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<?php 				$liked = ("SELETCT SUM(user_id) as totaluser_id FROM liked WHERE `user_id`='26762'");
			$liked_results = mysql_query($liked);
			$sum_liked = mysql_result($liked, 0); // <--- Error 
			echo $sum_liked;

?>

 

i got this error

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in

 

can sombady help?

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https://forums.phpfreaks.com/topic/251474-warning-mysql_result/
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$liked_results = mysql_query($liked) or trigger_error(mysql_error());

 

Do NOT use mysqli_error(), as you cannot mix mysql and mysqli.

 

Notice: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELETCT SUM(user_id) as totaluser_id FROM liked WHERE user_id = 26762' at line 1 in

<?php 
			$liked = ("SELECT SUM(user_id) as totaluser_id FROM liked WHERE `user_id`='26762'");
			$liked_results = mysql_query($liked);
			$sum_liked = mysql_result($liked, 0); // <-- line 15 
			echo $sum_liked;
			$liked_results = mysql_query($liked) or trigger_error(mysql_error());
?>

mate i got this error now 
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in mpla mpla on line 15

what is wrong now ?



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