redis基本操作

> set sayhi ‘hello world‘
OK
> get sayhi
"hello world"

> del username
(integer) 1
> get username
(nil)
127.0.

#方法1
> set username heboan EX 10  #10秒后过期
OK


#方法2
> setex username 10 heboan
OK

#方法3
> set username heboan
OK
> expire username 10
(integer) 1

> lpush books python php
(integer) 2
> lpush books java
(integer) 3
> lrange books 0 -1  #查看此列表所有元素
1) "java"
2) "php"
3) "python"


#["java", "php", "pyton"]

> rpush books c++ golang
(integer) 5
> lrange books 0 -1
1) "java"
2) "php"
3) "python"
4) "c++"
5) "golang"

#["java", "php", "python", "c++", "golang"]

#移除并返回左边的第一个元素
> lpop books
"java"

#移除并返回右边的第一个元素
> rpop books
"golang"

127.0.0.1:6379> lrange books 0 -1
1) "php"
2) "python"
3) "c++"

> llen books
(integer) 3

> lindex books 0
"php"
> lindex books 1
"python"
> lindex books 2
"c++"
> lindex books 3 #超过返回空
(nil)

> lrange books 0 -1
1) "php"
2) "php"
3) "golang"
4) "python"
5) "c++"
6) "python"
7) "php"
8) "java"


#count > 0：从表头开始向表尾搜索，移除与value相等的元素，数量为count
#删除2个php
> lrem books 2 php
(integer) 2
> lrange books 0 -1
1) "golang"
2) "python"
3) "c++"
4) "python"
5) "php"
6) "java"

#count < 0：从表尾开始向表头搜索，移除与 value相等的元素，数量为count的绝对值
#删除1个python
> lrem books -1 python
(integer) 1
> lrange books 0 -1
1) "golang"
2) "python"
3) "c++"
4) "php"
5) "java"

#count = 0：移除表中所有与value 相等的值

> sadd school_1 qinghua beida
(integer) 2

> smembers school_1
1) "beida"
2) "qinghua"

> sadd school_1 fudan
(integer) 1
 smembers school_1
1) "beida"
2) "fudan"
3) "qinghua"

> srem school_1 fudan   #删除元素fudan
(integer) 1
> smembers school_1
1) "beida"
2) "qinghua"

> smembers school_1
1) "beida"
2) "qinghua"

> scard school_1
(integer) 2

#交集就是取出集合之间共同拥有的元素

> smembers school_1
1) "beida"
2) "qinghua"

> smembers school_2
1) "fudan"
2) "qinghua"
3) "wangyi"

> sinter school_1 school_2
1) "qinghua"

#并集就是取出集合的所有元素，重复的元素只保留1个，因为集合不能有重复的元素

> smembers school_1
1) "beida"
2) "qinghua"

> smembers school_2
1) "fudan"
2) "qinghua"

> sunion schoo_1 school_2
1) "fudan"
2) "wangyi"
3) "qinghua"

> smembers school_1
1) "beida"
2) "qinghua"
> smembers school_2
1) "fudan"
2) "qinghua"
3) "wangyi"

> sdiff school_1 school2  #把school_2中不含有schoo_1中的元素去掉
1) "beida"
2) "qinghua"

> sdiff school_2 school1 #把school_1中不含有schoo_2中的元素去掉
1) "fudan"
2) "wangyi"
3) "qinghua"