PHP code warning SQL

Question

I'm trying to run a php file and I get two warnings, which I can't seem to place. The warnings:

PHP Warning: Directive 'safe_mode' is deprecated in PHP 5.3 and greater in Unknown on line 0
PHP Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /etc/baseconetrial2.php on line 19

The code:

<?php


//connectie database


mysql_connect('localhost','heregoestheusername','heregoesthepass');

@mysql_select_db('admin_subscriptions') or die( "Unable to select database");


$type = "proefabonnement";


$checksql = "SELECT * FROM subscriptions where (type_abonnement = '$type' AND  DATE(timestamp) = DATE_SUB( CURDATE( ) , INTERVAL 26))";

$checkresult = mysql_query($checksql);

while ($check = mysql_fetch_array($checkresult)) {





$mail_ontv = "[email protected]"; 

$_POST['onderwerp'] = "Verlopen Trial Account";

    // set datum 
    $datum = date("d.m.Y H:i"); 

    // set ip 
    $ip = $_SERVER['REMOTE_ADDR']; 

    $inhoud_mail .= $_SERVER['SCRIPT_URI'] . "\n\n";

    $inhoud_mail .= "Binnenkort verloopt er een trail account!\n\n\n";  
    $inhoud_mail .= "Bedrijfsnaam: " . $check['bedrijfsnaam'] . "\n\n"; 
    $inhoud_mail .= "Telefoonnummer: " . $check['telefoonnummer'] . "\n\n"; 
    $inhoud_mail .= "E-mail adres: " . $check['email'] . "\n\n"; 
    $inhoud_mail .= "Telefoonnummer contactpersoon: " . $check['telefoonnummercontact'] . "\n\n"; 
    $inhoud_mail .= "E-mail adres contactpersoon: " . $check['emailcontact'] . "\n\n"; 
    $inhoud_mail .= "Hieronder de link voor de klant:\n\n\n";
    $inhoud_mail .= "http://www.basecone.nl/upgrade1?key=".$check['unique']."\n\n"; 

    $inhoud_mail .= "Verstuurd op " . $datum . " via het ip " . $ip . "\n\n"; 

    $headers = "From: BaseconeWizard < [email protected] >";

    $headers = stripslashes($headers);
    $headers = str_replace("\n", "", $headers); // Verwijder \n 
    $headers = str_replace("\r", "", $headers); // Verwijder \r 
    $headers = str_replace("\"", "\\\"", str_replace("\\", "\\\\", $headers)); // Slashes van quotes 



    $_POST['onderwerp'] = str_replace("\n", "", $_POST['onderwerp']); // Verwijder \n 
    $_POST['onderwerp'] = str_replace("\r", "", $_POST['onderwerp']); // Verwijder \r 
    $_POST['onderwerp'] = str_replace("\"", "\\\"", str_replace("\\", "\\\\", $_POST['onderwerp'])); // Slashes van quotes 

    mail($mail_ontv, $_POST['onderwerp'], $inhoud_mail, $headers); 







}


?>

Thanks guys!

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — j0k
– j0k, Commented Mar 14, 2013 at 18:11

Martina · Accepted Answer · 2013-03-14 18:18:39Z

3

PHP Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /etc/baseconetrial2.php on line 19

$checksql = "SELECT * FROM subscriptions where (type_abonnement = '$type' AND  DATE(timestamp) = DATE_SUB( CURDATE( ) , INTERVAL 26 DAY))";

note the DAY parametre of DATE_SUB mysql function

I would also suggest to use mysql_real_escape_string() function to sanitize the "$type" variable

answered Mar 14, 2013 at 18:18

Martina

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5 Comments

user1555076 Over a year ago

I do understand your real_escape remark, though it's not clear to me what you mean by the other? 'note the DAY parametre of DATE_SUB mysql function'?

Martina Over a year ago

Hi hd1 i believe that what i wrote above is an answer to the problem (the fetch_array warning). There is an error in the mysql query causing mysql_query return "false" instead of resource. Fixing that will get rid of the warning.

Martina Over a year ago

Hi, Your SQL command goes like: .... DATE_SUB( CURDATE( ) , INTERVAL 26)) DATE_SUB function requires one more parametre - to say what the 26 is OF - eg. DAY, HOUR, MINUTE ...

user1555076 Over a year ago

Seriously? I didn't know that, do I place it before or after the number? DAY 26, or 26 DAY? Thanks!

user1555076 Over a year ago

Never mind, some Google-ing gave me the answer! And you put it in your own code I see. Thanks a lot!

Kermit · Accepted Answer · 2013-03-14 18:16:45Z

2

Some part of your code or your host is using safe_mode which has been deprecated.

The PHP safe mode is an attempt to solve the shared-server security problem. It is architecturally incorrect to try to solve this problem at the PHP level, but since the alternatives at the web server and OS levels aren't very realistic, many people, especially ISP's, use safe mode for now.

Please do not use @ to suppress errors. Use error handlers or enable full error reporting.

mysql_fetch_array() expects parameter 1 to be resource

Your query failed and is thus passing a boolean result to mysql_fetch_array. But since you're not catching errors, you won't know.

$checkresult = mysql_query($checksql) or die(mysql_error());

answered Mar 14, 2013 at 18:16

Kermit

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3 Comments

user1555076 Over a year ago

I deleted the '@' and added the last line of code you suggested. Which part of my code is using the safe_mode so I can alter it?

Kermit Over a year ago

@user1555076 Where is your PHP instance? On a host or locally? If local, you can edit your php.ini.

user1555076 Over a year ago

It's locally, I'll go and search for the PHP.ini. Thanks!

Muhammad Haseeb Khan · Accepted Answer · 2013-03-14 18:30:46Z

2

for

PHP Warning: Directive 'safe_mode' is deprecated in PHP 5.3 and greater in Unknown on line 0

Remove or comment

safe_mode = "On/Off" from your php.ini

For PHP Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /etc/baseconetrial2.php on line 19

Please Use

if (!$checksql) die(mysql_error());

it will show the problem in query

edited Mar 14, 2013 at 18:30

answered Mar 14, 2013 at 18:17

Muhammad Haseeb Khan

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1 Comment

user1555076 Over a year ago

Thanks, going to give this a try!

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