通过cron运行php脚本。 如何让它报告失败？

Question

I have a PHP script that I want to run on a cronjob. I am assuming if PHP fails due to a fatal error it will terminate the script which reports to cron of an error, however how do I force a failure reply that will instruct cron of an issue?

I will have some database calls and I want cron to report back an error (which fires off an email) if the database queries fail.

Answer 1

Here is what I do:

$query1 = "some query";
$results = mysql_query($query1);
$query1Error = mysql_error();

$query2 = "a different query";
$results = mysql_query($query2);
$query2Error = mysql_error();

if(!empty($query1Error) || !empty($query2Error) ) {
        $message = "An error occured updating the funnel. The details are below:"."

";
        $message = print_r($_POST,TRUE);
        $message .= "

Query 1: ".$query1;
        $message.="

 Query 1 Error: ".$query1Error; 
        $message .= "

Query 2: ".$query2;
        $message.="

 Query 2 Error: ".$query2Error; 
        mail("[e-mail]","Error updating funnel ".__FILE__." ",$message);  
   }

Of course, when i get around to it, I will replace the depreciated MySQL stuff.