Warning: mysql_result():

[ctcp]

<?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?

[floridaflatlander]

Did you use  or die("Error: ".mysqli_error($dbc)) ?

 

If I had to """guess""" one of you variables is bad.

[jcbones]

$liked_results = mysql_query($liked) or trigger_error(mysql_error());

 

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

[floridaflatlander]

  Quote

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

 

Thanks, I didn't notice that.

[PFMaBiSmAd]

The variable you put into the mysql_result() statement is not the variable you assigned the result resource from the mysql_query() statement to.

[ctcp]

  Quote

$liked_results = mysql_query($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

[PFMaBiSmAd]

You have a spelling error in your query.

[ctcp]

  Quote

You have a spelling error in your query.

 

i know this can you show me where is the error?

 

[Pikachu2000]

Read the error message. It's very apparent which word is misspelled.

[ctcp]

<?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 ?

[Pikachu2000]

You're passing the variable that holds the query string, rather than the result resource to mysql_result().