# SQL count e sum

## Question:

I have 6 SQL codes and they all search the same table, it looks like this:

``````+-----+-----+-----+
|  A  |  B  |  C  |
+-----+-----+-----+
|  1  |  1  |  1  |
+-----+-----+-----+
|  2  |  1  |  5  |
+-----+-----+-----+
|  3  |  2  |  3  |
+-----+-----+-----+
|  4  |  1  |  1  |
+-----+-----+-----+
|  5  |  1  |  4  |
+-----+-----+-----+
|  6  |  1  |  5  |
+-----+-----+-----+
|  7  |  9  |  1  |
+-----+-----+-----+
|  8  |  3  |  1  |
+-----+-----+-----+
``````

And the codes are these:

``````SELECT COUNT(*) as total, SUM(C) as soma FROM tabela WHERE B = 1;
SELECT COUNT(*) as total1 FROM tabela WHERE B = 1 AND C = 1;
SELECT COUNT(*) as total2 FROM tabela WHERE B = 1 AND C = 2;
SELECT COUNT(*) as total3 FROM tabela WHERE B = 1 AND C = 3;
SELECT COUNT(*) as total4 FROM tabela WHERE B = 1 AND C = 4;
SELECT COUNT(*) as total5 FROM tabela WHERE B = 1 AND C = 5;
``````

The returned values ​​are these:

``````total = 4
total1 = 2
total2 = 0
total3 = 0
total4 = 1
total5 = 2
soma = 16
``````

Everything works perfectly, but I would need everything to return in the same query, is that possible?

The intention is for it to count how many rows there are in the table, whose column B is equal to a value, then return in "total" and add all the values ​​in column C and return in "sum", but I need it to return the amount of times a value is repeated in the search.

What you need is something like this:

``````select
count(*) total,
sum(c) soma,
sum(case when C = 1 THEN 1 ELSE 0 END) Tot1,
sum(case when C = 2 THEN 1 ELSE 0 END) Tot2,
sum(case when C = 3 THEN 1 ELSE 0 END) Tot3
from tabela
where B = 1
``````

Projections in the format `sum(case when CONDICAO THEN 1 ELSE 0 END)` are very useful for this type of scenario. In CONDICAO you must put something that is rated True or False. In this case, you are testing C = 1, 2 or 3, but you could put C IN (1,2,3) , even combining with logical operators, example C = 1 AND B = 2 . Therefore, if the record being evaluated returns True, it adds 1, otherwise, it adds 0 (which makes no difference in the sum). Anyway, it's a way to count the records using SUM and not COUNT(*).

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