linux - Shell脚本+时间依赖

Question

我想为以下内容编写一个 shell 脚本:我想检查服务是否正在运行，如果正在运行则退出 1，否则在 5 分钟未运行后退出 -1。像这样的东西:

while(for 5 minutes) {
if service running, exit 1
}
exit -1   //service is not running even after 5 minutes, so exit -1.

我可以检查服务是否正在运行的情况，但无法添加时间限制部分。

这是我的尝试

if (( $(ps -ef | grep -v grep | grep tomcat7 | wc -l) > 0 )); 
then
echo "running"
else
echo "NOT running"
fi

Answer 1

你应该使用bash sleep命令。 man 页面的摘录:-

您可以在脚本中提供 sleep 5m 以等待 5 分钟并执行操作。

NAME
       sleep - delay for a specified amount of time

DESCRIPTION
       Pause  for  NUMBER  seconds.   SUFFIX  may be 's' for seconds (the default), 'm' for minutes, 'h' for hours or 'd' for days.  Unlike most implementations that require NUMBER be an
       integer, here NUMBER may be an arbitrary floating point number.  Given two or more arguments, pause for the amount of time specified by the sum of their values.

您的解决方案的正确方法是:-

#!/bin/bash
maxAttempts=0          
maxCounter=2     # Number of attempts can be controlled by this variable    

 while [ "$maxAttempts" -lt "$maxCounter" ]; do
    if ps -ef | grep -v grep | grep "tomcat7" > /dev/null
    then
        echo "tomcat7 service running, "
        exit 1
    else
        maxAttempts=$((maxAttempts+1))
        sleep 5m                           # The script waits for 5 minutes before exiting with error code '-1'
    fi
done

exit -1

if 条件在 ps -ef | grep -v grep | grep "tomcat7" 返回条件通过的命令成功错误代码。 >/dev/null 将所有标准 outpur(stdout, stderr) 抑制到 /dev/null 以便我们只能使用提供的命令的退出代码。