我必须构建一个应用程序，该应用程序应该将图片从手机发送到网络服务器。不幸的是，我真的不知道该怎么做。有人可以帮帮我吗？
【问题讨论】：
标签： android
使用网络服务来完成这项任务。
为了在android中使用网络服务，请访问此链接。
【讨论】：
这是我使用原始套接字将图像上传到远程服务器的代码。原始套接字优于 httpclient 的优点是您可以显示上传进度条。
免责声明：大部分代码主要来自stackoverflow。
/**
* Asynchronous task to upload file to server
*/
class UploadImageTask extends AsyncTask<File, Integer, Boolean> {
/** Upload file to this url */
private static final String UPLOAD_URL = "http://thibault-laptop:8080/report";
/** Send the file with this form name */
private static final String FIELD_FILE = "file";
private static final String FIELD_LATITUDE = "latitude";
private static final String FIELD_LONGITUDE = "longitude";
/**
* Prepare activity before upload
*/
@Override
protected void onPreExecute() {
super.onPreExecute();
setProgressBarIndeterminateVisibility(true);
mConfirm.setEnabled(false);
mCancel.setEnabled(false);
showDialog(UPLOAD_PROGRESS_DIALOG);
}
/**
* Clean app state after upload is completed
*/
@Override
protected void onPostExecute(Boolean result) {
super.onPostExecute(result);
setProgressBarIndeterminateVisibility(false);
mConfirm.setEnabled(true);
mDialog.dismiss();
if (result) {
showDialog(UPLOAD_SUCCESS_DIALOG);
} else {
showDialog(UPLOAD_ERROR_DIALOG);
}
}
@Override
protected Boolean doInBackground(File... image) {
return doFileUpload(image[0], UPLOAD_URL);
}
@Override
protected void onProgressUpdate(Integer... values) {
super.onProgressUpdate(values);
if (values[0] == 0) {
mDialog.setTitle(getString(R.string.progress_dialog_title_uploading));
}
mDialog.setProgress(values[0]);
}
/**
* Upload given file to given url, using raw socket
* @see http://stackoverflow.com/questions/4966910/androidhow-to-upload-mp3-file-to-http-server
*
* @param file The file to upload
* @param uploadUrl The uri the file is to be uploaded
*
* @return boolean true is the upload succeeded
*/
private boolean doFileUpload(File file, String uploadUrl) {
HttpURLConnection conn = null;
DataOutputStream dos = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
String separator = twoHyphens + boundary + lineEnd;
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
int sentBytes = 0;
long fileSize = file.length();
// The definitive url is of the kind:
// http://host/report/latitude,longitude
uploadUrl += "/" + mLocation.getLatitude() + "," + mLocation.getLongitude();
// Send request
try {
// Configure connection
URL url = new URL(uploadUrl);
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true);
conn.setDoOutput(true);
conn.setUseCaches(false);
conn.setRequestMethod("PUT");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
publishProgress(0);
dos = new DataOutputStream(conn.getOutputStream());
// Send location params
writeFormField(dos, separator, FIELD_LATITUDE, "" + mLocation.getLatitude());
writeFormField(dos, separator, FIELD_LONGITUDE, "" + mLocation.getLongitude());
// Send multipart headers
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"" + FIELD_FILE + "\";filename=\""
+ file.getName() + "\"" + lineEnd);
dos.writeBytes(lineEnd);
// Read file and create buffer
FileInputStream fileInputStream = new FileInputStream(file);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// Send file data
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
// Write buffer to socket
dos.write(buffer, 0, bufferSize);
// Update progress dialog
sentBytes += bufferSize;
publishProgress((int)(sentBytes * 100 / fileSize));
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);
dos.flush();
dos.close();
fileInputStream.close();
} catch (IOException ioe) {
Log.e(TAG, "Cannot upload file: " + ioe.getMessage(), ioe);
return false;
}
// Read response
try {
int responseCode = conn.getResponseCode();
return responseCode == 200;
} catch (IOException ioex) {
Log.e(TAG, "Upload file failed: " + ioex.getMessage(), ioex);
return false;
} catch (Exception e) {
Log.e(TAG, "Upload file failed: " + e.getMessage(), e);
return false;
}
}
private void writeFormField(DataOutputStream dos, String separator, String fieldName, String fieldValue) throws IOException
{
dos.writeBytes(separator);
dos.writeBytes("Content-Disposition: form-data; name=\"" + fieldName + "\"\r\n");
dos.writeBytes("\r\n");
dos.writeBytes(fieldValue);
dos.writeBytes("\r\n");
}
}
要开始上传，请使用以下命令：
new UploadImageTask().execute(new File(imagePath));
【讨论】：
我确实在 Android 中使用了 Rest webservice 和 DefaultHttpClient 类。要创建示例 REST Web 服务并在 Apache Tomcat 中部署，请按照教程Vogella
为了让rest服务接受图片，服务器端需要multipart content-type
@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
@Produces("application/json")
public String uploadFile(@FormDataParam("image") InputStream uploadedInputStream,
@FormDataParam("image") FormDataContentDisposition fileDetail) {
String uploadedFileLocation = "e://game/" + fileDetail.getFileName();
boolean response=false;
// save it
try{
OutputStream out = null;
int read = 0;
byte[] bytes = new byte[1024];
out = new FileOutputStream(new File(uploadedFileLocation));
while ((read = uploadedInputStream.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
return response=true;
}catch(IOException e){
e.printStackTrace();
}
return response;
}
在android端发送图片（我是在AsyncTask的doInBackground里面做的）
HttpClient httpClient = new DefaultHttpClient();
HttpPost postRequest = new HttpPost("http://"+ip+":8080/MiniJarvisFaceServer/image");
MultipartEntity reqEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
reqEntity.addPart("image", new FileBody(file));
postRequest.setEntity(reqEntity);
ResponseHandler<String> handler = new BasicResponseHandler();
String response = httpClient.execute(postRequest,handler);
Log.d("Response", response);
httpClient.getConnectionManager().shutdown();
【讨论】：
要执行 HTTP 请求，您可以使用 DefaultHttpClient 类和 HttpPost 类
HttpClient client = new DefaultHttpClient();
HttpPost post = new HttpPost("http://your.site/your/service");
// set some headers if needed
post.addHeader(....);
// and an eclosed entity to send
post.setEntity(....);
// send a request and get response (if needed)
InputStream responseStream = client.execute(post)..getEntity().getContent();
将实体添加到请求的方法取决于您的远程服务的工作方式。
【讨论】：
按照本指南进行操作。它在服务器端使用 PHP。我使用了 Android Studio 和 httpmime.4.3.6 并且工作得很好 http://www.androidhive.info/2014/12/android-uploading-camera-image-video-to-server-with-progress-bar/
它还支持视频，并展示了如何使用服务器的一些结果进行响应。唯一棘手的一点是确保您使用的是适用于 Android 的 HttClient 和正确版本的 HttpMime。现在 HttpMime 4.4.x 不工作，浪费了我一周的时间。使用 4.3.6
【讨论】：