如何从C#应用程序在Windows Photo Viewer中打开.jpg
图像?
而不是像这样在应用程序中,
FileStream stream = new FileStream("test.png", FileMode.Open, FileAccess.Read);
pictureBox1.Image = Image.FromStream(stream);
stream.Close();
发布于 2011-07-25 00:24:01
在新的Process中启动它
Process photoViewer = new Process();
photoViewer.StartInfo.FileName = @"The photo viewer file path";
photoViewer.StartInfo.Arguments = @"Your image file path";
photoViewer.Start();
发布于 2011-07-25 00:26:24
我认为你可以直接使用:
Process.Start(@"C:\MyPicture.jpg");
这将使用与.jpg文件相关联的标准文件查看器-默认情况下是windows图片查看器。
发布于 2019-02-16 01:06:19
该代码从ftp中获取照片,并在Windows照片查看器中显示照片。我希望它对你有用。
public void ShowPhoto(String uri, String username, String password)
{
WebClient ftpClient = new WebClient();
ftpClient.Credentials = new NetworkCredential(username,password);
byte[] imageByte = ftpClient.DownloadData(uri);
var tempFileName = Path.GetTempFileName();
System.IO.File.WriteAllBytes(tempFileName, imageByte);
string path = Environment.GetFolderPath(
Environment.SpecialFolder.ProgramFiles);
// create our startup process and argument
var psi = new ProcessStartInfo(
"rundll32.exe",
String.Format(
"\"{0}{1}\", ImageView_Fullscreen {2}",
Environment.Is64BitOperatingSystem ?
path.Replace(" (x86)", "") :
path
,
@"\Windows Photo Viewer\PhotoViewer.dll",
tempFileName)
);
psi.UseShellExecute = false;
var viewer = Process.Start(psi);
// cleanup when done...
viewer.EnableRaisingEvents = true;
viewer.Exited += (o, args) =>
{
File.Delete(tempFileName);
};
}
致以最良好的问候。
https://stackoverflow.com/questions/6808029
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