[英]Upload picture to server
我正在使用以下代码: http : //androidexample.com/Upload_File_To_Server_-_Android_Example/index.php?view=article_discription&aid=83&aaid=106
有用!
我想知道我要发送的照片是几秒钟前用相机拍摄的照片。 所以..我已经创建了一个简单的方法来拍照,然后此OnActivityResult:
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
if (requestCode == CAMERA_PIC_REQUEST) {
//2
Bitmap thumbnail = (Bitmap) data.getExtras().get("data");
//3
ByteArrayOutputStream bytes = new ByteArrayOutputStream();
thumbnail.compress(Bitmap.CompressFormat.JPEG, 100, bytes);
//4
File file = new File(Environment.getExternalStorageDirectory()+File.separator + "guasto.jpg");
try {
file.createNewFile();
FileOutputStream fo = new FileOutputStream(file);
//5
fo.write(bytes.toByteArray());
fo.close();
uploadFile(uploadFilePath + "" + uploadFileName); //this should call the working method to upload the picture
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
问题是上传无法正常工作,并且在执行以下代码行之前就停止了:
dos = new DataOutputStream(conn.getOutputStream());
我该怎么做才能解决问题...? 非常感谢你
尝试以下代码。 创建一个异步任务,并在doInBackground方法中添加代码。 您将获得InputStream的对象作为回报。
您应该使用Multipart实体。
File file = new File(selectedImage);
ContentBody cbFile = new FileBody(file, "image/*");
MultipartEntity reqEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
reqEntity.addPart(RequestParams.POST, cbFile);
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://yourwebserveraddress.com/apiname_or_whatever_path_it_is");
httpPost.setEntity(reqEntity);
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
InputStream is = httpEntity.getContent();
其中selectedImage是图像的路径。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.