使用 Shell 脚本每 5 秒运行一次 PHP 文件

Question

I want to run a php file every 5 seconds using Shell Script. But sometimes the script gets run every second or sometimes gets stop running. Do I need to use crontab also? Please help.

#!/bin/bash
while true; do
    begin=`date +%s`
    php /home/user/www/run.php
    end=`date +%s`
    if [ $(($end - $begin)) -lt 5 ]; then
        sleep $(($begin + 5 - $end))
    fi
done

Answer 1

The best thing would be to configure crontab to keep the execution or another program installed as a service.

The problem with contrab is that the minimum execution is every 1 minute. Therefore, you should create a script that executes every 5 seconds no more than 12 times. (12 x 5 seconds = 60 seconds)

Kill the process and re-run it with crontab.

Example

sript.sh

#!/bin/bash
# Do not run 12 times because this will same time as next crontab execution
for i in 1 2 3 4 5 6 7 8 9 11 
do 
    php /home/user/www/run.php
    sleep 5
done

On crontab

* * * * * /path/to/script.sh

Answer 2

Try this:

#!/bin/bash
while true; do
    sleep 5 &
    php /home/user/www/run.php &
    wait
done

wait (with no arguments) waits for all background jobs to complete, so the loop will block until the longer of the sleep and php processes completes.

---

Demonstrating the SECONDS variable

rand() { echo $(( RANDOM % $1 )); }

for i in {1..10}; do
    start=$SECONDS
    r=$(rand 10)
    echo "iteration $i, sleeping for $r seconds"
    sleep $r
    end=$SECONDS
    if ((end - start < 5)); then
        n=$((5 - (end - start)))
        echo "sleep for $n seconds"
        sleep $n
    fi
done

# or more simply by assigning to SECONDS
for i in {1..10}; do
    SECONDS=0
    r=$(rand 10)
    echo "iteration $i, sleeping for $r seconds"
    sleep $r
    duration=$SECONDS
    if ((duration < 5)); then
        n=$((5 - duration))
        echo "sleep for $n seconds"
        sleep $n
    fi
done

Certainly more complex than using wait