当前位置:网站首页>Redis publish subscribe command line implementation
Redis publish subscribe command line implementation
2022-07-05 06:07:00 【A light wind and light clouds】
- Open a client subscription channel1
SUBSCRIBE channel1

2、 Open another client , to channel1 Release the news hello
publish channel1 hello

Back to 1 Is the number of subscribers
Open the first client to see the sent message

notes : Published messages are not persistent , If the subscription client does not receive hello, Only messages published after subscription can be received
边栏推荐
- 网络工程师考核的一些常见的问题:WLAN、BGP、交换机
- 2022年贵州省职业院校技能大赛中职组网络安全赛项规程
- [rust notes] 14 set (Part 2)
- The connection and solution between the shortest Hamilton path and the traveling salesman problem
- Sword finger offer 05 Replace spaces
- How many checks does kubedm series-01-preflight have
- AtCoder Grand Contest 013 E - Placing Squares
- Simply sort out the types of sockets
- 快速使用Amazon MemoryDB并构建你专属的Redis内存数据库
- 个人开发的渗透测试工具Satania v1.2更新
猜你喜欢

可变电阻器概述——结构、工作和不同应用

【实战技能】非技术背景经理的技术管理

CCPC Weihai 2021m eight hundred and ten thousand nine hundred and seventy-five
![[jailhouse article] performance measurements for hypervisors on embedded ARM processors](/img/c0/4843f887f77b80e3b2329e12d28987.png)
[jailhouse article] performance measurements for hypervisors on embedded ARM processors

Implement an iterative stack

API related to TCP connection

Navicat连接Oracle数据库报错ORA-28547或ORA-03135

LeetCode 0108.将有序数组转换为二叉搜索树 - 数组中值为根,中值左右分别为左右子树

个人开发的渗透测试工具Satania v1.2更新

Overview of variable resistors - structure, operation and different applications
随机推荐
884. Uncommon words in two sentences
How to adjust bugs in general projects ----- take you through the whole process by hand
2020ccpc Qinhuangdao J - Kingdom's power
Brief introduction to tcp/ip protocol stack
对for(var i = 0;i < 5;i++) {setTimeout(() => console.log(i),1000)}的深入分析
Fried chicken nuggets and fifa22
“磐云杯”中职网络安全技能大赛A模块新题
【实战技能】如何做好技术培训?
LeetCode 0108.将有序数组转换为二叉搜索树 - 数组中值为根,中值左右分别为左右子树
927. Trisection simulation
【Rust 笔记】16-输入与输出(下)
EOJ 2021.10 E. XOR tree
Typical use cases for knapsacks, queues, and stacks
多屏电脑截屏会把多屏连着截下来,而不是只截当前屏
Appium基础 — 使用Appium的第一个Demo
leetcode-556:下一个更大元素 III
Full Permutation Code (recursive writing)
Common optimization methods
Implement an iterative stack
Codeforces Round #732 (Div. 2) D. AquaMoon and Chess