使用Polly从异步函数重试

Question

return await Policy
           .Handle<CustomException>()
           .RetryAsync(3, onRetryAsync: async (exception, retryCount, context) =>
           {
               return await runner.run(params);
           });

Task<IReadOnlyCollection<string>>

转换为任务返回委托的异步lambda表达式不能返回值

Answer 1

我认为在重试政策中运行你的逻辑是不寻常的--除非我误解了你的问题。更典型的是，通过调用运行逻辑的方法来执行策略。

类似于这样：

async Task Main()
{
    var polly = Policy
           .Handle<Exception>()        
           .RetryAsync(3, (exception, retryCount, context) => Console.WriteLine($"try: {retryCount}, Exception: {exception.Message}"));

    var result = await polly.ExecuteAsync(async () => await DoSomething());
    Console.WriteLine(result);
}

int count = 0;

public async Task<string> DoSomething()
{
    if (count < 3)
    {
        count++;
        throw new Exception("boom");
    }

    return await Task.FromResult("foo");
}

输出

try: 1, Exception: boom
try: 2, Exception: boom
try: 3, Exception: boom
foo