ProcessStartInfo，python.exe：无法打开文件 [Errno 22] 参数无效

Question

我有一个 python 脚本 'saudilp.py'（它分析车牌号），我尝试运行这个脚本在 c#

中

此 C# 代码：

// 1) create process info
        ProcessStartInfo start = new ProcessStartInfo();

        //cmd is full path to python.exe
        start.FileName = @"C:\Program Files (x86)\Microsoft Visual Studio\Shared\Python39_64\python.exe"; 


        // 2) Provide script and arguments
        string arg = "s10.jpeg";
        string pathScript = "D:/graduation project/pythonOCRtest/pythonOCRtest/OCR/saudilp.py";
        start.Arguments = $"\"{pathScript}\"\"{arg}\""; //args is path to .py file and any cmd line args

        // 3) process configuration
        start.UseShellExecute = false;
        start.CreateNoWindow = true; //do not create window 
        start.RedirectStandardOutput = true; //recive print lines from the script
        start.RedirectStandardError = true;

        // 4) Execute process and get output
        string result = "";
        string errors = "";

        using (Process process = Process.Start(start))
        {
            errors = process.StandardError.ReadToEnd();
            result = process.StandardOutput.ReadToEnd();
        }

        // 5) display output
        Console.WriteLine("ERRORS:");
        Console.WriteLine(errors);
        Console.WriteLine();
        Console.WriteLine("RESULTS:");
        Console.WriteLine(result);

控制台打印此错误：

C:\Program Files (x86)\Microsoft Visual Studio\Shared\Python39_64\python.exe: 无法打开文件 'D:\graduation project\pythonOCRtest\pythonOCRtest\OCR\saudilp.py"s10.jpeg': [Errno 22] 参数无效

Answer 1

你的论点似乎有一个额外的 \，试试

start.Arguments = $"\"{pathScript}\" {arg}\"";

您可能还需要 set-up 工作目录

start.WorkingDirectory = 路径

ProcessStartInfo，python.exe：无法打开文件 [Errno 22] 参数无效

ProcessStartInfo, python.exe: can't open file [Errno 22] Invalid argument

python

c#

.net

processstartinfo