统计多少个表
SELECT count(*) TABLES, table_schema FROM information_schema.TABLES WHERE table_schema = 'mk2' GROUP BY table_schema;
脚本如下:
#!/bin/bash
#数据库IP
dbserver='127.0.0.1'
#数据库用户名
dbuser='root'
#数据密码
dbpasswd='123456'
#数据库,如有多个库用空格分开
dbname='brdb_jtkg_0222 brdb_jtkg_0425'
#备份时间
backtime=`date +%Y%m%d-%H-%M`
#备份输出日志路径
logpath='/home/mysqlbackup/'
echo "################## ${backtime} #############################"
echo "开始备份"
#日志记录头部
echo "" >> ${logpath}/mysqlback.log
echo "-------------------------------------------------" >> ${logpath}/mysqlback.log
echo "备份时间为${backtime},备份数据库表 ${dbname} 开始" >> ${logpath}/mysqlback.log
#正式备份数据库
for table in $dbname; do
source=`mysqldump -h ${dbserver} -u ${dbuser} -p${dbpasswd} ${table} > ${logpath}/dump_${table}_${backtime}.sql` 2>> ${logpath}/mysqlback.log;
#备份成功以下操作
if [ "$?" == 0 ];then
cd ${logpath}
#删除七天前备份,也就是只保存7天内的备份
find $logpath -name "*.sql" -type f -mtime +7 -exec rm {} \; > ${logpath} 2>&1
echo "数据库表 ${dbname} 备份成功!!" >> ${logpath}/mysqlback.log
else
#备份失败则进行以下操作
echo "数据库表 ${dbname} 备份失败!!" >> ${logpath}/mysqlback.log
fi
done
echo "完成备份"
echo "################## ${backtime} #############################"
写到计划任务每天执行晚上11点59分备份数据库
# crontab -e
# crontab -l
[root@crm-mysql-master mysqlbak]# crontab -e
crontab: no changes made to crontab
[root@crm-mysql-master mysqlbak]# crontab -l
59 23 * * * /data/mysqlbak/mysqlbak.sh #分 时 天/几号 月 周 命令
参考:
解读“&1”、“&2”
https://www.cnblogs.com/liuchaogege/p/6124669.html