mysql query show only 1 result

i want to output the result based on todays date. the problem is, the output only show 1 result?

database report table:
id  |  r_amount | id_therapist | date | time | t_tanning | t_deep 

// this query works fine echoing all the result if i use while loop
$today = date('Y-m-d');
(1) $q = $db->query("SELECT * FROM report WHERE date='$today' ORDER BY date ASC")

// this query only show 1 output result?
(2) $q = $db->query("SELECT *, SUM(IF(t_tanning LIKE 'Pro Tan%', r_amount, 0)) AS totalProTan FROM report WHERE date='$today' ORDER BY date ASC")

while($r = $q->fetch_array(MYSQLI_ASSOC)) :
     // (1) echoing all result from database
     echo $r['r_amou开发者_运维技巧nt'].'<br>';
     // (2) echoing only 1 result????
     echo $r['totalProTan'].'<br>';
endwhile;

If the date field is of type datetime, you'll have to do something like

SELECT ... WHERE DATE(date)=CURDATE()

Notice that I'm using curdate() in the query. There's no need to generate the date value in PHP. MySQL is perfectly capable of doing that itself.

Try adding a GROUP BY statement to the second SQL statement.

• you should group by the key of the elemnts you want to be shown in the end result

The use of the aggregate function SUM will result in a single result. You are asking the database to get all the rows then sum up a value and give you the value.

To see the result for many groups of values you have to add a group by clause.