android video ?

 This post last edited by the kaly_liu on 2013-08-13 15:23:29

Android client is the use of socket camera data transmission , on another android device how to receive and play ?

 Socket tempSocket = new Socket(ipname, port);
                 outsocket = tempSocket.getOutputStream();
                 //写入头部数据信息
             	String msg=java.net.URLEncoder.encode("PHONEVIDEO|"+username+"|","utf-8");
                 byte[] buffer= msg.getBytes();
                 outsocket.write(buffer);
                 //将数据转成 数据流
                 ByteArrayInputStream inputstream = new ByteArrayInputStream(myoutputstream.toByteArray());
                 int amount;
                 //在while括号里面 读取数据的个数 amount
                 while ((amount = inputstream.read(byteBuffer)) != -1) {
                     outsocket.write(byteBuffer, 0, amount);//将 byteBuffer里的数据发送出去 ,一共有amount个
                 }
                 myoutputstream.flush();
                 myoutputstream.close();
                 tempSocket.close();           
 

 if(data!=null)
 		      {
 		        YuvImage image = new YuvImage(data,VideoFormatIndex, VideoWidth, VideoHeight,null);
 		        if(image!=null)
 		        {
 		        	ByteArrayOutputStream outstream = new ByteArrayOutputStream();
 		      	  	//在此设置图片的尺寸和质量 
 		      	  	image.compressToJpeg(new Rect(0, 0, (int)(VideoWidthRatio*VideoWidth), 
 		      	  		(int)(VideoHeightRatio*VideoHeight)), VideoQuality, outstream);  
 		      	  	outstream.flush();
 		      	  	//启用线程将图像数据发送出去
 		      	  	Thread th = new MySendFileThread(outstream,pUsername,serverUrl,serverPort);
 		      	  	th.start();  
 		        }
 

these two codes is part of the program the sender .
------ Solution ---------------------------------------- ----
socket too unstable , the landlord can do a set ?
------ Solution ---------------------------------------- ----
you are sending a jpeg types of data , I think you should define a tcp based on proprietary protocols, such as the first four bytes are the length jpeg , jpeg then followed by the actual data, and then received the first four bytes of the receiver to get the length , then the length of bytes received , and then converted to a bitmap in jpeg ImageView above show it. If it is video, then not so simple , requires both sides to negotiate video formats
------ Solution ------------------------- -------------------
generally use the RTP protocol packets issued
------ reference ----------- ----------------------------

anyway, you can send the data , the receiver did not get , do not know how effective amount · · Haha