I want to select distinct results from another select statement results
e.g;
select distinct from(select * from table)
following is result of inner select
testing department 9998901036 GOLD
testing department 9998901036 GOLD
I want to get distinct from above select result.
From your example, you could just do
select distinct * from table
But say you had some scenario where you wanted to distinct on some other results set, you could do
select distinct column1, column2 from (select * from table) T
Note that you have to alias your inner select
select distinct *
from
(select * from table) t
Works - You just need to give your sub select a table alias.
You can also use a CTE.
;WITH t AS
(
SELECT *
FROM table
)
SELECT DISTINCT *
FROM t
Related
I have two table with same name in two different schemas (old and new dump). I would like to know the difference between the two integration.
I have two queries, that gives old and new count:
select count(*) as count_old from(
SELECT
distinct id
FROM
schema1.compound)q1
select count(*) as count_new from(
SELECT
distinct id
FROM
schema2.compound)q2
I would like have the following output.
table_name count_new count_new diff
compound 4740 4735 5
Any help is appreciated. Thanks in advance
with counts as (
select
(select count(distinct id) from schema1.compound) as count_old,
(select count(distinct id) from schema2.compound) as count_new
)
select
'compound' as table_name,
count_old,
count_new,
count_old - count_new as diff
from counts;
I think you could do something like this:
SELECT 'compound' AS table_name, count_old, count_new, (count_old - count_new) AS diff FROM (
SELECT(
(SELECT count(*) FROM (SELECT DISTINCT id FROM schema1.compound)) AS count_old,
(SELECT count(*) FROM (SELECT DISTINCT id FROM schema2.compound)) AS count_new
)
It was probably answered already, but it is a subquery/nested query.
You can directly compute the COUNT on distinct values if you use the DISTINCT keyword inside your aggregation function. Then you can join the queries extracting your two needed values, and use them inside your query to get the output table.
WITH cte AS (
SELECT new.cnt AS count_new,
old.cnt AS count_old
FROM (SELECT COUNT(DISTINCT id) AS cnt FROM schema1.compound) AS old
INNER JOIN (SELECT COUNT(DISTINCT id) AS cnt FROM schema2.compound) AS new
ON 1 = 1
)
SELECT 'compound' AS table_name,
count_new,
count_old,
count_new = count_old AS diff
FROM cte
I need to get a count from a query with UNION ALL, but all my attempts so far have failed and I cannot get it to work.
My Initial SELECT (Which returns 8 unique departments)
select distinct Department
from ( select Department
from EFP_EmployeeFollowupManagerCommit
union all
select Department from EFP_EmploymentUser )
a order by Department;
I have tried different variations of the following
SELECT COUNT(Department) as "DepCount"
FROM ( select Department
from EFP_EmployeeFollowupManagerCommit
union all
select Department
from EFP_EmploymentUser );
Can anyone help?
UPDATE!
I finally got it to work .. thank you to all :-)
The final query:
select count(*) from (
select distinct Department
from (
select Department
from EFP_EmployeeFollowupManagerCommit
union all
select Department
from EFP_EmploymentUser ) as ddep) as depcount;
You can try the following query, it should work:
select count(*) from (select distinct Department
from ( select Department from EFP_EmployeeFollowupManagerCommit
union all
select Department from EFP_EmploymentUser ));
You have to write like this:
select count(distinct Department)
Wonder what would happen if i select both * and an aggregate together.
Will it be just one row or multiple rows?
example:
select *, count(*) from employee
Hello with this query I'm getting one result with four rows, how can I change it in order to get four named columns with their own result every one?
SELECT COUNT(*) FROM vehicles WHERE cus=1
UNION ALL
SELECT COUNT(*) FROM user WHERE cus=1
UNION ALL
SELECT COUNT(*) FROM vehicle_events WHERE cus=1
UNION ALL
SELECT COUNT(*) FROM vehicle_alerts WHERE cus=1
Thanks in advance.
SELECT a.ct veh_count, b.ct user_count, c.ct event_count, d.ct alert_count
FROM
( SELECT COUNT(*) ct FROM vehicles WHERE cus=1 ) a,
( SELECT COUNT(*) ct FROM user WHERE cus=1 ) b,
( SELECT COUNT(*) ct FROM vehicle_events WHERE cus=1 ) c,
( SELECT COUNT(*) ct FROM vehicle_alerts WHERE cus=1 ) d;
UNION only adds rows; it has no effect on the columns.
Columns, which define the "shape" of the row tuples, must appear as selected columns1.
For example:
SELECT
(SELECT COUNT(*) FROM vehicles WHERE cus=1) as veh_count
,(SELECT COUNT(*) FROM users WHERE cus=1) as user_count
..
1 There are other constructs that can allow this, see crosstab for example - but the columns are fixed by the query command. It takes dynamic SQL to get a variable number of columns.
I have a field called "Users", and I want to run SUM() on that field that returns the sum of all DISTINCT records. I thought that this would work:
SELECT SUM(DISTINCT table_name.users)
FROM table_name
But it's not selecting DISTINCT records, it's just running as if I had run SUM(table_name.users).
What would I have to do to add only the distinct records from this field?
Use count()
SELECT count(DISTINCT table_name.users)
FROM table_name
SQLFiddle demo
This code seems to indicate sum(distinct ) and sum() return different values.
with t as (
select 1 as a
union all
select '1'
union all
select '2'
union all
select '4'
)
select sum(distinct a) as DistinctSum, sum(a) as allSum, count(distinct a) as distinctCount, count(a) as allCount from t
Do you actually have non-distinct values?
select count(1), users
from table_name
group by users
having count(1) > 1
If not, the sums will be identical.
You can see for yourself that distinct works with the following example. Here I create a subquery with duplicate values, then I do a sum distinct on those values.
select DistinctSum=sum(distinct x), RegularSum=Sum(x)
from
(
select x=1
union All
select 1
union All
select 2
union All
select 2
) x
You can see that the distinct sum column returns 3 and the regular sum returns 6 in this example.
You can use a sub-query:
select sum(users)
from (select distinct users from table_name);
SUM(DISTINCTROW table_name.something)
It worked for me (innodb).
Description - "DISTINCTROW omits data based on entire duplicate records, not just duplicate fields." http://office.microsoft.com/en-001/access-help/all-distinct-distinctrow-top-predicates-HA001231351.aspx
;WITH cte
as
(
SELECT table_name.users , rn = ROW_NUMBER() OVER (PARTITION BY users ORDER BY users)
FROM table_name
)
SELECT SUM(users)
FROM cte
WHERE rn = 1
SQL Fiddle
Try here yourself
TEST
DECLARE #table_name Table (Users INT );
INSERT INTO #table_name Values (1),(1),(1),(3),(3),(5),(5);
;WITH cte
as
(
SELECT users , rn = ROW_NUMBER() OVER (PARTITION BY users ORDER BY users)
FROM #table_name
)
SELECT SUM(users) DisSum
FROM cte
WHERE rn = 1
Result
DisSum
9
If circumstances make it difficult to weave a "distinct" into the sum clause, it will usually be possible to add an extra "where" clause to the entire query - something like:
select sum(t.ColToSum)
from SomeTable t
where (select count(*) from SomeTable t1 where t1.ColToSum = t.ColToSum and t1.ID < t.ID) = 0
May be a duplicate to
Trying to sum distinct values SQL
As per Declan_K's answer:
Get the distinct list first...
SELECT SUM(SQ.COST)
FROM
(SELECT DISTINCT [Tracking #] as TRACK,[Ship Cost] as COST FROM YourTable) SQ