如何在执行Polly策略时包含返回语句？

Question

下面是我在Oracle应用程序中的代码，其中处理了与C#、FTP和Null引用异常的连接：

public Result Execute()
{
    Result result = null;
    string errorMessage = string.Empty;

    var retryTimes = 1000;
    var retryableErrorCodes = new[] { "ORA-03113", "ORA-03114", "ORA-12543",
        "ORA-12170", "ORA-12154", "ORA-12541", "ORA-12560", "ORA-03135",
        "Connection request timed out" };
    var retryExceptionError = new[] { "Object reference not set to an instance of an object" };

    RetryPolicy retryPolicyFTP = Policy
        .Handle<Xceed.Ftp.FtpInvalidStateException>().Or<Xceed.Ftp.FtpIOException>()
        .WaitAndRetry(retryTimes, _ => TimeSpan.FromSeconds(1));

    RetryPolicy retryPolicyOracle = Policy
        .Handle<OracleException>(ex => retryableErrorCodes.Any(errorCode => ex.Message.ToString().Contains(errorCode)))
        .WaitAndRetry(retryTimes, _ => TimeSpan.FromSeconds(1));

    RetryPolicy retryException = Policy
        .Handle<Exception>(ex => retryExceptionError.Any(errorCode => ex.Message.ToString().Contains(errorCode)))
        .WaitAndRetry(retryTimes, _ => TimeSpan.FromSeconds(1));

    Policy.Wrap(retryPolicyFTP, retryPolicyOracle, retryException).Execute(() =>
    {
        //few lines of C# Code like fetching details from Database
        if(some condition)
        {
            //Some Operations
            return new Result(ResultType.Failure, "This function has Failed");
        }
        if(some other condition)
        {
            //Some Operations
            return new Result(ResultType.Success, "This function is Successful");
        }
        //Some more lines of C# Code
    });
    return Result.Successful;
}

使用这段代码，我不能在函数的中间使用return关键字，因为Polly框架不允许它这样做。

请您建议在函数中间处理return关键字的更好方法是什么？

Answer 1

在Polly中，您可以为方法和函数定义装饰器。

对于方法，重试策略应该定义如下：

RetryPolicy retryPolicyFTP = Policy
    .Handle<Xceed.Ftp.FtpInvalidStateException>().Or<Xceed.Ftp.FtpIOException>()
    .WaitAndRetry(retryTimes, _ => TimeSpan.FromSeconds(1));

对于函数，应该像这样定义重试策略：

RetryPolicy<Result> retryPolicyFTP = Policy<Result>
    .Handle<Xceed.Ftp.FtpInvalidStateException>().Or<Xceed.Ftp.FtpIOException>()
    .WaitAndRetry(retryTimes, _ => TimeSpan.FromSeconds(1));

您应该在这里发现的<Result>部件在lhs和rhs以及。

有了这些知识，您的方法可以重写如下：

public Result Execute()
{
    Result result = null;
    string errorMessage = string.Empty;

    var retryTimes = 1000;
    var retryableErrorCodes = new[] { "ORA-03113", "ORA-03114", "ORA-12543", "ORA-12170", "ORA-12154", "ORA-12541", "ORA-12560", "ORA-03135", "Connection request timed out" };
    var retryExceptionError = new[] { "Object reference not set to an instance of an object" };

    RetryPolicy<Result> retryPolicyFTP = Policy<Result>
        .Handle<Xceed.Ftp.FtpInvalidStateException>().Or<Xceed.Ftp.FtpIOException>()
        .WaitAndRetry(retryTimes, _ => TimeSpan.FromSeconds(1));

    RetryPolicy<Result> retryPolicyOracle = Policy<Result>
        .Handle<OracleException>(ex => retryableErrorCodes.Any(errorCode => ex.Message.ToString().Contains(errorCode)))
        .WaitAndRetry(retryTimes, _ => TimeSpan.FromSeconds(1));

    RetryPolicy<Result> retryException = Policy<Result>
        .Handle<Exception>(ex => retryExceptionError.Any(errorCode => ex.Message.ToString().Contains(errorCode)))
        .WaitAndRetry(retryTimes, _ => TimeSpan.FromSeconds(1));

    Result res = Policy.Wrap(retryPolicyFTP, retryPolicyOracle, retryException).Execute(() =>
    {
        if (some condition)
        {
            return new Result(ResultType.Failure, "This function has Failed");
        }
        if (some other condition)
        {
            return new Result(ResultType.Success, "This function is Successful");
        }
        return Result.Successful;
    });
    return res;
}

因为您的Execute必须返回一个Result，所以Result.Successful可以移动到Execute块中。

我还建议将战略声明和执行分开如下：

public Result Execute()
{
    ...

    var strategy = Policy.Wrap(retryPolicyFTP, retryPolicyOracle, retryException)
    return strategy.Execute(() =>
    {
        if (some condition)
        {
            return new Result(ResultType.Failure, "This function has Failed");
        }
        if (some other condition)
        {
            return new Result(ResultType.Success, "This function is Successful");
        }
        return Result.Successful;
    });
}