PHP array_sum not working - php

Question

I am trying to query an SQL table, and then have all the values from one column (it's a "tinyint(1)") added up and then printed, using array_sum. (Currently there are only two rows).
When both rows have a "1" in that column, it gives 2, the correct answer. And when they are both "0" it gives 0, correct again. But when one is "1" and one "0" it gives 2, not 1.
This is my code
$con = mysql_connect("XXX","XXX","XXX");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("XXX_test", $con);
$day = $_GET["day"];
$query = mysql_query("SELECT $day FROM sites");
if (!$query)
{
die('Could not query: ' . mysql_error());
}
$rows = mysql_fetch_array($query);
echo array_sum($rows);
mysql_close($con);
?>
Thanks for any help.
--UPDATE--
I just solved this myself.
With this, if you want to know:
$query = mysql_query("SELECT name, SUM($day) AS Alerts FROM sites");
while($row = mysql_fetch_array($query))
{
echo $row['Alerts'];
echo "<br />";
}

Answer 1

This in fact never works correctly.
$rows = mysql_fetch_array($query);
That line only ever fetches the first row. The second row is never retrieved.
The reason it gives 2 using array_sum is that mysql_fetch_array gets the result in two different formats, with keys both by name and number. So you'll get something like this:
array(
0 => '1',
'Tuesday' => '1'
)
Obviously this will always be 2 or 0.
You should almost certainly do a count on the DB server:
SELECT COUNT(*) FROM sites WHERE $day=1
Note, however, that your biggest current problem is the enormous gaping SQL injection hole. Sanitise your input.

Answer 2

mysql_fetch_array() only fetches one row at a time. You need to fetch the rows in a while loop. Then you can either append the results to an array and sum those, or sum them in the loop. You can also SUM in mysql anyway, so you might look into using that.
I'm also obligated to suggest that you use PDO or some other DB wrapper instead of mysql_. You also need to escape variables you use in queries under most circumstances, especially if they come from _GET.

Answer 3

I would suggest doing it in the mysql query:
SELECT SUM($day) FROM sites
also, please escape the $day variable.

Answer 4

More of a comment:
array_sum can not work with the array you're passing to it. You need to pass an array for what the function actually works, like array(1, 2, 3) which will give you 6 then.
Next to that, let MySQL itself calculate the SUM (or COUNT?), so it's easier to handle in your script as others have posted.

Answer 5

I just solved this myself.
With this, if you want to know:
$query = mysql_query("SELECT name, SUM($day) AS Alerts FROM sites");
while($row = mysql_fetch_array($query))
{
echo $row['Alerts'];
echo "<br />";
}