C# capturing image from webcam

ccapturefileimagewebcam

Last two days I've been looking for a way to capture an image from the webcam using C#. I'm pretty new in c# and I DO NOT want to use external third party libs, so I found two nice ways, but both seems to return almost the same error. I couldn't get any of them to work, so it would be really nice if you help me to get one of them running or help me to find alternative.

So the first way I found is using Windows WIA. I found the following code:

CommonDialogClass dialog = new CommonDialogClass();
Device camera = dialog.ShowSelectDevice(WiaDeviceType.CameraDeviceType, true, false);
        // take the photo 
Item item = camera.ExecuteCommand(CommandID.wiaCommandTakePicture);
ImageFile image = (ImageFile)item.Transfer(FormatID.wiaFormatJPEG);
        // filename and saving 
image.SaveFile("Test.jpg");

this code seems to be exacly what I'm looking for, but I can't get it running, because I'm getting the following error on the second line:

Exception from HRESULT: 0x80210015

The second way I found is using the Avicap32.dll with the following sample:

http://www.timvw.be/wp-content/code/csharp/testavicap32.zip

but I'm getting on this code:

Image image = ((CaptureDevice)cboDevices.SelectedItem).Capture();
image.Save(@"c:\capture.png", ImageFormat.Png);

the following exception:
NullReferenceException:
Object reference not set to an instance of an object.

I think that both solution are causing problems because they can't find my camera, but I can use my camera in skype without any problems.

Best Answer

I tried multiple approaches, and the easiest approach for me was Emgu.cv(nuget package).

        VideoCapture capture = new VideoCapture(); //create a camera capture
        Bitmap image = capture.QueryFrame().Bitmap; //take a picture

That's it (as of API version 3.3.0)

Old API Approach

        Capture capture = new Capture(); //create a camera captue
        Bitmap image = capture.QueryFrame().Bitmap; //take a picture

Related Solutions

Java – Capturing image from webcam in java

This JavaCV implementation works fine.

Code:

import org.bytedeco.javacv.*;
import org.bytedeco.opencv.opencv_core.IplImage;

import java.io.File;

import static org.bytedeco.opencv.global.opencv_core.cvFlip;
import static org.bytedeco.opencv.helper.opencv_imgcodecs.cvSaveImage;

public class Test implements Runnable {
    final int INTERVAL = 100;///you may use interval
    CanvasFrame canvas = new CanvasFrame("Web Cam");

    public Test() {
        canvas.setDefaultCloseOperation(javax.swing.JFrame.EXIT_ON_CLOSE);
    }

    public void run() {

        new File("images").mkdir();

        FrameGrabber grabber = new OpenCVFrameGrabber(0); // 1 for next camera
        OpenCVFrameConverter.ToIplImage converter = new OpenCVFrameConverter.ToIplImage();
        IplImage img;
        int i = 0;
        try {
            grabber.start();

            while (true) {
                Frame frame = grabber.grab();

                img = converter.convert(frame);

                //the grabbed frame will be flipped, re-flip to make it right
                cvFlip(img, img, 1);// l-r = 90_degrees_steps_anti_clockwise

                //save
                cvSaveImage("images" + File.separator + (i++) + "-aa.jpg", img);

                canvas.showImage(converter.convert(img));

                Thread.sleep(INTERVAL);
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static void main(String[] args) {
        Test gs = new Test();
        Thread th = new Thread(gs);
        th.start();
    }
}

There is also post on configuration for JavaCV

You can modify the codes and be able to save the images in regular interval and do rest of the processing you want.

Java – How to create a Java string from the contents of a file

Read all text from a file

Java 11 added the readString() method to read small files as a String, preserving line terminators:

String content = Files.readString(path, StandardCharsets.US_ASCII);

For versions between Java 7 and 11, here's a compact, robust idiom, wrapped up in a utility method:

static String readFile(String path, Charset encoding)
  throws IOException
{
  byte[] encoded = Files.readAllBytes(Paths.get(path));
  return new String(encoded, encoding);
}

Read lines of text from a file

Java 7 added a convenience method to read a file as lines of text, represented as a List<String>. This approach is "lossy" because the line separators are stripped from the end of each line.

List<String> lines = Files.readAllLines(Paths.get(path), encoding);

Java 8 added the Files.lines() method to produce a Stream<String>. Again, this method is lossy because line separators are stripped. If an IOException is encountered while reading the file, it is wrapped in an UncheckedIOException, since Stream doesn't accept lambdas that throw checked exceptions.

try (Stream<String> lines = Files.lines(path, encoding)) {
  lines.forEach(System.out::println);
}

This Stream does need a close() call; this is poorly documented on the API, and I suspect many people don't even notice Stream has a close() method. Be sure to use an ARM-block as shown.

If you are working with a source other than a file, you can use the lines() method in BufferedReader instead.

Memory utilization

The first method, that preserves line breaks, can temporarily require memory several times the size of the file, because for a short time the raw file contents (a byte array), and the decoded characters (each of which is 16 bits even if encoded as 8 bits in the file) reside in memory at once. It is safest to apply to files that you know to be small relative to the available memory.

The second method, reading lines, is usually more memory efficient, because the input byte buffer for decoding doesn't need to contain the entire file. However, it's still not suitable for files that are very large relative to available memory.

For reading large files, you need a different design for your program, one that reads a chunk of text from a stream, processes it, and then moves on to the next, reusing the same fixed-sized memory block. Here, "large" depends on the computer specs. Nowadays, this threshold might be many gigabytes of RAM. The third method, using a Stream<String> is one way to do this, if your input "records" happen to be individual lines. (Using the readLine() method of BufferedReader is the procedural equivalent to this approach.)

Character encoding

One thing that is missing from the sample in the original post is the character encoding. There are some special cases where the platform default is what you want, but they are rare, and you should be able justify your choice.

The StandardCharsets class defines some constants for the encodings required of all Java runtimes:

String content = readFile("test.txt", StandardCharsets.UTF_8);

The platform default is available from the Charset class itself:

String content = readFile("test.txt", Charset.defaultCharset());

Note: This answer largely replaces my Java 6 version. The utility of Java 7 safely simplifies the code, and the old answer, which used a mapped byte buffer, prevented the file that was read from being deleted until the mapped buffer was garbage collected. You can view the old version via the "edited" link on this answer.