Question

单击按钮时，我试图保存picEmployeeImage的图像。 PicEmployeeImage会传输视频，因此在按下照片时会将其保存到文件中。目前，当我尝试它时，它说＆＃34;在GDI +＆＃34;

中发生了一般性错误

我该如何解决这个问题？

请帮助。

以下是代码：

public void TakePhoto()
        {
            Bitmap bm = new Bitmap(picEmployeeImage.Width, picEmployeeImage.Height);
            bm = (Bitmap) picEmployeeImage.Image;
            bm.Save(@"\Images\" + currentlySelectedId + ".bmp", ImageFormat.Bmp);

        }


private void btnCaptureNewImage_Click(object sender, EventArgs e)
        {
            if (btnCaptureNewImage.Text == "Capture New Image")
            {
                VideoShow();
                btnCaptureNewImage.Text = "Take Photo";
            } else
            {
                TakePhoto();
                btnCaptureNewImage.Text = "Capture New Image";
            }
        }

 void VideoShow()
        {
            VideoCaptureDevices = new FilterInfoCollection(FilterCategory.VideoInputDevice);
            FinalVideoSource = new VideoCaptureDevice(VideoCaptureDevices[0].MonikerString);
            FinalVideoSource.NewFrame += new NewFrameEventHandler(FinalVideoSource_NewFrame);
            FinalVideoSource.Start();
        }

 void FinalVideoSource_NewFrame(object sender, NewFrameEventArgs eventArgs)
            {
                image = (Bitmap)eventArgs.Frame.Clone();
                image = CropBitmap(image, 150, 100, 300, 200);
                picEmployeeImage.Image = image;
            }

更新：错误信息：

System.Runtime.InteropServices.ExternalException (0x80004005): A generic error occurred in GDI+.
   at System.Drawing.Image.Save(String filename, ImageCodecInfo encoder, EncoderParameters encoderParams)
   at System.Drawing.Image.Save(String filename, ImageFormat format)
   at BusinessSoftware.frmLoginPage.TakePhoto() in C:\Users\....cs:line 721
   at BusinessSoftware.frmLoginPage.btnCaptureNewImage_Click(Object sender, EventArgs e) in C:\Users\....cs:line 705
   at System.Windows.Forms.Control.OnClick(EventArgs e)
   at System.Windows.Forms.Button.OnClick(EventArgs e)
   at System.Windows.Forms.Button.OnMouseUp(MouseEventArgs mevent)
   at System.Windows.Forms.Control.WmMouseUp(Message& m, MouseButtons button, Int32 clicks)
   at System.Windows.Forms.Control.WndProc(Message& m)
   at System.Windows.Forms.ButtonBase.WndProc(Message& m)
   at System.Windows.Forms.Button.WndProc(Message& m)
   at System.Windows.Forms.Control.ControlNativeWindow.OnMessage(Message& m)
   at System.Windows.Forms.Control.ControlNativeWindow.WndProc(Message& m)
   at System.Windows.Forms.NativeWindow.Callback(IntPtr hWnd, Int32 msg, IntPtr wparam, IntPtr lparam)

Answer 1

这是因为GDI + Image类不是线程安全的。

创建一个新的位图图像并绘制您在此新Bitmap对象上捕获的图像并保存此新对象。代码如下：

var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
var i = arr.length;
var partLen = i / 3;
var parentArr = [],
  temp = [];

function shuffle(array) {
  var currentIndex = array.length,
    temporaryValue, randomIndex;

  // While there remain elements to shuffle...
  while (0 !== currentIndex) {

    // Pick a remaining element...
    randomIndex = Math.floor(Math.random() * currentIndex);
    currentIndex -= 1;

    // And swap it with the current element.
    temporaryValue = array[currentIndex];
    array[currentIndex] = array[randomIndex];
    array[randomIndex] = temporaryValue;
  }

  return array;
}
var shuffled = shuffle(arr);
while (i--) {
  if (temp.length >= partLen) {
    parentArr.unshift(temp);
    temp = [];
  }
  temp.push(shuffled[i]);
}
if (temp.length) {
  parentArr.unshift(temp);
  temp = [];
}
console.log(JSON.stringify(parentArr, null, 4));

另外，我建议你在点击“拍照”按钮时停止FinalVideoSource

public void TakePhoto()
{
    Bitmap bm = new Bitmap(picEmployeeImage.Width, picEmployeeImage.Height);
    bm = (Bitmap)picEmployeeImage.Image;

    Bitmap bmp = new Bitmap(bm.Width, bm.Height);
    Graphics g = Graphics.FromImage(bmp);
    g.DrawImage(bm, 0, 0, bmp.Width, bmp.Height);
    bmp.Save(@"\Images\" + currentlySelectedId + ".bmp", ImageFormat.Bmp);
}

HTH。

保存图像时GDI +中的一般错误

1 个答案: