lpush key value1 [value2] ……
rpush key value1 [value2] ……
lrange key start stop
lindex key index llen key
lpop key
rpop key
127.0.0.1:6379> lpush list1 one
(integer) 1
127.0.0.1:6379> lpush list1 two
(integer) 2
127.0.0.1:6379> lpush list1 three
(integer) 3
127.0.0.1:6379> lrange list1 0 2
1) "three"
2) "two"
3) "one"
127.0.0.1:6379> rpush list2 one two three
(integer) 3
127.0.0.1:6379> lrange list2 0 2
1) "one"
2) "two"
3) "three"
127.0.0.1:6379> lrange list2 0 -1 # 和python的列表一样
1) "one"
2) "two"
3) "three"
127.0.0.1:6379> lpush list2 0 # 将一个值或多个值,插入到列表头部
(integer) 4
127.0.0.1:6379> rpush list2 4 # 将一个值或多个值,插入到列表尾部
(integer) 5
127.0.0.1:6379> lrange list2 0 -1
1) "0"
2) "one"
3) "two"
4) "three"
5) "4"
127.0.0.1:6379> lindex list2 0 # 通过下标获取list中的某一个值
"one"
127.0.0.1:6379> lindex list2 2
"three"
127.0.0.1:6379> llen list1 # 返回列表的长度
(integer) 3
127.0.0.1:6379> lpop list2 # 移除列表的第一个元素
"0"
127.0.0.1:6379> rpop list2 # 移除列表的最后一个元素
"4"
127.0.0.1:6379> lrange list2 0 -1
1) "one"
2) "two"
3) "three"
blpop key1 [key2] timeout
brpop key1 [key2] timeout brpoplpush source destination timeout
127.0.0.1:6379> lpush list5 a b
(integer) 2
127.0.0.1:6379> lpop list5
"b"
127.0.0.1:6379> lpop list5
"a"
127.0.0.1:6379> lpop list5
(nil)
127.0.0.1:6379> blpop list5 10
(nil)
(10.10s)
127.0.0.1:6379> lpush list5 d
(integer) 1
127.0.0.1:6379> blpop list5 10
1) "list5"
2) "d"
(5.73s)
127.0.0.1:6379> rpush list5 e
(integer) 1
127.0.0.1:6379> lrange list5 0 -1
1) "e"
127.0.0.1:6379> brpop list5 5
1) "list5"
2) "e"
127.0.0.1:6379> lrange list5 0 -1
(empty array)
ltrim key start stop
127.0.0.1:6379> rpush mylist a b c d
(integer) 4
127.0.0.1:6379> lrange mylist 0 -1
1) "a"
2) "b"
3) "c"
4) "d"
127.0.0.1:6379> ltrim mylist 1 2 # 通过下标截取指定的长度 这个list已经被改变了
OK
127.0.0.1:6379> lrange mylist 0 -1 # 只剩下截取后的元素
1) "b"
2) "c"
rpoplpush key key2
127.0.0.1:6379> rpush mylist1 a b c d
(integer) 4
127.0.0.1:6379> lrange mylist1 0 -1
1) "a"
2) "b"
3) "c"
4) "d"
127.0.0.1:6379> rpoplpush mylist1 mylist2 # # 移除列表的最后一个元素,并将其移动到新的列表
"d"
127.0.0.1:6379> lrange mylist1 0 -1
1) "a"
2) "b"
3) "c"
127.0.0.1:6379> lrange mylist2 0 -1 # # 查看目标列表中,确实存在该值
1) "d"
lset key 下标 new value
127.0.0.1:6379> exists list # 判断这个列表是否存在
(integer) 0
127.0.0.1:6379> lset list 0 mao # 如果不存在列表 我们去更新会报错
(error) ERR no such key
127.0.0.1:6379> lpush list mao
(integer) 1
127.0.0.1:6379> lrange list 0 -1
1) "mao"
127.0.0.1:6379> lset list 0 zhu # 如果存在,更新当前下标的值
