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搜索到 4 篇与 MYSQL 的结果
2022-05-13

Mysql实现排行榜sql语句
Mysql实现排行榜sql语句
Mysql实现排行榜sql语句直接上代码-- 按余额查询排行榜 SELECT m.id, m.nickname, m.openid, m.avatar, @rownum := @rownum + 1 AS rownum FROM ( SELECT @rownum := 0 ) r, ( SELECT * FROM `tbl_member` WHERE credit> 0 ORDER BY credit DESC, created_at DESC ) AS m -- 查询某个用户的余额 排行名词 SELECT * FROM ( SELECT m.id, m.nickname, m.openid, m.avatar, @rownum := @rownum + 1 AS rownum FROM ( SELECT @rownum := 0 ) r, ( SELECT * FROM `tbl_member` WHERE credi...
2022-05-13

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2022年05月13日
2,174 阅读
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2021-09-03

Mysql5.7主从配置
Mysql5.7主从配置
环境说明搭建的环境1主1从：Mysql_1作为主服务器ip：192.168.254.129，Mysql_2作为从服务器ip：192.168.254.130首先准备好2台安装好Mysql5.7的服务器，安装过程可以查看这个链接：https://www.dcr163.cn/519.html这里我是使用的VMware Workstation 这个虚拟机软件，首先在centos7上安装好1台mysql5.7之后，这里命名虚拟机为Mysql_1，克隆1份Mysql_1虚拟机出来，命名为Mysql_2。依次打开这2台mysql服务器，因为是克隆出来的，Mysql_2的Mysql软件UUID是和第1台一样的，这里我们需要更改一下Mysql_2的UUID。使用SSH工具连接到Mysql_2服务器上，先连接进入mysql服务[root@mysql1 ~]# mysql -uroot –p查询出自己的UUIDmysql> SELECT UUID();上面可以看到，这个是本机的UUID，复制这一串UUID，我们退出mysqlmysql> quit修改Mysql_2的UUID进入到安装目录下...
2021-09-03

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2021年09月03日
2,179 阅读
0 评论
2021-09-01

centos7源码安装mysql5.7
centos7源码安装mysql5.7
Centos7 Mysql5.7源码安装下载Mysql5.7下载地址：https://downloads.mysql.com/archives/get/p/23/file/mysql-boost-5.7.34.tar.gz登陆进入服务器后输入下面命令，下载mysql5.7源码：wget https://downloads.mysql.com/archives/get/p/23/file/mysql-boost-5.7.34.tar.gz安装CMake编译Mysql5.7需要用到CMake，没有CMake的先下载安装，官网https://cmake.org/download/，输入：wget https://github.com/Kitware/CMake/releases/download/v3.21.1/cmake-3.21.1.tar.gz 下载好后，解压tar -zxvf cmake-3.21.1.tar.gz [root@localhost ~]# tar -zxvf cmake-3.21.1.tar.gz 进入cmake-3.21.1目录，配置：[root@localh...
2021-09-01

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2021年09月01日
2,227 阅读
0 评论
2021-01-26

两张表分组查询之后合并表，合并两张表
两张表分组查询之后合并表，合并两张表
两张表分组查询之后并且合并数据结果集，条件是两张表的查询结果字段需要一模一样，才能合并两张表-- 两张表分组查询之后 再合并结果 SELECT * FROM ((SELECT SUM(money) total_money,agentid user_id FROM dcr_declaration GROUP BY agentid) a1 ) UNION ALL SELECT * FROM ((SELECT SUM(agent_level_money) total_money,user_id FROM dcr_user_recommend_log GROUP BY user_id) a2 ); -- 两张表分组查询之后 再合并结果并且分组查询 SELECT SUM(total_money) total,user_id FROM ( SELECT * FROM ((SELECT SUM(money) total_money,agentid user_id FROM dcr_declaration GROUP BY agentid) a1 ) UNION ALL SELECT *...
2021-01-26

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2021年01月26日
3,752 阅读
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