C# process.start 参数

Question

我们可以创建一个.exe 文件的快捷方式。

如您所知，快捷方式文件可以包含参数。

我想构建一个使用C #代码自动分配参数的程序。

这是我的代码：

    private void button4_Click(object sender, EventArgs e)
    {
        HttpWebRequest req =
        WebRequest.Create("http://test.com/contact_ip.txt") as HttpWebRequest;

        HttpWebResponse resp = req.GetResponse() as HttpWebResponse;

        using (StreamReader sr = new StreamReader(resp.GetResponseStream()))
        {
            string ResponseText = sr.ReadToEnd();
            //ResponseText

            System.Diagnostics.Process.Start(".\\folder\\test.exe");
            // e.g.)) test.exe 127.0.0.1(ResponseText) 1234(port)
        }
    }

如何在test.exe 中包含 IP 地址和端口？

e.g.)) .\folder\test.exe 127.0.0.1 4455

e.g.2)) Var ResponseText = ip Address

Answer 1

你可以试试这个：

private void button4_Click(object sender, EventArgs e)
{
    HttpWebRequest req =
    WebRequest.Create("http://test.com/contact_ip.txt") as HttpWebRequest;

    HttpWebResponse resp = req.GetResponse() as HttpWebResponse;

    using (StreamReader sr = new StreamReader(resp.GetResponseStream()))
    {
        string ResponseText = sr.ReadToEnd();
        //ResponseText

        System.Diagnostics.Process proc = new System.Diagnostics.Process();
        proc.StartInfo.FileName = @".\\folder\\test.exe";
        proc.StartInfo.Arguments = "127.0.0.1 4455";
        proc.Start();
    }
}