OK
127.0.0.1:6379> lrange list 0 -1
1) "zhu"
127.0.0.1:6379> lset list 1 gou # 如果不存在下标 我们去更新会报错
(error) ERR index out of range
insert key before/after value new-value
127.0.0.1:6379> rpush mylist 'hello'
(integer) 1
127.0.0.1:6379> rpush mylist 'world'
(integer) 2
127.0.0.1:6379> linsert mylist before 'world' 'niu'
(integer) 3
127.0.0.1:6379> lrange mylist 0 -1
1) "hello"
2) "niu"
3) "world"
127.0.0.1:6379> linsert mylist after 'world' 'zhu'
(integer) 4
127.0.0.1:6379> lrange mylist 0 -1
1) "hello"
2) "niu"
3) "world"
4) "zhu"
应用场景 微信朋友圈点赞,要求按照点赞顺序显示点赞好友信息,如果取消点赞,移除对应好友信息
解决方案
lrem key count value
127.0.0.1:6379> rpush friend a b c d e # 朋友圈点赞顺序 a b c d e
(integer) 5
127.0.0.1:6379> lrange friend 0 -1
1) "a"
2) "b"
3) "c"
4) "d"
5) "e"
127.0.0.1:6379> lrem friend 1 d # 移除d
(integer) 1
127.0.0.1:6379> lrange friend 0 -1
1) "a"
2) "b"
3) "c"
4) "e"
127.0.0.1:6379> rpush friend d
(integer) 5
127.0.0.1:6379> lrange friend 0 -1
1) "a"
2) "b"
3) "c"
4) "e"
5) "d"
127.0.0.1:6379> lpush letter a b a c d a o a u e a
(integer) 11
127.0.0.1:6379> lrange letter 0 -1
1) "a"
2) "e"
3) "u"
4) "a"
5) "o"
6) "a"
7) "d"
8) "c"
9) "a"
10) "b"
11) "a"
127.0.0.1:6379> lrem letter 3 a # 删除前三个a
(integer) 3
127.0.0.1:6379> lrange letter 0 -1
1) "e"
2) "u"
3) "o"
4) "d"
5) "c"
6) "a"
7) "b"
8) "a"
解决方案
我们使用三个redis客户端
第一台服务器产生日志
127.0.0.1:6379> rpush logs a1..
(integer) 1
127.0.0.1:6379> rpush logs a1...
(integer) 3
第二台服务器产生日志
127.0.0.1:6379> rpush logs b1..
(integer) 2
127.0.0.1:6379> rpush logs b1...
(integer) 5
第三台服务器产生日志
127.0.0.1:6379> rpush logs c1..
(integer) 4
127.0.0.1:6379> rpush logs c1...
(integer) 6
日志顺序
127.0.0.1:6379> lrange logs 0 -1
1) "a1.."
2) "b1.."
3) "a1..."
4) "c1.."
5) "b1..."
6) "c1..."
sadd key member1 [member2]
smembers key
srem key member1 [member2]
127.0.0.1:6379> sadd myset xiaotian # set集合中添加元素
(integer) 1
127.0.0.1:6379> sadd myset lovexiaotian
(integer) 1
127.0.0.1:6379> sadd myset maomao
(integer) 1
127.0.0.1:6379> smembers myset # 查看指定set的所有值
1) "maomao"
2) "xiaotian"
3) "lovexiaotian"
127.0.0.1:6379> srem myset maomao
(integer) 1
127.0.0.1:6379> smembers myset
1) "xiaotian"
2) "lovexiaotian"
scard key
sismember key member
127.0.0.1:6379> scard myset
(integer) 2
127.0.0.1:6379> sismember myset xiaotian # 判断某一个值是不是在set集合中
(integer) 1
127.0.0.1:6379> sismember myset zhuzhu
(integer) 0
业务场景 每位用户首次使用微博时会设置3项爱好的内容,但是后期为了增加用户的活跃度、兴趣点,必须让用户对其他信息类别逐渐产生兴趣,增加客户留存度,如何实现?
