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  1. POhow to do a looping to retrieve streaming image?
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    <p>how can i get all the streaming jpeg to create an avi?</p> <pre><code>Bitmap img = eventArgs.Frame; for (int i = 0; i &lt; 1000; i++) { //img.SetPixel(i, i, Color.Red); writer.AddFrame(img); } writer.Close(); </code></pre> <p>i got this example from AForge, <a href="http://www.aforgenet.com/framework/docs/html/5fedbbbe-6d28-5f69-35a2-9d7119544a2d.htm" rel="nofollow">http://www.aforgenet.com/framework/docs/html/5fedbbbe-6d28-5f69-35a2-9d7119544a2d.htm</a> but when i test it, it create a video file with 1 same frame from beginning to the end of the video.</p> <p>how can i execute the writer.AddFrame(img); to create video file with new frame everytime?<br></p> <p><br> EDIT:</p> <pre><code>using System; using System.Collections.Generic; using System.Linq; using System.Web; using System.Web.UI; using System.Drawing; using System.IO; using AForge.Video; using AForge.Video.VFW; using AForge.Video.DirectShow; namespace VidRec { public partial class LiveRecording : System.Web.UI.Page { AVIWriter writer = new AVIWriter("MSVC"); //video with compression protected void Page_Load(object sender, EventArgs e) { //camera source string streamingSource = "http://e888.example.com:81/snapshot.cgi"; // create new AVI file and open it writer.Open("c:\\video.avi", 640, 480); //IPcam streaming source to be recorded JPEGStream JPEGSource = new JPEGStream(streamingSource); //MJPEGStream JPEGSource = new MJPEGStream(streamingSource); //login name and password JPEGSource.Login = "login"; JPEGSource.Password = "password"; //set NewFrame event handler JPEGSource.NewFrame += new NewFrameEventHandler(video_NewFrame); //start streaming the image files JPEGSource.Start(); } private void video_NewFrame(object sender, NewFrameEventArgs eventArgs) //event handler for NewFrame { //get frame Bitmap image = eventArgs.Frame; //add the image as a new frame of video file writer.AddFrame(image); } protected void Button1_Click(object sender, EventArgs e) { writer.Close(); } } </code></pre> <p>}</p>
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