从 crontab 执行 php 脚本的正确方法

Question

I have a .php script that includes other php scripts and as final result a db is updated and a report file has been created.

if I manually call it from a browser all works fine.

I tried to call it from the crontab with the following syntax

/usr/local/bin/php /home/PATH/myscript.php

and at the planned time, it seems to stop at the first include(); and die. I added a mail(); sending after each include();

If I call the file from the browser I receive all the 5 mails, when it's called from the crontab, I receive only the first email

Where am I doing wrong?

*edit: I tried to call always the first script and I received all the mail. It seems the second script contains something that is stopping the code but ONLY is it's called by the cron

Answer 1

Use this instead

php -q /home/PATH/myscript.php

As per this answer by St. Woland , the -q flag:

The -q flag suppresses HTTP header output.

Answer 2

You can run a php script like this:

php -f script.php

If you want to use another php.ini (what often happens with console apps), use

php -c path/where/php.ini -f script.php

type

php --help

for more options.