Trying to Sum Distinct Values SQL

Getting sum of a column that needs a distinct value from other column

I guess this is a job for a subquery. So let's take your problem step by step.

I'm trying to find all the rows in the balance column that are the same and have the same date,

This subquery gets you that, I believe. It give the same result as SELECT DISTINCT but it also counts the duplicated rows.

                SELECT COUNT(*) num_same_rows, balance, date
                  FROM `table`
                 WHERE a.datum BETWEEN '2021-01-01' AND '2021-09-01'
                 GROUP BY date, balance

and then find the sum of the balance column.

Nest the subquery like this.

SELECT SUM(balance) summed_balance, date
  FROM (
                SELECT COUNT(*) num_same_rows, balance, date
                  FROM `table`
                 WHERE a.datum BETWEEN '2021-01-01' AND '2021-09-01'
                 GROUP BY date, balance
       ) subquery
 GROUP BY date

If you only want to consider rows that actually have duplicates, change your subquery to

                SELECT COUNT(*) num_same_rows, balance, date
                  FROM `table`
                 WHERE a.datum BETWEEN '2021-01-01' AND '2021-09-01'
                 GROUP BY date, balance
                HAVING COUNT(*) >= 1

Be careful here, though. You didn't tell us what you want to do, only how you want to do it. The way you described your problem calls for discarding duplicated data before doing the sums. Is that right? Do you want to discard data?

SQL query with distinct and sum

SELECT color, fruit, sum(rating)
FROM medleys
GROUP BY color, fruit

Distinct is used to select distinct elements, nothing more, while you want to aggregate and for that you need GROUP BY and aggregation functions (SUM).

SQL Server SUM() for DISTINCT records

Use count()

SELECT count(DISTINCT table_name.users)
FROM table_name

SQLFiddle demo

Sum distinct by separate ID column

Get the distinct combinations of ID and Value in a subquery and then the sum of Values:

SELECT SUM(Value) sum_value
FROM (SELECT DISTINCT ID, Value FROM tablename) t

Another way to do it is with SUM() window function:

SELECT DISTINCT SUM(MAX(Value)) OVER() sum_value
FROM tablename
GROUP BY ID

See the demo.

How do I SUM DISTINCT Rows?

If you just want an overall figure for it try

select sum(PopEstimate2005), sum(EstimatesBase2000)
from(
    SELECT  Distinct
        Zipcodes.CountyID, 
        us_co_est2005_allData.PopEstimate2005, 
        us_co_est2005_allData.EstimatesBase2000, 
        users_link_territory.userID 
    FROM 
        Zipcodes Inner Join 
        Users_link_territory ON zipcodes.CountyID = Users_link_territory.CountyID Inner Join 
        us_co_est2005_alldata ON zipcodes.FIPS = us_co_est2005_alldata.State AND zipcodes.code = us_co_est2005_alldata.County 
    WHERE 
        (users_link_territory.userid = 4)
) as foo

Get sum of column with distinct id in SQL

You can use two levels of aggregation:

select sum(avg_amount)
from (select ext_id, avg(amount) as avg_amount
      from t
      group by ext_id
     ) x

SUM(DISTINCT) Based on Other Columns

select sum (rate)
from yourTable
group by first_name, last_name

Edit

If you want to get all sum of those little "sums", you will get a sum of all table..

Select sum(rate) from YourTable

but, if for some reason are differents (if you use a where, for example)
and you need a sum for that select above, just do.

select sum(SumGrouped) from 
(    select sum (rate) as 'SumGrouped'
    from yourTable
    group by first_name, last_name) T1

SQL Server Sum Distinct Group by

With a long version, but easy to understand query. The below query using CTE should help you

with records as
(select distinct id, date, amt from table_name)
select sum (amt), id
from records
group by 
id;

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