如何检查系统在C#中是否具有AMD或NVIDIA?

问题描述

我正在尝试使用C#创建一个以太坊挖矿客户端，并且我需要检查系统是否具有AMD或NVIDIA.这是因为程序需要知道是否应该通过CUDA或OpenCL挖掘以太坊.

I'm trying to make an Ethereum mining client using C#, and I need to check whether the system has AMD or NVIDIA. This is because the program needs to know whether it should mine Ethereum via CUDA or OpenCL.

推荐答案

您需要使用System.Management命名空间(可以在引用/程序集下找到)

You need to use System.Management Namespace (You can find under references/Assemblies)

添加名称空间后，需要导航ManagementObject的所有属性，并导航propertydata的所有属性，直到在name属性上创建描述为止.

After adding namespace you need to navigate all properties of ManagementObject and navigate all properties of propertydata till founding description on name property.

我为控制台应用程序编写了此解决方案.您可以调整您的解决方案.

I wrote this solution for console app. You can adapt your solution.

尝试一下:

 using System;
 using System.Management;

 namespace ConsoleApp1
 {
 class Program
 {
    static void Main(string[] args)
    {
        ManagementObjectSearcher searcher = new
 ManagementObjectSearcher("SELECT * FROM Win32_DisplayConfiguration");

        string gc = "";
        foreach (ManagementObject obj in searcher.Get())
        {
            foreach (PropertyData prop in obj.Properties)
            {
                if (prop.Name == "Description")
                {
                    gc = prop.Value.ToString().ToUpper();

                    if (gc.Contains("INTEL") == true)
                    {
                      Console.WriteLine("Your Graphic Card is Intel");
                    }
                    else if (gc.Contains("AMD") == true)
                    {
                        Console.WriteLine("Your Graphic Card is AMD");
                    }
                    else if (gc.Contains("NVIDIA") == true)
                    {
                        Console.WriteLine("Your Graphic Card is NVIDIA");
                    }
                    else
                    {
                        Console.WriteLine("Your Graphic Card cannot recognized.");

                    }
                    Console.ReadLine();
                }
            }
        }
    }
}
}

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