Is there a way to select the sum of a column and other columns at the same time in SQL?
Example:
SELECT sum(a) as car,b,c FROM toys
How about:
select sum(a) over(), b, c from toy;
or, if it's required:
select sum(a) over(partition by b), b, c from toy;
try add GROUP BY
SELECT sum(a) as car,b,c FROM toys
GROUP BY b, c
Since you do not give much context, I assume you mean one of the following :
SELECT (SELECT SUM(a) FROM Toys) as 'car', b, c FROM Toys;
or
SELECT SUM(a) as Car, b, c FROM Toys GROUP BY b, c;
SELECT b,
c,
(SELECT sum(a) FROM toys) as 'car'
FROM toys
Related
In CrateDB is there a way to avoid to re-select the same column in nested SELECT statement, to show the value in the results?
e.i. in the following query, is there any way to avoid re-selecting A and B through the nested SELECT? Ideally would be nice to select just once in the first
SELECT
A,
B,
AB,
A * B * AB AS ABAB,
A / AB::DECIMAL AS AAB,
FROM (
SELECT
A,
B,
(A + B) AS AB,
FROM (
SELECT
(SELECT count(*) FROM schema.table_01 WHERE process_state IN ('State_1')) AS A,
(SELECT count(*) FROM schema.table_01 WHERE process_state IN ('State_2')) AS B,
) alias_for_subquery_01
) alias_for_subquery_02
Thanks
count(*) OVER (PARTITION BY a, b ORDER BY a, b, c) * 10
This produces the same result as:
dense_rank() OVER (PARTITION BY a, b ORDER BY a, b, c) * 10
Used in a query like this:
SELECT
dense_rank() OVER (ORDER BY a, b) ,
a || b,
count(*) OVER (
PARTITION BY a, b
ORDER BY a, b, c
) * 10 ,
a2,
b1,
c1,
cc1,
c2,
FROM
join ....
ORDER BY 1, 6;
I'm happy with my query result.
But should I appreciate one approach over the other and why?
After PARTITION BY a, b there is no point in adding aor b to ORDER BY, like David commented.
So we simplify to:
count(*) OVER (PARTITION BY a, b ORDER BY c) * 10
dense_rank() OVER (PARTITION BY a, b ORDER BY c) * 10
These two only happen to be equivalent while c is UNIQUE. Else they are not.
You'd need to define exactly what the number is supposed to signify, and show your table definition, and the exact query because joins can introduce duplicates and NULL values.
row_numer() or rank() are similar window functions ...
Performance is practically the same for all of them.
I want to select the same columns from different tables that look the same
(daily tables).
I saw this SELECT from multiple tables with the same structure answer but if I'm going according to this I'm ending up with a huge query.
this code is similar to what I have, according to answer above I need to do the following:
select a, b, c
from (
select a, b, c, d, e from hourly.16
union all
select a, b, c, d, e from hourly.15
)
isn't there an option to do something like:
select a, b, c
from (
select a, b, c, d, e from (hourly.16 union all hourly.15)
)
so I won't end up with huge queries?
#standardSQL
SELECT a, b, c
FROM (
SELECT a, b, c, d, e
FROM `project.hourly.*`
WHERE _TABLE_SUFFIX BETWEEN '15' AND '16'
)
Above is assuming that hourly is your dataset
I have a table that has millions of records.
So I might have these columns
a, b, c, d
I need to select all the distinct records based on columns a and b.
But I need to select columns a, b, c and d not just a and b.
Can I do this?
edit
Data might be
1,1,frog,green
1,1,frog,brown
2,1,cat,black
2,4,dog,white
so i need;
1,1,frog,green
2,1,cat,black
2,4,dog,white
SQL Server supports Common Table Expression and Window Function. The query below uses ROW_NUMBER() which ranks the record according to group. It sorts by c ASC, d ASC (just play with it).
WITH records
AS
(
SELECT a, b, c, d,
ROW_NUMBER() OVER(PARTITION BY a, b ORDER BY c, d) rn
FROM TableName
)
SELECT a, b, c, d
FROM records
WHERE rn = 1
SQLFiddle Demo
TSQL Ranking Functions
partition by is your man
SELECT a, b, c, d FROM (
SELECT a, b, c, d, ROW_NUMBER() OVER (PARTITION BY a, b ORDER BY a, b) rn
FROM table
) sq
where rn = 1
Please try:
select *
From(
select
row_number() over (partition by a, b order by a, b) RNum,
*
from
YourTable
)x
where RNum=1
Sample
select * From(
select row_number() over (partition by a, b order by a, b) RNum, *
from(
select 1 a, 1 b, 'frog' c, 'green' d union all
select 1 a, 1 b, 'frog' c, 'brown' d union all
select 2 a, 1 b, 'cat' c, 'black' d union all
select 2 a, 4 b, 'dog' c, 'white' d)x
)y
where RNum=1
Suppose i have this table
table (a,b,c,d). Datatypes are not important.
I want to do this
select a as a1,b as b1,c as c1,
(select sum(d) from table where a=a1 and b=b1) as total
from table
group by a,b,c
...but I can't find a way (sqldeveloper keeps complaining with "from clause not found".)
Is there a way? Is it possible?
SELECT a as a1,b as b1,c as c1,
(
SELECT SUM(d)
FROM mytable mi
WHERE mi.a = mo.a
AND mi.b= mo.b
) as total
FROM mytable mo
GROUP BY
a, b, c
It's much more simple and efficient to rewrite it as this:
SELECT a AS a1, B AS b1, c AS c1, SUM(SUM(d)) OVER (PARTITION BY a, b) AS total
FROM mytable
GROUP BY
a, b, c
Note the SUM(SUM(d)) here.
The innermost SUM is the aggregate function. It calculates the SUM(d) a-b-c-wise.
The outermost SUM is the analytic function. It sums the precalculated SUM(d)'s a-b-wise, and returns the value along with each row.
Du you mean something like this?
select a as a1,
b as b1,
c as c1,
sum(sum(d)) OVER (PARTITION BY a, b) AS total
from table
group by a,b,c
you can do it with aliases:
SELECT a AS a1, b AS b1, c AS c1,
(SELECT SUM(d)
FROM test_t t_in
WHERE t_in.a = t.a
AND t_in.b = t.b) AS total
FROM test_t t
GROUP BY a, b, c