09 2022 档案

lua脚本实现redis命令decrby减至最小为0
摘要：local current = redis.call('get', KEYS[1]); if current == false then return nil; end if tonumber(ARGV[1]) > tonumber(current) then return redis.call(' 阅读全文
posted @ 2022-09-19 18:09 Gyoung 阅读(354) 评论(0) 推荐(0)