Question

我有以下ServiceStack Web服务

public class RetreadServices : Service
{
    public IRepository Repository { get; set; }

    //http://stackoverflow.com/questions/18902135/servicestack-no-server-side-async-support

    public async Task<ValidationResponse> Get(RetreadSsnInfoAsync request)
    {
        return await Task.Factory.StartNew(() =>
        {
            Tenant.SetTenant(request.TenantId);
            return new ValidationResponse { Result = Repository.CannotRetreadSsn(request.Ssn) };                    
        });
    }

    public async Task<ValidationResponse> Get(RetreadEmailInfoAsync request)
    {
        return await Task.Factory.StartNew(() =>
        {
            Tenant.SetTenant(request.TenantId);
            return new ValidationResponse { Result = Repository.CannotRetreadEmail(request.Emails) };
        });
    }

    //more methods constructed in the same fashion

}

然后在我的应用程序的一个单独的类中我有这个方法（基于来自@StephenCleary的代码）

private async Task<ValidationResponse[]> DoOperationsConcurrentlyAsync(string tenantId, string[] emails, string ssn)
{
    Task<ValidationResponse>[] tasks =
    {
        ResolveService<RetreadServices>().Get(new RetreadEmailInfoAsync { TenantId = tenantId, Emails = emails }), 
        ResolveService<RetreadServices>().Get(new RetreadSsnInfoAsync { TenantId = tenantId, Ssn = ssn })
    };

    await Task.WhenAll(tasks);

    ValidationResponse[] result = new ValidationResponse[tasks.Length];

    return result;
}

并像这样被调用

synchronous code...
synchronous code...

var result = DoOperationsConcurrentlyAsync(request.TenantId, request.Emails, request.Ssn);

synchronous code...
synchronous code...

所有这一切的目标是等待任务并行处理并希望减少我的全部时间......但是DoOperationsConcurrentlyAsync需要阻止，我该怎么做？然后作为奖励，如何使用async修饰符将返回值从各个任务传递回方法并退出方法？

Answer 1

我希望我能正确理解这一点。但是，您只是需要添加await关键字吗？

synchronous code...
synchronous code...

var result = await DoOperationsConcurrentlyAsync(request.TenantId, request.Emails, request.Ssn);

synchronous code...
synchronous code...

编辑：奖金问题：再次，我希望我能正确理解你......

但我会替换它：

await Task.WhenAll(tasks);

ValidationResponse[] result = new ValidationResponse[tasks.Length];

return result;

......简单地说：

return await Task.WhenAll(tasks);

这应该按你想要的方式工作。

如何在同步代码.NET 4.5中等待所有并发任务完成？

1 个答案: