#!/bin/bash

mysql_c="mysql -uroot -p123456"

$mysql_c -e "show slave status\G" > /usr/local/mysql_s.log

n1=`wc -l /usr/local/mysql_s.log |awk '{print $1}'`

if [ $n1 -gt 0 ]
then
        y1=`grep 'Slave_IO_Running:' /usr/local/mysql_s.log |awk -F : '{print $2}' |sed 's/ //'`
        y2=`grep 'Slave_SQL_Running' /usr/local/mysql_s.log |awk -F : '{print $2}' |sed 's/ //'`

        if [ $y1 == "No" ] || [ $y2 == "No" ]
        then
                echo '有问题'
        else
                echo '没问题'
        fi
fi

 

相关文章:

  • 2021-04-17
  • 2021-06-30
  • 2022-01-09
  • 2021-06-30
  • 2021-08-18
  • 2022-12-23
  • 2021-10-13
猜你喜欢
  • 2022-12-23
  • 2020-01-09
  • 2021-08-31
  • 2021-08-08
  • 2022-12-23
  • 2022-12-23
相关资源
相似解决方案