I have 50+ cronjobs like the one given below running in my Centos 7 server.
curl -s https://url.com/file.php
This runs every 10 minutes. When running manually from the shell it only takes 1-2 minutes. It also is working fine using cronjob. The problem is that it does not exit after the execution. When i check my processes using ps command, it shows many cronjobs of previous dates(even 10 days before) which accumulates the total proccesses in my server.
Line in crontab :-
*/10 * * * * user curl -s https://url.com/file.php > /dev/null 2>&1
Is there any reasion for this? If i rememmber correctly this happened after latest patch update.
Please help.
Modify your command to store the logs in log files instead of dumping it to /dev/null.
Options
--max-time
--connect-timeout
--retry
--retry-max-time
can be used to control the curl command behaviour.
I need to run a curl request to locahost at least once in every 30 mins. So the command will be curl http://localhost:8080.
The catch here is, I want to select a time randomly between 5min - 30 min and then execute the curl command. The pseudo-code might look like this
while(true)
n = random number between 5-30
run curl http://localhost:8080 after 'n' minutes
A detailed answer would be nice since I don't have much knowledge about linux.
while true; do
sleep $(((RANDOM%25+5)*60))
curl http://localhost:8080
done
If you run above script, you must run as background process and make sure it will not be killed by something (OS, other users,...)
Another way is use cronjob to trigger automatically but the script is more complex.
Cronjob setting:
* * * * * bash test_curl.sh >> log_file.log
Shell script test_curl.sh:
#!/bin/bash
# Declare some variable
EXECUTE_TIME_FILE_PATH="./execute_time"
# load expected execute time
EXPECTED_EXECUTE_TIME=$(cat $EXECUTE_TIME_FILE_PATH)
echo "Start at $(date)"
# calculate current time and compare with expected execute time
CURRENT_MINUTE_OF_TIME=$(date +'%M')
if [[ "$EXPECTED_EXECUTE_TIME" == "$CURRENT_MINUTE_OF_TIME" ]];
then
curl http://localhost:8080
# Random new time from 5 -> 30
NEXT_RANDOM=$((RANDOM%25+5))
# Get current time
CURRENT_TIME=$(date +'%H:%M')
# Calculate next expected execute time = Current Time + Next Random
NEXT_EXPECTED_EXECUTE_TIME=$(date -d "$CURRENT_TIME $NEXT_RANDOM minutes" +'%M')
# Save to file
echo -n $NEXT_EXPECTED_EXECUTE_TIME > $EXECUTE_TIME_FILE_PATH
echo "Next Executed time is $(date -d "$CURRENT_TIME $NEXT_RANDOM minutes" +'%H:%M')"
else
echo "This $(date +'%H:%M') is not expected time to run test"
fi
echo "End at $(date)"
I commented out in line so you can read it easily.
**
Update: Importance: file execute_time must has initial value. For
example, the current minute of first time you execute.
**
I need to setup simple cronejob to ping my website url every 2 minutes, what i did for now make no effect, can somebody help me write simple ping cronjob?
Here is what i have for now
2 * * * * /usr/bin/wget -O - -q -t 1 http://example.com/api/get.php?refresh=yes
This might help:
*/2 * * * * wget -q -O - "http://example.com/api/get.php?refresh=yes" >/dev/null 2>&1
Every 2 minutes (15 in the article), ping the specified file without
any output whatsoever. Just a silent ping to execute my script,
nothing more, nothing less.
Source: https://wp-mix.com/cron-notes-ping-external-url-with-no-output/
I am trying to set up Recurring Invoices as documented on http://www.fusioninvoice.com/support/recurring-invoices but I don't understand and cant work out what to do.
I have found a alphanumeric 'key' but not sure where or what to do?
Do I need to createa file or something?
If curl is available on your system, you can schedule it to ping the URL.
Use crontab -e to edit the cron table and create a new entry. For example, this will ping the URL every day at midnight.
0 0 * * * /usr/bin/curl --silent http://your-server.com/index.php/invoices/cron/recur/your-cron-key-here &>/dev/null
I want to create cron job that runs a script every 5 seconds. Seeing that cron jobs only allows increments of minutes 0-59 and so on.
I thought to create another script that calls my original script written below.
#!/bin/bash
while true
do
# script in the same directory as this script. is this correct?
bash makemehappy.sh
sleep 1
done
I now, need to know how to run this script every time i boot my computer and for it to start itself if it isn't running for some reason.
I am also aware that running this script every minute wouldn't be a good thing. :)
if there is an easier way to run a script every 5 seconds please advise.
Please and thank you.
I wouldn't use cron for this. I would use that bash script (use an absolute path, unless you want it to be portable and know that the directory structure will be preserved).
Instead, I would just sleep 5, just like you did (only 5 seconds instead of 1).
As far as starting it with your system, that depends on the system. On (some) Linux distros, there's a file called /etc/rc.local in which you can add scripts to run when the system starts. Well... I shouldn't be so general, the distros that I have used have this. If you're running Ubuntu, there is no longer an inittab, they use upstart, btw.
So if you have an endless loop and an entry in /etc/rc.local, then you should be golden for it to run endlessly (or until it encounters a problem and exits).
Try using anacron or, better yet, an init script to start when the computer starts.
If you want the script to "restart itself", you'll need to run something every few minutes to check the original is still running. This can be done in inittab (/etc/inittab) or, on Ubuntu, /etc/event.d. Try man 5 inittab, looking at the 'respawn' option.
Some crons have an #reboot time specifier (this covers all the time and date fields). If yours does, you can use that. If this is a "system" service (rather than something running for yourself), the other solutions here are probably better.
Init scripts are fine at boot, but don't detect if a process fails and has to be restarted. supervisord does a great job of detecting failed processes and restarting them. I'd recommend a script with a 5-second loop like #Tim described, but wrap supervisord around it to make sure it keeps running.
As explained in detail in my answer to a similar question, you can use SystemD timer units with whatever schedule that you want - down to a theoretical 1 nanosecond schedule with no sleep kludges
Quick overview:
Setup a SystemD service to run what you want - this can be as simple as:
/home/myusuf3/.config/systemd/user/makemehappy.service
[Unit]
Description=Make me happy
[Service]
ExecStart=/home/myusuf3/.local/bin/makemehappy.sh
Setup a SystemD timer with the schedule that you want, as documented in man systemd.timer:
/home/myusuf3/.config/systemd/user/makemehappy.timer
[Unit]
Description=Schedule to make me happy
[Timer]
OnBootSec=5
OnUnitActiveSec=5
AccuracySec=1
Enable and start the timer:
:
systemctl --user daemon-reload
systemctl --user enable makemehappy.timer
systemctl --user start makemehappy.timer
(after you enable it, it will autostart every time you start the computer, but you probably want to start it now anyway).
To answer the question in the title, this is how to run a cronjob every 5 seconds :
* * * * * /path/to/script.sh
* * * * * sleep 5 && /path/to/script.sh
* * * * * sleep 10 && /path/to/script.sh
* * * * * sleep 15 && /path/to/script.sh
* * * * * sleep 20 && /path/to/script.sh
* * * * * sleep 25 && /path/to/script.sh
* * * * * sleep 30 && /path/to/script.sh
* * * * * sleep 35 && /path/to/script.sh
* * * * * sleep 40 && /path/to/script.sh
* * * * * sleep 45 && /path/to/script.sh
* * * * * sleep 50 && /path/to/script.sh
* * * * * sleep 55 && /path/to/script.sh
It's not beautiful, but this solution comes with no extra tools nor dependencies. So if you have working cron jobs, this should work right away.