业务分析
解决方案
srandmember key [count]
spop key [count]
127.0.0.1:6379> sadd news youxi # set集合中添加元素
(integer) 1
127.0.0.1:6379> sadd news tiyu
(integer) 1
127.0.0.1:6379> sadd news yule
(integer) 1
127.0.0.1:6379> sadd news kexue
(integer) 1
127.0.0.1:6379> srandmember news 1 # 随机抽选出一个元素
1) "yule"
127.0.0.1:6379> srandmember news 1
1) "youxi"
127.0.0.1:6379> srandmember news 1
1) "kexue"
127.0.0.1:6379> scard news
(integer) 4
127.0.0.1:6379> srandmember news 3 # # 随机抽选出指定个数的元素
1) "kexue"
2) "tiyu"
3) "yule"
127.0.0.1:6379> srandmember news 3
1) "youxi"
2) "kexue"
3) "yule"
127.0.0.1:6379> spop news # 随机删除一些set合集中的元素
"kexue"
127.0.0.1:6379> smembers news
1) "youxi"
2) "tiyu"
3) "yule"
127.0.0.1:6379> spop news 2 # 随机指定个数的set合集中的元素
1) "tiyu"
2) "youxi"
127.0.0.1:6379> smembers news
1) "yule"
业务场景
解决方案
sinter key1 [key2]
sunion key1 [key2] sdiff key1 [key2]
sinterstore destination key1 [key2]
sunionstore destination key1 [key2] sdiffstore destination key1 [key2]
smove source destination member
127.0.0.1:6379> sadd user1 mao # 添加用户1的朋友
(integer) 1
127.0.0.1:6379> sadd user1 zhu
(integer) 1
127.0.0.1:6379> sadd user1 niu
(integer) 1
127.0.0.1:6379> sadd user2 zhu # 添加用户2的朋友
(integer) 1
127.0.0.1:6379> sadd user2 gou
(integer) 1
127.0.0.1:6379> sinter user1 user2 # 共同朋友 交集
1) "zhu"
127.0.0.1:6379> sunion user1 user2 # 并集
1) "mao"
2) "gou"
3) "niu"
4) "zhu"
127.0.0.1:6379> sdiff user1 user2 # 差集
1) "mao"
2) "niu"
127.0.0.1:6379> sdiff user2 user1 # 差集
1) "gou"
127.0.0.1:6379> sinterstore user3 user1 user2 # 将用户1和用户2的交集存入用户3
(integer) 1
127.0.0.1:6379> smembers user3
1) "zhu"
127.0.0.1:6379> sunionstore user4 user1 user2 # 将用户1和用户2的并集存入用户4
(integer) 4
127.0.0.1:6379> smembers user4
1) "mao"
2) "gou"
3) "niu"
4) "zhu"
127.0.0.1:6379> sdiffstore user5 user1 user2 # 将用户1和用户2的差集存入用户5
(integer) 2
127.0.0.1:6379> smembers user5
1) "mao"
2) "niu"
127.0.0.1:6379> smove user2 user1 gou # 将user2的元素移动到user1里
(integer) 1
127.0.0.1:6379> smembers user1
1) "mao"
2) "gou"
3) "niu"
4) "zhu"
127.0.0.1:6379> smembers user2
1) "zhu"
集团公司共具有12000名员工,内部OA系统中具有700多个角色,3000多个业务操作,23000多种数据,每位员工具有一个或多个角色,如何快速进行业务操作的权限校验?
解决方案
127.0.0.1:6379> sadd rid:001 getall getById # 添加角色权限
(integer) 2
127.0.0.1:6379> sadd rid:002 getCount getall insert
(integer) 3
127.0.0.1:6379> sunionstore uid:007 rid:001 rid:002 # 将uid为007的员工权限合并
(integer) 4
127.0.0.1:6379> smembers uid:007
1) "insert"
2) "getById"
3) "getCount"
4) "getall"
127.0.0.1:6379> sismember uid:007 insert
(integer) 1
解决方案
127.0.0.1:6379> sadd ips 192.168.188.10 # 一个独立ip访问
(integer) 1
127.0.0.1:6379> sadd ips 192.168.188.20 # 第二个独立ip访问
(integer) 1
127.0.0.1:6379> sadd ips 192.168.188.10 # 相同的ip访问则不增加
(integer) 0
127.0.0.1:6379> scard ips
(integer) 2
快速获取信息,个别特种行业网站信息通过爬虫获取分析后,可以转换成商业机密进行出售。例如第三方火车票、机票、酒店刷票代购软件,电商刷评论、刷好评。 + 同时爬虫带来的伪流量也会给经营者带来错觉,产生错误的决策,有效避免网站被爬虫反复爬取成为每个网站都要考虑的基本问题。在基于技术层面区分出爬虫用户后,需要将此类用户进行有效的屏蔽,这就是==黑名单==的典型应用。 + ps:不是说爬虫一定做摧毁性的工作,有些小型网站需要爬虫为其带来一些流量。
解决方案
添加数据
zadd key score1 member1 [score2 member2]
获取全部数据
zrange key start stop [WITHSCORES]
zrevrange key start stop [WITHSCORES]
zrem key member [member ...]
