MySQL SUM() all the values of a Column for an ID(User) & echo total

Question

Successfully connected to my database. I have a table like,

stid |       date          | bf | ln | dn | meal_total
------------------------------------------------------
01   | 2015-11-27 13:12:01 | 1  | 2  | 3  |    6
01   | 2015-11-27 11:12:01 | 0  | 3  | 2  |    5
02   | 2015-11-27 12:12:01 | 1  | 1  | 1  |    3
03   | 2015-11-27 11:12:01 | 1  | 1  | 1  |    3

I just need to echo the SUM of all the values of the meal_total column recorded in current month for a stid like,

Total Meal for `stid` in current month : 11.

I am trying this code like,

 $sql = "select sum(meal_total) 
         from meal 
         where stid='$_SESSION[stid]'  
           AND date >= DATE_ADD(CURDATE(), INTERVAL -DAY(CURDATE())+1 DAY) ";

And I am trying to show the result in a html table like,

<? php 
echo sum(meal_total); 
?>

Now, Sir, please help me to solve. Thanks in advance.

[I have tried my best to clarify my problem.]

Answer 1

Since you didn't provide any valid PHP code, I'm just going to guess that you don't know how to access the SUM field from the result set.

When you SELECT SUM(meal_total) you should alias the field in order to access it from your PHP result array like so SELECT SUM(meal_total) AS meal_sum and you can echo $resultArray['meal_sum'] to get the value in question.

Answer 2

您需要在查询中使用Group by语句： http : //dev.mysql.com/doc/refman/5.7/en/group-by-functions.html

Answer 3

If you name your result, you can get it from your results a bit easier. I suggest that you use DATE_FORMAT , which lets you easily compare parts of dates.

$sql = "select sum(meal_total) AS meals 
        from meal 
        where stid='$_SESSION[stid]' 
          AND DATE_FORMAT( `date`, '%Y-%m' ) = DATE_FORMAT( NOW(), '%Y-%m' )";

# get result from sql
...
#

echo $result['meals'];

Offtopic: Even though I am a 'Sir', please beware that there are many (great) female programmers aswel.

Answer 4

Try

select
  m.stdid,
  sum(m.meal_total) as month_meals
from 
  meal as m
where 
   year(m.date)= year(month(CURRENT_DATE()) 
AND 
   month(m.date) = month(month(CURRENT_DATE())
group by m.stdid