bash脚本，执行PHP文件每5秒 [英] Bash script that executes php file every 5 seconds

问题描述

Bash脚本：

  $猫test.sh
＃！/斌/庆典
而真正的
做
 /home/user/public_html/website.com/test.php
 睡眠5
DONE
 

我试着用cron来调用它：

  * * * * * /home/user/public_html/website.com/test.sh
 

哪里是我的错？不要紧，其中.SH所在？

编辑：
我固定从上面的所有问题。我也创造了这个cron作业prevent多个proccesses：

  * * * * *羊群-n /tmp/test.lock /home/user/public_html/website.com/test.sh
 

现在的问题是，在bash执行每5秒的PHP，这将创建类似的多个proccesses：

 用户12:31 /home/user/public_html/website.com/test.php
用户12:31 /home/user/public_html/website.com/test.php
用户12:31 /home/user/public_html/website.com/test.php
 

如何控制呢？我需要改变bash脚本，所以如果在执行PHP它可以检测。

这是正确的：

 ＃！/斌/庆典
而真正的
如果[$（的pidof test.php的）]
然后
回声暗战
其他
做
 /home/user/public_html/website.com/test.php＆安培;
 睡眠5
科幻
DONE
 

解决方案

这是另一种答案是运行你只有12次的脚本，并把它放在crontab中

 ＃！/斌/庆典
在1 2 3 4 5 6 7 8 9 0 1 2环;做
  /home/user/public_html/website.com/test.php＆安培;
  睡眠5
DONE
 

- 或 -

 ＃！/斌/庆典
循环= 0
而[$循环-lt 12]。做
  /home/user/public_html/website.com/test.php＆安培;
  睡眠5
  循环= $（（$循环+ 1））
DONE
 

- 更新 -

如果你的目标是不是有多个test.php的，使用这个脚本：（只运行一次，不要把它在crontab）：

 ＃！/斌/庆典而真实的;做
    开始=`日期+％s`
    /home/user/public_html/website.com/test.php
    结束=`日期+％s`
    如果[$（（$结束 -  $开头））-lt 5];然后
        睡眠$（（$开头+ 5  -  $结束））
    科幻
DONE
 

说明：这个脚本调用test.php的，并等待它终止（因为它剂量没有＆安培; 到底）。然后，它测量时间：如果它已经过去了5秒，就马上打电话test.php的。否则，休眠的剩余时间，使下一个test.php的将在从previous test.php的开头

开始5秒钟后称为

Bash script:

$ cat test.sh
#!/bin/bash
while true
do
 /home/user/public_html/website.com/test.php
 sleep 5
done

Im trying to call it with cron:

* * * * * /home/user/public_html/website.com/test.sh

Where is my mistake? Does it matter where the .sh is located?

EDIT: I fixed all the problems from above. I also created this cron job to prevent multiple proccesses:

* * * * * flock -n /tmp/test.lock /home/user/public_html/website.com/test.sh

The problem now is that the bash executes the php every 5 seconds and this creates multiple proccesses like:

user 12:31 /home/user/public_html/website.com/test.php
user 12:31 /home/user/public_html/website.com/test.php
user 12:31 /home/user/public_html/website.com/test.php

How to control this? I need to change the bash script so it can detect if the PHP is being executed.

Is this correct:

#!/bin/bash
while true
if [ "$(pidof test.php)" ]
then
echo "Running"
else
do
 /home/user/public_html/website.com/test.php &
 sleep 5
fi
done

解决方案

An alternative answer is to run you scripts for 12 times only, and put it in crontab

#!/bin/bash
for loop in 1 2 3 4 5 6 7 8 9 0 1 2; do
  /home/user/public_html/website.com/test.php &
  sleep 5
done

-- OR --

#!/bin/bash
loop=0
while [ $loop -lt 12 ]; do
  /home/user/public_html/website.com/test.php &
  sleep 5
  loop=$(($loop+1))
done

--update--

If your goal is not to have more than one test.php, use this script: (Run it once only, do not putting it in the crontab):

#!/bin/bash

while true; do
    begin=`date +%s`
    /home/user/public_html/website.com/test.php
    end=`date +%s`
    if [ $(($end - $begin)) -lt 5 ]; then
        sleep $(($begin + 5 - $end))
    fi
done

Explanation: This script calls test.php, and wait for it to terminate (since it dose not have & in the end). Then it measure the time: if it's already passed 5 seconds, it call test.php right away. otherwise, it sleeps for the remaining time so that next test.php will be called at the next 5 seconds starting from the beginning of the previous test.php

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