AVG() fuction always returning "Array"

Question

This is what I got, UPDATED

$avg_query = "SELECT AVG(rating) AS avg_rating FROM kicks WHERE userid='$userid'";
                        $avg_result = mysqli_query($cxn, $avg_query) or die("Couldn't execute query.");

                        $row = mysqli_fetch_assoc($avg_result);
                        $average = $row['avg_rating'];

Every time I echo it out, it always returns array. Never used this function before, but i don't understand why it would do that.

Answer 1

Shouldn't you be assigning to $row here?

$row = mysqli_fetch_assoc($avg_result);
$average = $row['AVG(rating)'];

I'd also suggest using an alias when you are selecting the result of a function call:

$avg_query = "SELECT AVG(rating) AS avg_rating FROM kicks WHERE userid='$userid'"
...
$average = $row['avg_rating'];

If this still doesn't give any result it is probably because:

• There are no rows in that table for that user.
• There is a row in that table where the rating is set to the value NULL.

In the second case you can fix it by filtering out those rows:

SELECT AVG(rating) AS avg_rating
FROM kicks
WHERE userid='$userid'
AND rating IS NOT NULL

Answer 2

$avg = mysqli_fetch_assoc($avg_result);
$average = $row['AVG(rating)'];

where did you get $row?

Answer 3

Btw, this should be done something like this:

$avg_query = "SELECT AVG(id) as average_id FROM unfollowr_users";
$avg_result = mysql_query($avg_query) or die("Couldn't execute query.");

$row = mysql_fetch_assoc($avg_result);
$average = $row['average_id'];
echo "$average\n";

AVG is not an PHP function it's SQL function. We just add better name when we SELECT.

Answer 4

The string "Array" is what you get when PHP thinks you want it to convert an array to a string. If you see the word "Array" where you expected more meaningful output, it means that you tried to treat a variable containing an array as though it actually contained a string.

When you see this behaviour from an echo command, one thing you could try is to place the following before the echo, as a debugging measure (let's say your variable's name is $var):

echo '<pre>';
var_dump($var);
echo '</pre>';

You will then see something like

array(0) {
0 => string(3) 'cat'
1 => string(6) 'doggie'
}

This tells you what structure your array has and what its keys are. You should then look back in your code and try to work out why your variable contains an array when you had expected a string - perhaps you had misunderstood the behaviour of a standard library function, or something like that.