这是我的android代码:
try
{
FileInputStream fileInputStream = new FileInputStream(new File(pathToOurFile3) );
URL url = new URL(urlServer);
connection = (HttpURLConnection) url.openConnection();
// Allow Inputs & Outputs
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
// Enable POST method
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
outputStream = new DataOutputStream( connection.getOutputStream() );
outputStream.writeBytes(twoHyphens + boundary + lineEnd);
outputStream.writeBytes("Content-Disposition: form-data; name=\"profile\";filename=\"" + pathToOurFile3 +"\"" + lineEnd);
outputStream.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// Read file
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(lineEnd);
outputStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
int serverResponseCode = connection.getResponseCode();
String serverResponseMessage = connection.getResponseMessage();
Log.d("serverResponseCode"+"", serverResponseCode +"");
Log.d("serverResponseMessage", serverResponseMessage);
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex)
{
Log.d("upload", ex.toString());//Exception handling
}
这是php代码:
<?php
$target_path = "./";
$target_path = $target_path . basename( $_FILES['profile']['name']);
if(move_uploaded_file($_FILES['profile']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['profile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>
该文件可以上传到服务器。然而,响应消息与我的预期不同
从Logcat,serverResponseCode is 200
和serverResponseMessage is OK
但是从php,我想得到像" XXX has been upload"
这样的消息
有谁知道如何获取消息??
答案 0 :(得分:1)
我终于得到了答案:
InputStream is = connection.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
String stringResponse = sb.toString();
Log.d("response string:", stringResponse);