无法打开输入文件:filename.php

Question

在.bat文件中包含代码PHP filename.php我打开bat文件后收到错误消息无法找到输入文件:filename.php.任何人都可以提出建议

private void button5_Click_1(object sender, EventArgs e)
{
    try
    {
        string bat = "D:\\folder1\\file1.bat";
        if (File.Exists(bat))
        {
            if (radioButton1.Enabled == true)
            {
                MessageBox.Show(bat.ToString());
                System.Diagnostics.Process.Start(bat);
            }
        }
        catch (Exception ex)
        {
            MessageBox.Show(ex.ToString());
        }
    }
}

Answer 1

我猜这个过程不是在同一个文件夹中启动的filename.php.指定bat文件中的完整路径或设置启动文件夹:

Process process = new Process();
process.StartInfo.FileName = bat;
process.StartInfo.WorkingDirectory = "D:\\folder1";  // or whatever is appropriate
process.Start();