127.0.0.1:6379> zadd scores 100 maomao 60 xiaotian 94 zhuzhu 47 pancheng # 添加四个学生和成绩
(integer) 4
127.0.0.1:6379> zrange scores 0 -1 # 获取全部数据从小到大
1) "pancheng"
2) "xiaotian"
3) "zhuzhu"
4) "maomao"
127.0.0.1:6379> zrange scores 0 -1 withscores # withscores 可以获取姓名和成绩
1) "pancheng"
2) "47"
3) "xiaotian"
4) "60"
5) "zhuzhu"
6) "94"
7) "maomao"
8) "100"
127.0.0.1:6379> zrevrange scores 0 -1 withscores # 反转 从大到小排序
1) "maomao"
2) "100"
3) "zhuzhu"
4) "94"
5) "xiaotian"
6) "60"
7) "pancheng"
8) "47"
127.0.0.1:6379> zrem scores pancheng # 删除数据
(integer) 1
127.0.0.1:6379> zrevrange scores 0 -1 withscores
1) "maomao"
2) "100"
3) "zhuzhu"
4) "94"
5) "xiaotian"
6) "60"
zrangebyscore key min max [WITHSCORES] [LIMIT]
zrevrangebyscore key max min [WITHSCORES]
zremrangebyrank key start stop
zremrangebyscore key min max
127.0.0.1:6379> zrangebyscore scores 50 100 withscores # 从小到大
1) "xiaotian"
2) "60"
3) "pancheng"
4) "88"
5) "zhuzhu"
6) "94"
7) "maomao"
8) "100"
127.0.0.1:6379> zrevrangebyscore scores 100 50 withscores # 从大到小
1) "maomao"
2) "100"
3) "zhuzhu"
4) "94"
5) "pancheng"
6) "88"
7) "xiaotian"
8) "60"
127.0.0.1:6379> zrange scores 0 -1 withscores
1) "xiaotian"
2) "60"
3) "pancheng"
4) "88"
5) "zhuzhu"
6) "94"
7) "maomao"
8) "100"
127.0.0.1:6379> zremrangebyrank scores 0 1 # 根据下标删除
(integer) 2
127.0.0.1:6379> zrange scores 0 -1 withscores
1) "zhuzhu"
2) "94"
3) "maomao"
4) "100"
127.0.0.1:6379> zrange scores 0 -1 withscores
1) "xiaotian"
2) "66"
3) "pancheng"
4) "88"
5) "zhuzhu"
6) "94"
7) "maomao"
8) "100"
127.0.0.1:6379> zremrangebyscore scores 70 90 # 根据条件来删除
(integer) 1
127.0.0.1:6379> zrange scores 0 -1 withscores
1) "xiaotian"
2) "66"
3) "zhuzhu"
4) "94"
5) "maomao"
6) "100"
==注意:==
zcard key
zcount key min max
zinterstore destination numkeys key [key ...]
zunionstore destination numkeys key [key ...]
127.0.0.1:6379> zadd s1 50 aa 60 bb 70 cc
(integer) 3
127.0.0.1:6379> zadd s2 60 aa 40 bb 90 dd
(integer) 3
127.0.0.1:6379> zadd s3 70 aa 20 bb 100 dd
(integer) 3
127.0.0.1:6379> zinterstore ss 3 s1 s2 s3
(integer) 2
127.0.0.1:6379> zrange ss 0 -1 withscores
1) "bb"
2) "120"
3) "aa"
4) "180"
127.0.0.1:6379> zinterstore sss 3 s1 s2 s3 aggregate max # 求最大值
(integer) 2
127.0.0.1:6379> zrange sss 0 -1 withscores
1) "bb"
2) "60"
3) "aa"
4) "70"
业务场景
业务分析
解决方案
zrank key member
zrevrank key member
zscore key member
zincrby key increment member
127.0.0.1:6379> zadd music 800 suis 750 ikura 700 aca # 三个歌手的播放量
(integer) 3
127.0.0.1:6379> zrank music suis # 从小到大的排名
(integer) 2
127.0.0.1:6379> zrank music aca
(integer) 0
127.0.0.1:6379> zrank music ikura
(integer) 1
127.0.0.1:6379> zrevrank music suis # 从大到小查看排名
(integer) 0
127.0.0.1:6379> zscore music aca # 获取播放量
"700"
127.0.0.1:6379> zincrby music 10 aca # 增加
"710"
候要慎重
复==覆盖==,保留最后一次修改的结果
127.0.0.1:6379> zadd test1 11 aa
(integer) 1
127.0.0.1:6379> zrange test1 0 -1 withscores
1) "aa"
2) "11"
127.0.0.1:6379> zadd test1 22 aa
(integer) 0 # 返回失败 但是数值仍然改变
127.0.0.1:6379> zrange test1 0 -1 withscores
