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[英]Uploading an image to server in android without changing in multipart form
[英]Android Uploading Video as Multipart to Server
我正在開發Android應用程序,用戶將視頻上傳到服務器。 我是android開發的新手,所以我無法掌握上傳視頻的概念。 我知道的是,我需要創建一個意圖,要求用戶選擇視頻,然后將其轉換為uri文件。 我打算在httpurlconnection中使用asynctask。 任何人都可以幫助我如何處理asynctask?
我認為這可以幫助您上傳視頻
public static int upLoad2Server(String sourceFileUri) {
String upLoadServerUri = "your remote server link";
// String [] string = sourceFileUri;
String fileName = sourceFileUri;
HttpURLConnection conn = null;
DataOutputStream dos = null;
DataInputStream inStream = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
String responseFromServer = "";
File sourceFile = new File(sourceFileUri);
if (!sourceFile.isFile()) {
Log.e("Huzza", "Source File Does not exist");
return 0;
}
try { // open a URL connection to the Servlet
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(upLoadServerUri);
conn = (HttpURLConnection) url.openConnection(); // Open a HTTP connection to
the URL
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("uploaded_file", fileName);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data;
name=\"uploaded_file\";filename=\""+ fileName + "\"" + lineEnd);
dos.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available(); // create a buffer of maximum size
Log.i("Huzza", "Initial .available : " + bytesAvailable);
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();
Log.i("Upload file to server", "HTTP Response is : " + serverResponseMessage + ": " + serverResponseCode);
// close streams
Log.i("Upload file to server", fileName + " File is written");
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
ex.printStackTrace();
Log.e("Upload file to server", "error: " + ex.getMessage(), ex);
} catch (Exception e) {
e.printStackTrace();
}
//this block will give the response of upload link
try {
BufferedReader rd = new BufferedReader(new InputStreamReader(conn
.getInputStream()));
String line;
while ((line = rd.readLine()) != null) {
Log.i("Huzza", "RES Message: " + line);
}
rd.close();
} catch (IOException ioex) {
Log.e("Huzza", "error: " + ioex.getMessage(), ioex);
}
return serverResponseCode; // like 200 (Ok)
} // end upLoad2Server
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