无法打开从套接字接收到的文件 android java
Can't open file received from socket android java
我正在制作一个带有套接字的应用程序,我想在其中共享来自同一 wifi 中的两个设备的数据。使用我的代码,我可以在两个设备之间成功共享文件的确切大小,并以特定名称保存到设备存储中,但是当我尝试打开它时,它无法在文件管理器中打开。我已经用 mp4 文件和 apk 文件试过了
这是发件人的代码
@Override
public void run() {
//File file = new File(src);
File file = new File(Environment.getExternalStorageDirectory() + "/c.mp4");
byte[] bytes = new byte[(int) file.length()];
BufferedInputStream bis;
try {
bis = new BufferedInputStream(new FileInputStream(file));
DataInputStream dis = new DataInputStream(bis);
OutputStream os = socket.getOutputStream();
DataOutputStream dos = new DataOutputStream(os);
dos.writeUTF("anything");
dos.writeLong(bytes.length);
int read;
while ((read = dis.read(bytes)) != -1){
dos.write(bytes,0,read);
}
//os.write(bytes, 0, bytes.length); //commented
//os.flush(); //commented
socket.close();
final String sentMsg = "File sent to: " + socket.getInetAddress();
MainActivity.this.runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(MainActivity.this, sentMsg, Toast.LENGTH_LONG).show();
}});
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} finally {
try {
socket.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
这是接收器
@Override
public void run() {
Socket socket = null;
int bytesRead;
InputStream in; //changed
int bufferSize=0;
try {
socket = new Socket(dstAddress, dstPort);
bufferSize = socket.getReceiveBufferSize();
in = socket.getInputStream();
DataInputStream clientData = new DataInputStream(in);
File file = new File(Environment.getExternalStorageDirectory(), "c.mp4");
OutputStream output = new FileOutputStream(file);
byte[] buffer = new byte[bufferSize];
int read;
while ((read = clientData.read(buffer)) != -1){
output.write(buffer, 0 , read);
}
//bos.close(); //commented
socket.close();
MainActivity2.this.runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(MainActivity2.this, "Finished", Toast.LENGTH_LONG).show();
}});
} catch (IOException e) {
e.printStackTrace();
final String eMsg = "Something wrong: " + e.getMessage();
MainActivity2.this.runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(MainActivity2.this,
eMsg,
Toast.LENGTH_LONG).show();
}});
} finally {
if(socket != null){
try {
socket.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
您需要像写入套接字输出流一样读取客户端数据(DataInputStream)。
uft-8 字符串,
长期价值,
序号字节数
您的 mp4 文件内容将以“任何内容”和长度为前缀。所以它将被视为损坏的文件。
不保存文件(“任何”和长度)
我正在制作一个带有套接字的应用程序,我想在其中共享来自同一 wifi 中的两个设备的数据。使用我的代码,我可以在两个设备之间成功共享文件的确切大小,并以特定名称保存到设备存储中,但是当我尝试打开它时,它无法在文件管理器中打开。我已经用 mp4 文件和 apk 文件试过了
这是发件人的代码
@Override
public void run() {
//File file = new File(src);
File file = new File(Environment.getExternalStorageDirectory() + "/c.mp4");
byte[] bytes = new byte[(int) file.length()];
BufferedInputStream bis;
try {
bis = new BufferedInputStream(new FileInputStream(file));
DataInputStream dis = new DataInputStream(bis);
OutputStream os = socket.getOutputStream();
DataOutputStream dos = new DataOutputStream(os);
dos.writeUTF("anything");
dos.writeLong(bytes.length);
int read;
while ((read = dis.read(bytes)) != -1){
dos.write(bytes,0,read);
}
//os.write(bytes, 0, bytes.length); //commented
//os.flush(); //commented
socket.close();
final String sentMsg = "File sent to: " + socket.getInetAddress();
MainActivity.this.runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(MainActivity.this, sentMsg, Toast.LENGTH_LONG).show();
}});
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} finally {
try {
socket.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
这是接收器
@Override
public void run() {
Socket socket = null;
int bytesRead;
InputStream in; //changed
int bufferSize=0;
try {
socket = new Socket(dstAddress, dstPort);
bufferSize = socket.getReceiveBufferSize();
in = socket.getInputStream();
DataInputStream clientData = new DataInputStream(in);
File file = new File(Environment.getExternalStorageDirectory(), "c.mp4");
OutputStream output = new FileOutputStream(file);
byte[] buffer = new byte[bufferSize];
int read;
while ((read = clientData.read(buffer)) != -1){
output.write(buffer, 0 , read);
}
//bos.close(); //commented
socket.close();
MainActivity2.this.runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(MainActivity2.this, "Finished", Toast.LENGTH_LONG).show();
}});
} catch (IOException e) {
e.printStackTrace();
final String eMsg = "Something wrong: " + e.getMessage();
MainActivity2.this.runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(MainActivity2.this,
eMsg,
Toast.LENGTH_LONG).show();
}});
} finally {
if(socket != null){
try {
socket.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
您需要像写入套接字输出流一样读取客户端数据(DataInputStream)。
uft-8 字符串, 长期价值, 序号字节数
您的 mp4 文件内容将以“任何内容”和长度为前缀。所以它将被视为损坏的文件。
不保存文件(“任何”和长度)