Question

I already read (this), but couldn’t figure out a way to implement it to my specific problem. I know SUM() is an aggregate function and it doesn’t make sense not to use it as such, but in this specific case, I have to SUM() all of the results while maintaining every single row.

Here’s the table:

--ID-- --amount--
  1        23
  2        11
  3        8
  4        7

I need to SUM() the amount, but keep every record, so the output should be like:

--ID-- --amount--
  1        49
  2        49
  3        49
  4        49

I had this query, but it only sums each row, not all results together:

SELECT 
    a.id,
    SUM(b.amount)

FROM table1 as a 
JOIN table1 as b ON a.id = b.id
GROUP BY id

Without the SUM() it would only return one single row, but I need to maintain all ID’s…

Note: Yes this is a pretty basic example and I could use php to do this here,but obviously the table is bigger and has more rows and columns, but that’s not the point.

Answer 1

SELECT a.id, b.amount
FROM table1 a
CROSS JOIN
(
    SELECT SUM(amount) amount FROM table1
) b

You need to perform a cartesian join of the value of the sum of every row in the table to each id. Since there is only one result of the subselect (49), it basically just gets tacked onto each id.