Redis Python - 如何根据特定模式删除所有键在python中，无需python迭代 [英] Redis Python - how to delete all keys according to a specific pattern In python, without python iterating

问题描述

我正在编写一个 django 管理命令来处理我们的一些 redis 缓存.基本上，我需要选择所有键，确认某个模式(例如:前缀:*")并删除它们.

I'm writing a django management command to handle some of our redis caching. Basically, I need to choose all keys, that confirm to a certain pattern (for example: "prefix:*") and delete them.

我知道我可以使用 cli 来做到这一点:

I know I can use the cli to do that:

redis-cli KEYS "prefix:*" | xargs redis-cli DEL

但我需要在应用程序中执行此操作.所以我需要使用 python 绑定(我使用的是 py-redis).我曾尝试将列表输入到删除中，但失败了:

But I need to do this from within the app. So I need to use the python binding (I'm using py-redis). I have tried feeding a list into delete, but it fails:

from common.redis_client import get_redis_client
cache = get_redis_client()
x = cache.keys('prefix:*') 

x == ['prefix:key1','prefix:key2'] # True

# 现在

cache.delete(x) 

# 返回 0 .什么都没有删除

# returns 0 . nothing is deleted

我知道我可以遍历 x:

I know I can iterate over x:

for key in x:
   cache.delete(key)

但这将失去 redis 惊人的速度并滥用其功能.py-redis 是否有 pythonic 解决方案，无需迭代和/或 cli?

But that would be losing redis awesome speed and misusing its capabilities. Is there a pythonic solution with py-redis, without iteration and/or the cli?

谢谢！

推荐答案

我认为

 for key in x: cache.delete(key)

非常好且简洁.delete 真的要一次一个键，所以你必须循环.

is pretty good and concise. delete really wants one key at a time, so you have to loop.

否则，这个之前的问答 为您指明了一个基于 lua 的解决方案.

Otherwise, this previous question and answer points you to a lua-based solution.

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