c# - 无法从线程任务 转换为系统操作

Question

我在按钮中运行任务时遇到问题问题是:无法从 threading.task.task 转换为系统操作我在按钮中标记了行

         private async void BtnStart_Click(object sender, EventArgs e)         {
        if (String.IsNullOrEmpty( txtProxy.Text) || lstviewcomp.Items.Count==0)
        {
            MessageBox.Show("Please uploads files");
            return;
        }
        proxies = txtProxy.Text.Split('\n');
        proxycount = proxies.Length;
        foreach (string item in lstviewcomp.Items)
        {
            proxycount++;
            if (proxyCounter> proxycount)
            {
                proxyCounter = 0;
            }
            ProxyInfo = proxies[proxyCounter].Split(',');
            var result = await Task.Run( MainAsync("", "", "", "", "", "")).GetAwaiter().GetResult();// Problem in this line 


            // proxyCounter++;
        }
    }
公共(public)异步任务 MainAsync(字符串 instausername，字符串 pass，字符串 proxyip，字符串 proxyport，字符串 proxyusername，字符串 proxypass)        {            尝试            {               //创建用户 session 数据并提供登录详细信息                var userSession = new UserSessionData                {                    UserName = instausername,                    密码=通过                };               //创建代理处理器                var httpHndler = new HttpClientHandler();                IWebProxy proxy = new WebProxy(proxyip,Convert.ToInt32(proxyport));                proxy.Credentials = new NetworkCredential(proxyusername, proxypass);                httpHndler.Proxy = 代理；               //使用 Builder 创建新的 InstaApi 实例                _instaApi = 新 InstaApiBuilder()                    .SetUser(用户 session )                    .UseHttpClientHandler(httpHndler)                    .UseLogger(logger: new DebugFileLogger())//将记录器用于请求和调试消息                    .SetRequestDelay(TimeSpan.FromSeconds(1))//设置请求之间的延迟                    .Build();
//登录
            txtLog.Text = txtLog.Text + $"以 {userSession.UserName} 身份登录"+"在 "+DateTime.Now+"\n";
            var logInResult = await _instaApi.LoginAsync();
            如果(!logInResult.Succeeded)
            {
                txtLog.Text = txtLog.Text + $"无法登录:{logInResult.Info.Message}"+ "At "+ DateTime.Now + "\n";

            }
            别的
            {
                txtLog.Text = txtLog.Text + $"登录成功:{userSession.UserName}"+ "At "+ DateTime.Now + "\n";

            }
        }
         catch (异常前)
        {
            控制台.WriteLine(ex);
        }
        最后
        {
            var logoutResult = Task.Run(() => _instaApi.LogoutAsync()).GetAwaiter().GetResult();
            if (logoutResult.Succeeded) txtLog.Text = txtLog.Text + "注销成功\n";

        }
        返回假；
    }

Answer 1

public async Task MainAsync

应该改为

public async Task<bool> MainAsync

然后代替

var result = await Task.Run( MainAsync("", "", "", "", "", "")).GetAwaiter().GetResult();

你可以使用

var result = await MainAsync("", "", "", "", "", "");

此外，您还需要在以下行中使用 async:

var logoutResult = Task.Run(() => _instaApi.LogoutAsync()).GetAwaiter().GetResult();

变成:

var logoutResult = await _instaApi.LogoutAsync();