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returning no results from a SELECT WHERE ORDER by LIMIT mysql query


Go to solution Solved by Barand,

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i am trying to getting the results from mysql database 'userdatabase' where the user_type column match and is_approved field is 1.

 

i am trying to get the top 5 rows with the highest average_ratings.

 

here is my code:

$mysqli = new mysqli("localhost", "levelDbUser", "honey", "levelDb");
$outpostlevel = (int) $array[user_type];
    $sql = "SELECT id, image_type, total_ratings, average_rating FROM userdatabase WHERE user_type = ? && is_approved = 1 ORDER by average_rating ASC LIMIT 5";
    $statement = $mysqli->prepare($sql);
    $statement->bind_param("i", $outpostlevel);
    $statement->execute();
	$statement->bind_result($id, $image_type, $total_ratings, $average_rating);
    $statement->store_result();
    while ($statement->fetch()) {
	$topoutposts = array(
		'id' 			  => $id,
		'image_type'	  => $image_type,
	    'total_ratings'	  => $total_ratings,
		'average_rating'  => $average_rating
	    );
    }
print_r ($topoutposts);	 

but the print_r array is returning nothing (blank)

Edited by james909

it is only returning the first  mysql row in the results,

like this:

Array ( [id] => 1 [image_type] => jpg [total_ratings] => 17 [average_rating] => 8.6 )

 

 

 

how do i get the results to be a multi-dimensional array

like this:

Array ( [1] => Array(  [id] => 1 [image_type] => jpg [total_ratings] => 17 [average_rating] => 8.6 )  

[2] => Array(  [id] => 2 [image_type] => jpeg [total_ratings] => 10 [average_rating] => 7.2 )

[3] => Array(  [id] => 3 [image_type] => png [total_ratings] => 8 [average_rating] => 9.3 )

[4] => Array(  [id] => 4 [image_type] => jpg [total_ratings] => 12 [average_rating] => 5.2 )

[5] => Array(  [id] => 5 [image_type] => png [total_ratings] => 15 [average_rating] => 8.1 ))

 

so that all 5 rows from the results are displayed

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