Related to ProcessStartInfo() method in C#
<p>I want to give command name and file path in ProcessStartInfo() method in C#.</p>

<p>So I have a command name("F:\AndroidProjects\AndProj3>) and file path("F:\Android\apache-ant-1.8.2-bin\apache-ant-1.8.2\bin\ant debug") just like that but it is not working and process can not be started.</p>

<p>Please give me a solution for starting the process because command name first execute and after that file path will be execute.
How I can pass the both argument in ProcessStartInfo() method?</p>

<pre><code>public static string BuildAndroidProject()
    {
        string result="";
      //  string ProjNameNDLocation = ProjectLocation + "\\" + ProjectName + "&gt;";
        try
        {

            System.Diagnostics.ProcessStartInfo androidBuildProj = new System.Diagnostics.ProcessStartInfo("F:\\AndroidProjects\\AndProj3 F:\\Android\\apache-ant-1.8.2-bin\\apache-ant-1.8.2\\bin\\ant debug");//ProjNameNDLocation, Program.ANDROIDDEBUGGCMD);
            androidBuildProj.RedirectStandardOutput = true;
            androidBuildProj.UseShellExecute = false;
            androidBuildProj.CreateNoWindow = true;
            System.Diagnostics.Process androidProcess = new System.Diagnostics.Process();
            androidProcess.StartInfo = androidBuildProj;
            androidProcess.Start();
           result = androidProcess.StandardOutput.ReadToEnd();
            androidProcess.Close();

        }
        catch (Exception e)
        {

        }
        return result;
    }
</code></pre>

<p>Problem is in the <code>ProcessInfoStart</code> Function.  How can I run this command?</p>
