如何将图像从SD卡上传到服务器上的文件夹
问题描述：
我想从SD卡上传图像到服务器中的文件夹。我有以下信息 http://mywebsite.myworks.com/parent/webservices/myuploader.ashx（POST）如何将图像从SD卡上传到服务器上的文件夹
Content-Disposition: form-data; name="Filename"
FilName.jpg
Content-Disposition: form-data; name="password"
mypassword
Content-Disposition: form-data; name="action"
upload
Content-Disposition: form-data; name="parentNodeId"
1297
Content-Disposition: form-data; name="replaceExisting"
0
Content-Disposition: form-data; name="username"
admin
Content-Disposition: form-data; name="Filedata"; filename="2012-11-07 11-03-37.jpg"
Content-Type: application/octet-stream
[DATA]
基本上我想要的是上传图片到我已经明白迄今上传到其parentNodeId的文件夹folder.What是1297年，我看到一些教程关于如何将图片上传到php服务器。基于此我写了一个函数。林分享下面
private void doFileUpload() {
HttpURLConnection conn = null;
DataOutputStream dos = null;
DataInputStream inStream = null;
String exsistingFileName = "/sdcard/r.jpg";
// Is this the place are you doing something wrong.
String lineEnd = "rn";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
String responseFromServer = "";
String urlString = "http://mywebsite.myworks.com/parent/webservices/myuploader.ashx";
try {
// ------------------ CLIENT REQUEST
Log.e("UploadMe", "Inside second Method");
FileInputStream fileInputStream = new FileInputStream(new File(exsistingFileName));
// open a URL connection to the Servlet
URL url = new URL(urlString);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
// Allow Inputs
conn.setDoInput(true);
// Allow Outputs
conn.setDoOutput(true);
// Don't use a cached copy.
conn.setUseCaches(false);
// Use a post method.
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
// conn.setRequestProperty("Content-Type",
// "multipart/form-data;boundary="+boundary);
conn.setRequestProperty("Content-Type", "application/octet-stream");
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"Filename\";filename=\""
+ exsistingFileName + "\"" + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"Filename\";filename=\""
+ "r.jpg" + "\"" + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"password\";filename=\""
+ "mypassword" + "\"" + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"action\";filename=\""
+ "upload" + "\"" + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"parentNodeId\";filename=\""
+ "-1" + "\"" + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"replaceExisting\";filename=\""
+ "0" + "\"" + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"username\";filename=\""
+ "admin" + "\"" + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"Filedata\";filename=\""
+ "2012-11-07 11-03-37.jpg" + "\"" + lineEnd);
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes(lineEnd);
Log.e("UploadMe", "Headers are written");
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
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);
// close streams
Log.e("UploadMe", "File is written");
fileInputStream.close();
dos.flush();
dos.close();
}
catch (MalformedURLException ex)
{
Log.e("UploadMe", "error: " + ex.getMessage(), ex);
}
catch (IOException ioe)
{
Log.e("UploadMe", "error: " + ioe.getMessage(), ioe);
}
// ------------------ read the SERVER RESPONSE
try {
inStream = new DataInputStream(conn.getInputStream());
String str;
while ((str = inStream.readLine()) != null) {
Log.e("UploadMe", "Server Response" + str);
}
inStream.close();
}
catch (IOException ioex) {
Log.e("UploadMe", "error: " + ioex.getMessage(), ioex);
}
}
程序而不进行碰撞并即时得到这些从我的日志猫output.But我couldnot看到我的服务器文件夹的任何运行它。
11-22 10:26:39.546: E/UploadMe(3363): Inside second Method
11-22 10:26:39.843: E/UploadMe(3363): Headers are written
11-22 10:26:39.851: E/UploadMe(3363): File is written
11-22 10:26:40.820: E/UploadMe(3363): Server Response<?xml version="1.0"?>
11-22 10:26:40.820: E/UploadMe(3363): Server Response<MediaResponse xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" success="false" />
因为我不知道上传文件到服务器，这是上传文件的方式吗？
答
一个lib中是可用的呼叫android-async-http-1.4.2.jar利用这一点，你可以上传文件，后弦的数据也
这里是示例代码
RequestParams params = new RequestParams();
params.put("fname", "name");// here write your parameter name and its value
params.put("file", new File(filePath)); // Upload a File
// params.put("profile_picture2", someInputStream); // Upload an
// InputStream
// params.put("profile_picture3", new ByteArrayInputStream(someBytes));
// // Upload some bytes
AsyncHttpClient client = new AsyncHttpClient();
client.post(urlServer, params, new AsyncHttpResponseHandler() {
@Override
public void onSuccess(String arg0) {
super.onSuccess(arg0);
Log.v("from response", arg0);
}
});
这里的jar下载链接
答
看到这个
链接how to Upload image into server in Android?
在这个例子中，他们想同您上传影像服务器，但店文件夹是 Web服务开发者的任务，他实际上不存储图像。
你的任务只是发送图像到服务器。
试试这个http://*.com/questions/13383833/upload-picture-to-server – Ronnie