1) "aa"
2) "22"
127.0.0.1:6379> zadd test1 33 aa
(integer) 0
127.0.0.1:6379> zrange test1 0 -1 withscores
1) "aa"
2) "33"
VIP体验、云盘下载体验VIP、数据查看体验VIP。当VIP体验到期后,如果有效管理此类信息。即便对于正式VIP用户也存在对应的管理方式。
解决方案
对于基于时间线限定的任务处理,将处理时间记录为score值,利用排序功能区分处理的先后顺序
记录下一个要处理的时间,当到期后处理对应任务,移除redis中的记录,并记录下一个要处理的时间
当新任务加入时,判定并更新当前下一个要处理的任务时间
为提升sorted_set的性能,通常将任务根据特征存储成若干个sorted_set。例如1小时内,1天内,周内,月内,季内,年度等,操作时逐级提升,将即将操作的若干个任务纳入到1小时内处理的队列中
获取当前系统时间
time
127.0.0.1:6379> zadd ts 1509802345 uid:001 1509802390 uid:002 151038284 uid:003
(integer) 3
127.0.0.1:6379> zrange ts 0 -1 withscores
1) "uid:003"
2) "151038284"
3) "uid:001"
4) "1509802345"
5) "uid:002"
6) "1509802390"
127.0.0.1:6379> zrem ts uid:003
(integer) 1
127.0.0.1:6379> zrange ts 0 -1 withscores
1) "uid:001"
2) "1509802345"
3) "uid:002"
4) "1509802390"
127.0.0.1:6379> zadd ts 1510382841 uid:003
(integer) 1
127.0.0.1:6379> zrange ts 0 -1 withscores
1) "uid:001"
2) "1509802345"
3) "uid:002"
4) "1509802390"
5) "uid:003"
6) "1510382841"
127.0.0.1:6379> time # 时间戳
1) "1618647056"
2) "262066"
127.0.0.1:6379> time
1) "1618647057"
2) "779953"
127.0.0.1:6379> time
1) "1618647059"
2) "242339"
当任务或者消息待处理,形成了任务队列或消息队列时,对于高优先级的任务要保障对其优先处理,如何实现任务权重管理。
解决方案
127.0.0.1:6379> zadd order 102004 order:id:1
(integer) 1
127.0.0.1:6379> zadd order 101008 order:id:2
(integer) 1
127.0.0.1:6379> zrevrange order 0 -1 withscores # 订单优先级
1) "order:id:1"
2) "102004"
3) "order:id:2"
4) "101008"
127.0.0.1:6379> zrange order 0 -1
1) "order:id:2"
2) "order:id:1"
业务场景 人工智能领域的语义识别与自动对话将是未来服务业机器人应答呼叫体系中的重要技术,百度自研用户评价语义识别服务,免费开放给企业试用,同时训练百度自己的模型。现对试用用户的使用行为进行限速,限制每个用户每分钟最多发起10次调用
解决方案
到达10就进行限制
127.0.0.1:6379> get uid:00415
(nil)
127.0.0.1:6379> setex uid:00415 60 1
OK
127.0.0.1:6379> get uid:00415
"1"
127.0.0.1:6379> incr uid:00415
(integer) 2
127.0.0.1:6379> incr uid:00415
(integer) 3
127.0.0.1:6379> get uid:00415
"3"
127.0.0.1:6379> incrby uid:00415 7
(integer) 10
127.0.0.1:6379> get uid:00415
(nil)
解决方案改良
127.0.0.1:6379> get uid:00415
(nil)
127.0.0.1:6379> setex uid:00415 60 9223372036854775797
OK
127.0.0.1:6379> get uid:00415 # 设定一个最大值-10的值
"9223372036854775797"
127.0.0.1:6379> incr uid:00415
(integer) 9223372036854775798
127.0.0.1:6379> incrby uid:00415 9
(integer) 9223372036854775807
127.0.0.1:6379> incr uid:00415 # 再次增加 发现数值超范围了 然后告诉用户超次了
(error) ERR increment or decrement would overflow
业务场景 使用微信的过程中,当微信接收消息后,会默认将最近接收的消息置顶,当多个好友及关注的订阅号同时发送消息时,该排序会不停的进行交替。同时还可以将重要的会话设置为置顶。一旦用户离线后,再次打开微信时,消息该按照什么样的顺序显示?
解决方案
127.0.0.1:6379> lrem 100 1 200
(integer) 0
127.0.0.1:6379> lpush 100 200
(integer) 1
127.0.0.1:6379> lrem 100 1 300
(integer) 0
127.0.0.1:6379> lpush 100 300
(integer) 2
127.0.0.1:6379> lrem 100 1 400
(integer) 0
127.0.0.1:6379> lpush 100 400
(integer) 3
127.0.0.1:6379> lrem 100 1 200
(integer) 1
127.0.0.1:6379> lpush 100 200
(integer) 3
127.0.0.1:6379> lrem 100 1 300
(integer) 1
127.0.0.1:6379> lpush 100 300
(integer) 3
127.0.0.1:6379> lrange 100 0 -1 # 最后接收消息的顺序
1) "300"
2) "200"
3) "400"