Question

我正在使用Azure Redis缓存进行开发，并想验证我处理异常的方式。根据最佳实践，有可能面对RedisConnectionExceptions并解决此问题，我们必须处置旧的ConnectionMultiplexer并创建一个新的ConnectionMultiplexer。如果abortConnect设置为false，那么多路复用器将在不引发错误的情况下静默重试连接。因此，如果引发异常，则只有在尝试重新连接后仍会失败。我对此的理解正确吗？这是我的连接字符串-

cachename.redis.cache.windows.net:6380,password=Password,ssl=True,abortConnect=False

我相信仅当您尝试在多路复用器上调用GetConnection（）时，连接异常才会发生。在下面找到我的代码-

    static Lazy<ConnectionMultiplexer> multiplexer = CreateMultiplexer();

    public static ConnectionMultiplexer GetConnection() => multiplexer.Value;

    private static Lazy<ConnectionMultiplexer> CreateMultiplexer()
    {
        return new Lazy<ConnectionMultiplexer>(() => ConnectionMultiplexer.Connect(connectionString));
    }

    private static void CloseMultiplexer(Lazy<ConnectionMultiplexer> oldMultiplexer)
    {
        if (oldMultiplexer != null)
        {
            oldMultiplexer.Value.Close();
        }
    }

    public static void Reconnect()
    {
        var oldMultiplexer = multiplexer;
        CloseMultiplexer(multiplexer);
        multiplexer = CreateMultiplexer();
    }

我正在下面的另一堂课中使用它-

    public class RedisCacheManager
    {
        private static IDatabase _cache;

        private TimeSpan expiry = new TimeSpan(hours: 6, minutes: 0, seconds: 0);

        public RedisCacheManager()
        {
            try
            {
                _cache = RedisCacheHelper.GetConnection().GetDatabase();
            }
            catch(RedisConnectionException)
            {
                RedisCacheHelper.Reconnect();
                new RedisCacheManager();
            }
        }

        public async Task<RedisValue[]> GetFromCacheAsync(List<string> keys)
        {
            var cacheValues = await _cache.StringGetAsync(keys.Select(k => (RedisKey)k).ToArray());
            return cacheValues;
        }

        public async Task SaveInCacheAsync<TValue>(Dictionary<string, TValue> kvps)
        {
            var tasks = new List<Task>();

            foreach(var kvp in kvps)
            {
                tasks.Add(_cache.StringSetAsync(kvp.Key, JsonConvert.SerializeObject(kvp), expiry));
            }
            await Task.WhenAll(tasks);
        }
        }

我不确定id在catch块中调用构造函数是否是一个好习惯。还有在调用StringGetAsync和StringSetAsync时需要处理的其他任何异常吗？

Answer 1

CacheManager如下所示：

using Newtonsoft.Json;
using StackExchange.Redis;
using System;
using System.Collections.Generic;
using System.Linq;
using System.Threading.Tasks;

public sealed class RedisCacheManager : IDisposable
{
    private readonly TimeSpan _expiry;

    private readonly Lazy<ConnectionMultiplexer> _lazyConnection;

    private ConnectionMultiplexer Connection { get => _lazyConnection.Value; }

    public RedisCacheManager(string connectionString, TimeSpan expiry)
    {
        _expiry = expiry;

        _lazyConnection = new Lazy<ConnectionMultiplexer>(() => ConnectionMultiplexer.Connect(connectionString));
    }

    public async Task<RedisValue[]> GetFromCacheAsync(IEnumerable<string> keys)
    {
        var cacheValues = await Connection.GetDatabase()
          .StringGetAsync(keys.Select(key => (RedisKey)key).ToArray()).ConfigureAwait(false);
        return cacheValues;
    }

    public async Task SaveInCacheAsync<TValue>(Dictionary<string, TValue> kvps)
    {
        var tasks = kvps
            .Select(kvp => Connection.GetDatabase().StringSetAsync(kvp.Key, JsonConvert.SerializeObject(kvp), _expiry))
            .ToArray();

        await Task.WhenAll(tasks).ConfigureAwait(false);
    }

    public void Dispose()
    {
        if (_lazyConnection.IsValueCreated)
        {
            _lazyConnection.Value.Dispose();
        }
    }
}

使用：

public readonly static RedisCacheManager RedisCacheManager = new RedisCacheManager("connection string", TimeSpan.FromHours(6));

备注：

旨在 abortConnect = false （这意味着即使未建立与Azure Redis缓存的连接，调用也会成功），并且不应从构造函数中抛出任何异常Redis例外
从GetDatabase返回的对象是cheap pass-thru object，不需要存储。
GetFromCacheAsync / SaveInCacheAsync 方法可以向外部抛出异常，这是可以的。您可以应用Retry-policy来解决瞬态故障。
如果您有任何IoC容器，则应使用单个实例范围（例如Autofac registration）创建 RedisCacheManager

处理Azure Redis缓存异常

1 个答案: