CODEWITHSUNDEEP
phpcron
emma
emma

Reputation: 759

Cron job could not open input file

I have a cron job on my hosting server that is supposed to execute a phpscript every 30 minutes. This php script is used to scrap a csv file and uptate that data into a database (i'm using MySQL for that). When executed manually that script works - it takes it about 45 seconds to finish the process.

Now when that cron job runs i'm getting this message: 

Could not open input file.

At first i thought that this might be because it takes over 30 seconds to execute that script so i decided to set the max execution time at the begging of the php script:

ini_set('max_execution_time', 300);

But again the same message came up. This is my cron job:

/usr/local/bin/php /home/emmaein/domain.com/folder/script.php?token=d8cn3j

P.S: that token get variable is sort of a password so it can't be executed by everybody so basically my php script looks like this:

if($_GET['token'] === 'd8cn3j'){
    //open csv
    //get data
    //update db
} else{
    exit('I see you >:)');
}

Upvotes: 1

Views: 2495

Answers (1)

Sean Bright
Sean Bright

Reputation: 120644

When PHP is not running inside a web server, you can't access $_GET variables (there's no such thing). Instead you should use command line arguments:

<?php
if ($argc > 1 && $argv[1] === 'd8cn3j') {
    // Do stuff
}

And then your crontab becomes:

/usr/local/bin/php /home/emmaein/domain.com/folder/script.php d8cn3j

The concept of $argc (the number of arguments) and $argv (an array of the arguments) is fairly standard among CLI programs, and is documented on PHP's website.

Upvotes: 5

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