【发布时间】:2012-03-08 09:37:15
【问题描述】:
我想将文件从客户端上传到服务器。
客户端:Java 使用 HTTP 发布 服务器:Java Servlet
我在这里添加了客户端编码。但是,我对服务器端处理一无所知。请帮我写一个代码sn-p。
private String Tag = "UPLOADER";
private String urlString = "http://192.168.42.140:8080/sampweb";
HttpURLConnection conn;
String exsistingFileName = "/sdcard/log.txt";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
try {
// ------------------ CLIENT REQUEST
Log.e(Tag, "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);
DataOutputStream dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos
.writeBytes("Content-Disposition: post-data; name=uploadedfile;filename="
+ exsistingFileName + "" + lineEnd);
dos.writeBytes(lineEnd);
Log.e(Tag, "Headers are written");
// create a buffer of maximum size
int bytesAvailable = fileInputStream.available();
int maxBufferSize = 1000;
// int bufferSize = Math.min(bytesAvailable, maxBufferSize);
byte[] buffer = new byte[bytesAvailable];
// read file and write it into form...
int bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);
while (bytesRead > 0) {
dos.write(buffer, 0, bytesAvailable);
bytesAvailable = fileInputStream.available();
bytesAvailable = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);
}
// send multipart form data necessary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// close streams
Log.e(Tag, "File is written");
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
Log.e(Tag, "error: " + ex.getMessage(), ex);
}
catch (IOException ioe) {
Log.e(Tag, "error: " + ioe.getMessage(), ioe);
}
try {
BufferedReader rd = new BufferedReader(new InputStreamReader(conn
.getInputStream()));
String line;
while ((line = rd.readLine()) != null) {
Log.e("Dialoge Box", "Message: " + line);
}
rd.close();
} catch (IOException ioex) {
Log.e("MediaPlayer", "error: " + ioex.getMessage(), ioex);
}
}
我是服务器编程的新手。如果我犯了任何错误,请澄清我。谢谢!
【问题讨论】:
-
标题具有误导性,但问题本身很清楚,试图在 Java 中实现 server 端,而不是客户端。当前链接的副本描述了客户端实现。
-
你误用了
available(),这里根本不需要使用。
标签: java file-upload