在一天内的特定时间段内调用旋度

I have to execute a URL using curl and if the output contains a "hello" string, then I will exit successfully out of the shell script otherwise I keep retrying till 8 AM in the morning and then exit with an error message if it still doesn't contain that string.

我必须使用curl执行URL，如果输出包含一个“hello”字符串，那么我将成功地从shell脚本中退出，否则我将继续重试，直到早上8点，如果仍然不包含该字符串，则以错误消息退出。

I got below script but I am not able to understand how I can run while loop till 8 AM and if still curl output doesn't contain "hello" string?

我得到了下面的脚本，但是我无法理解如何在8点之前运行while循环，如果curl输出不包含“hello”字符串?

#!/bin/bash

while true
do
    curl -s -m 2  "some_url" 2>&1 | grep "hello"
    sleep 15m
done

So if it is after 3 PM then start making curl call until 8 AM and if it is successful with that curl call giving "hello" string, exit successfully otherwise after 8AM exit with error message.

如果是在下午3点以后，那么开始调用curl直到早上8点，如果这个curl调用成功地发出了“hello”字符串，那么在早上8点之后将成功退出，并发出错误消息。

And if it is before 3 PM then it will keep sleeping until it is passed 3 PM.

如果是在下午3点之前，它会一直睡到下午3点。

I have to add this logic within script and I can't use cron here.

我必须在脚本中添加这个逻辑，我不能在这里使用cron。

2 个解决方案

#1

You can use the script as follows, tested with GNU date

您可以使用下面的脚本，使用GNU date进行测试

#/bin/bash

retCode=0                                      # Initializing return code to of the piped commands
while [[ "$(date +"%T")" < '08:00:00' ]];      # loop from current time to next occurence of '08:00:00'
do
    curl -s -m 2  "some_url" 2>&1 | grep "hello" 
    retCode=$?                                 # Storing the return code
    [[ $retCode ]] && break                    # breaking the loop and exiting on success           
    sleep 15m                                  
done

[[ $retCode -eq 1 ]] && echo "String not found" >> /dev/stderr  # If the search string is not found till the last minute, print the error message

#2

I think you can use date +%k to retrieve current hour and compare with 8 AM and 13 PM. Code may like this

我认为您可以使用日期+%k来检索当前小时，并与上午8点和晚上13点进行比较。这样的代码可能

hour=`date +%k`
echo $hour
if [[ $hour -gt 15 || $hour -lt 8 ]]; then
    echo 'in ranage'
else
    echo 'out of range'
fi

#1