【发布时间】:2017-12-12 14:07:07
【问题描述】:
我想通过 android 将图片上传到 Perticular Url(www.myUrl.com/myFolderName/) 并检索该 url 的地址。
这是我的代码
class SaveFile extends AsyncTask<String,String,String>
{
FileInputStream fileInputStream;
HttpURLConnection connection;
URL url;
DataOutputStream outputStream;
int bytesRead,bytesAvailable,bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
File selectedFile = new File(selectedImagePath);
//selectedImagePath = path of Image in phonememory
@Override
protected String doInBackground(String... params)
{
Log.v(TAG,"Image path is " + selectedImagePath);
try
{
fileInputStream = new FileInputStream(selectedFile);
url = new URL(SEND_IMAGE);
connection= (HttpURLConnection)url.openConnection();
Log.v(TAG,"connection established");
connection.setDoOutput(true);
connection.setDoInput(true);
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection","Keep-Alive");
connection.setRequestProperty("ENCTYPE","multipart/form-data");
connection.setRequestProperty("Content-Type","multipart/form-data;boundary=*****");
//connection.setRequestProperty("image1",selectedImagePath);
//connection.setRequestProperty("user_id","1");
outputStream = new DataOutputStream(connection.getOutputStream());
outputStream.writeBytes("--*****\r\n");
outputStream.writeBytes("Content-Disposition:form-data;name=\"uploaded_file\";filename=\""+selectedImagePath+"\""+"\r\n");
outputStream.writeBytes("\r\n");
//outputStream.writeBytes("--*****\r\n");
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable,maxBufferSize);
buffer = new byte[bufferSize];
bytesRead = fileInputStream.read(buffer,0,bufferSize);
while (bytesRead>0)
{
outputStream.write(buffer,0,bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable,maxBufferSize);
bytesRead = fileInputStream.read(buffer,0,bufferSize);
}
outputStream.writeBytes("\r\n");
outputStream.writeBytes("--*****--\r\n");
int serverResponseCode = connection.getResponseCode();
String serverResponseMessage = connection.getResponseMessage();
if(serverResponseCode == HttpURLConnection.HTTP_OK) {
String line;
StringBuilder sb = new StringBuilder();
BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
while ((line = reader.readLine()) != null) {
sb.append(line);
}
Log.v(TAG,"Response is " +sb.toString());
}
Log.v("UploadImage", "Server Response is: " + serverResponseMessage + ": " + serverResponseCode);
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception e)
{
Log.v("UploadImage"," " + e.toString());
}
return null;
}
@Override
protected void onPostExecute(String s) {
super.onPostExecute(s);
}
}
但我收到 serverResponse 404
我尝试了Uploading Image to Server - Android 之类的解决方案,但没有奏效。
我搜索了很多,但找不到解决方案
请帮忙!!
【问题讨论】:
-
没有帖子正文的设备会收到什么?
-
我没听懂你的意思,你能解释一下你想说什么吗??
标签: android httpclient androidhttpclient