使用网络服务?
答
基本上你需要两面。第一个是在Android上。你必须发送(在ASyncTask e.x.中执行)数据到你的web服务。在这里,我做你一个小方法,它发送一个文件和一些额外的POST值的网址:
private boolean handleFile(String filePath, String mimeType) {
HttpURLConnection connection = null;
DataOutputStream outStream = null;
DataInputStream inStream = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
String urlString = "http://your.domain.com/webservice.php";
try {
FileInputStream fileInputStream = null;
try {
fileInputStream = new FileInputStream(new File(filePath));
} catch(FileNotFoundException e) { }
URL url = new URL(urlString);
connection = (HttpURLConnection) url.openConnection();
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
SharedPreferences prefs = PreferenceManager.getDefaultSharedPreferences(getBaseContext());
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
outStream = new DataOutputStream(connection.getOutputStream());
// Some POST values
outStream.writeBytes(addParam("additional_param", "some value");
outStream.writeBytes(addParam("additional_param 2", "some other value");
// The file with the name "uploadedfile"
outStream.writeBytes(twoHyphens + boundary + lineEnd);
outStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + filePath +"\"" + lineEnd + "Content-Type: " + mimeType + lineEnd + "Content-Transfer-Encoding: binary" + lineEnd);
outStream.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
outStream.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
outStream.writeBytes(lineEnd);
outStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
fileInputStream.close();
outStream.flush();
outStream.close();
} catch (MalformedURLException e) {
Log.e("APP", "MalformedURLException while sending\n" + e.getMessage());
} catch (IOException e) {
Log.e("APP", "IOException while sending\n" + e.getMessage());
}
// This part checks the response of the server. If its "UPLOAD OK" the method returns true
try {
inStream = new DataInputStream(connection.getInputStream());
String str;
while ((str = inStream.readLine()) != null) {
if(str=="UPLOAD OK") {
return true;
} else {
return false;
}
}
inStream.close();
} catch (IOException e){
Log.e("APP", "IOException while sending\n" + e.getMessage());
}
return false;
}
现在,我们已经得到了对方:http://your.domain.com/webservice.php 您的服务器。它需要一些逻辑(例如PHP)来处理发送的POST请求。 像这样的工作:
<?php
$target_path = "videos/";
$target_path = $target_path.'somename.3gp';
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
exit("UPLOAD OK");
} else {
exit("UPLOAD NOT OK");
}
?>
嘿感谢一吨!这是很好的给我一个踢开始。我会与它合作,如果有任何问题将再次联系 – NickHalden 2010-08-09 07:35:51
谢谢,并没有问题:) 与ASyncTask的东西是重要的。想象一下,你有一个5MiB文件要上传,用户在这个屏幕上被拦截,没有任何反应,甚至没有加载图标等等。 – 2010-08-09 08:47:34