) {
file.mkdirs();
}
File outputFile = new File(file, fileName);
FileOutputStream fos = new FileOutputStream(outputFile);
InputStream is = c.getInputStream();
byte[] buffer = new byte[1024];
int len1 = 0;
while ((len1 = is.read(buffer)) != -1) {
fos.write(buffer, 0, len1);
}
fos.flush();
fos.close();
is.close();
} catch (IOException e) {
Log.e("Abhan", "Error: " + e);
}
Log.i("Abhan", "Check Your File.");
}
}
回答by AlikElzin-kilaka
Just use apache's copy method (Apache Commons IO) - the advantage of using Java!
只需使用 apache 的复制方法（Apache Commons IO）——使用 Java 的优势！
IOUtils.copy(is, os);
Do not forget to close the streams in a finally block:
不要忘记在 finally 块中关闭流：
try{
...
} finally {
IOUtils.closeQuietly(is);
IOUtils.closeQuietly(os);
}
回答by grepit
I fixed the code based on previous feedbacks on this thread. I tested using eclipse and multiple large files. It is working fine. Just have to copy and paste this to your environment and change the http path and the location which you would like the file to be downloaded to.
我根据之前对该线程的反馈修复了代码。我使用 eclipse 和多个大文件进行了测试。它工作正常。只需将其复制并粘贴到您的环境中，然后更改 http 路径和您希望将文件下载到的位置。
try {
//this is the file you want to download from the remote server
String path ="http://localhost:8080/somefile.zip";
//this is the name of the local file you will create
String targetFileName
boolean eof = false;
URL u = new URL(path);
HttpURLConnection c = (HttpURLConnection) u.openConnection();
c.setRequestMethod("GET");
c.setDoOutput(true);
c.connect();
FileOutputStream f = new FileOutputStream(new File("c:\junk\"+targetFileName));
InputStream in = c.getInputStream();
byte[] buffer = new byte[1024];
int len1 = 0;
while ( (len1 = in.read(buffer)) > 0 ) {
f.write(buffer,0, len1);
}
f.close();
} catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Good luck Alireza Aghamohammadi
祝阿里雷扎·阿加莫哈马迪好运
回答by njzk2
new DefaultHttpClient().execute(new HttpGet("http://www.path.to/a.mp4?video"))
.getEntity().writeTo(
new FileOutputStream(new File(root,"Video.mp4")));
