Question

我正在创建Process类的实例，并将StartInfo.FileName路径传递给MS Paint，并作为文件的参数路径：

Process paint = new Process();
paint.StartInfo.FileName = @"C:\Windows\System32\mspaint.exe";
SchemaPath = @"E:\directory\Alisa\0-01\0-01 (1).jpg";
Console.WriteLine(SchemaPath);
paint.StartInfo.Arguments = SchemaPath;
paint.Start();

我的问题是 - 当我的文件名包含空格时，mspaint完全忽略该文件并抛出错误。例如，当我的文件有路径时：

E:\directory\Alisa\0-01\0-01abc.jpg

一切正常，但当我有路径文件时：

E:\directory\Alisa\0-01\0-01 (1).jpg

Paint打开但会抛出消息：

E:\directory\Alisa\0-01\0-01.png was not found.

有谁知道如何处理它？</ p>

Answer 1

尝试引用类似

的路径

paint.StartInfo.Arguments = string.Format("\"{0}\"", SchemaPath);

Answer 2

fname.Replace(@" ", "");

使用上述方法，您可以在处理之前替换空间

            Process paint = new Process();
            string fname = @"C:\ Windows\System32\mspaint.exe";
            string k = fname.Replace(@" ", "");
            paint.StartInfo.FileName =k; 
            Console.WriteLine(SchemaPath);
            paint.StartInfo.Arguments = SchemaPath;
            paint.Start();

Answer 3

Dim bmpFile As String = "C:\Users\Public\ScreenShot.jpg"
Using BMP As New Bitmap(AxAcroPDF1.Width - 20, AxAcroPDF1.Height) '20 IS ATTEMPT TO CLIP OFF THE SCROLL BAR.
            Using G As Graphics = Graphics.FromImage(BMP)
                G.CopyFromScreen(0, 0, 0, 0, BMP.Size)
                BMP.Save(bmpFile, Imaging.ImageFormat.Jpeg)
            End Using
End Using
Shell("cmd /c mspaint /p " & bmpFile) 'WORKS!

进程无法在mspaint中打开文件

3 个答案: