I try to make a sum of multiple select count distinct with case function. For example:
SELECT id_dept,
count(DISTINCT case when e.statut='pub' then id_patients end) AS nb_patients_pub,
count(DISTINCT case when e.statut='priv' then id_patients end) AS nb_patients_priv
FROM venues
I would like to make of these two results in only one columns.
Is it possible?
I think that you want in:
SELECT
id_dept,
COUNT(DISTINCT CASE WHEN e.statut IN ('pub', 'priv') THEN id_patients END) AS nb_patients_pub_and_venues
FROM venues
GROUP BY id_dept
Note that I added a GROUP BY clause to the query, which was initially missing (this is a syntax error in almost all databases).
Depending on your data, this might not do exactly what you want; if a given id_patient has both statuses, then it will be counted only once, whereas your code counted it once in each count(distinct ...). If so, then you can just keep the two separated counts, and sum them:
SELECT
id_dept,
COUNT(DISTINCT CASE WHEN e.statut IN = 'pub' THEN id_patients END)
+ COUNT(DISTINCT CASE WHEN e.statut IN = 'priv' THEN id_patients END)
AS nb_patients_pub_and_venues
FROM venues
GROUP BY id_dept
If you're happy with current code, then either sum (using +) those counts, or use that query as a CTE (or an inline view) and
with test as
(SELECT id_dept,
count(DISTINCT case when e.statut='pub' then id_patients end)
AS nb_patients_pub,
count(DISTINCT case when e.statut='priv' then id_patients end)
AS nb_patients_priv
FROM venues
GROUP BY id_dept
)
select id, nb_patients_pub + nb_patients_priv as result
from test;
Related
I have a query that includes two subqueries with similar column 'day'. I would like to show values in a following way:
day cnt1 cnt_total
But in a query I have it does not recognize that the day column is similar and makes a multiplication of all rows in nested statement one by all rows in nested statement two.
Is there a way to make it recognize that the day column is similar?
The query looks as follows:
SELECT p1.day, p1.count AS cnt1, p2.count AS cnt_total
FROM
(
SELECT day, COUNT(DISTINCT id) AS count FROM table
WHERE 1=1
AND service="service"
AND action="action"
AND path LIKE "%search%"
AND year="2021"
GROUP BY day
) p1,
(
SELECT day, COUNT(DISTINCT id) AS count FROM table
WHERE 1=1
AND service="service"
AND action="action"
AND year="2021"
GROUP BY day
) p2;
You should be able to do this with conditional aggregation, so only one SELECT is needed:
SELECT day,
COUNT(DISTINCT CASE WHEN action = 'mousedown' AND data["path"] LIKE '%go-to-latest-search%' THEN gsid END) AS count,
COUNT(DISTINCT CASE WHEN action = 'impress' THEN gsid END) as cnt_total
FROM hit
WHERE service = 'sauto' AND
year = '2021' AND
month = '07'
GROUP BY day
Is it possible to write one query, where I would group by 2 columns in a table to get the count of total members plus get a sum of one column in that same table, but grouped by one column?
For example, the data looks like this
I want to get a count on distinct combinations of columns "OHID" and "MemID" and get the SUM of the "Amount" column grouped by OHID. The result is supposed to look like this
I was able to get the count correct using this query below
SELECT count(*) as TotCount
from (Select DISTINCT OHID, MemID
from #temp) AS TotMembers
However, when I try to use this query below to get all the results together, I am getting a count of 15 and a totally different total sum.
SELECT t.OHID,
count(TotMembers.MemID) as TotCount,
sum(t.Amount) as TotalAmount
from (Select DISTINCT OHID, MemID
from #temp) AS TotMembers
join #temp t on t.OHID = TotMembers .OHID
GROUP by t.OHID
If I understand correctly, you want to consider NULL as a valid value. The rest is just aggregation:
select t.ohid,
(count(distinct t.memid) +
(case when count(*) <> count(t.memid) then 1 else 0 end)
) as num_memid,
sum(t.amount) as total_amount
from #temp t
group by t.ohid,
The case logic might be a bit off-putting. It is just adding 1 if any values are NULL.
You might find this easier to follow with two levels of aggregation:
select t.ohid, count(*), sum(amount)
from (select t.ohid, t.memid, sum(t.amount) as amount
from #temp t
group by t.ohid, t.memid
) t
group by t.ohid
I want to retrieve different information in one statement from the same table and they have different number of rows.
The first select has five rows in the result and the second select has three rows because some prices have null value. I thought maybe if I can put zero instead of null so they will match the same number of rows but I don't know how to do that, or is there another solution?
select count(ID), Land
from Film_ha2911
group by Land
union
select count(ID)
from Film_ha2911
where Price is not null
group by Land;
The use of UNION implies that the number and type of columns in select must corresponding
so in your case you should use null for not select columns
select count(ID), Land
from Film_ha2911
group by Land
union
select count(ID), null
from Film_ha2911
where Price is not null
group by Land;
But in this case seems you need a left join on the subquery for land
select t1.count1, t1.land , t2.count2
from (
select count(ID) count1, Land
from Film_ha2911
group by Land
) t1
left join (
select count(ID) count2, land
from Film_ha2911
where Price is not null
group by Land;
) t2 on t1.land = t2.land
The desired result can be achieved by single SELECT without UNION.
Extra column: PriceNotNull to differentiate is Price value filled or not:
SELECT
Land,
CASE WHEN Price IS NOT NULL THEN 'True' ELSE 'False' END PriceNotNull,
COUNT(ID) AS Count_ID
FROM Film_ha2911
GROUP BY Land, CASE WHEN Price IS NOT NULL THEN 'True' ELSE 'False' END
You can just use count():
select Land, count(*) as total_rows,
count(price) as total_with_price
from Film_ha2911
group by Land;
count() counts the number of non-NULL values, so no special logic is needed to count non-NULL values. By count(id) I assume you want to count all the rows. count(*) is more explicit -- as would count(1) which some people prefer.
If you actually want this on separate rows, I would add an indicator for what the count means:
select Land, 'total rows' as which, count(*) as total_rows
from Film_ha2911
group by Land
union all
select Land, 'with price', count(price)
from Film_ha2911
group by Land;
However, I think the first version with two separate columns is more useful.
I have some E_ids which are linked to a couple of d_ids and with o_count in any of (1,0,null).
So if any of the E_IDs have an O_count = 1, I have to club it into one row and write the O_count = 1 for that E_ID else 0.
But when I do the below, I get all the rows without the grouping, i.e, I get two rows of the same e_ids. Is there any other way to do the same?
SELECT DISTINCT E_ID, status
(CASE WHEN o_count = any(1) THEN 1
WHEN o_count = any(0) THEN 0
ELSE null END
) Ocount
FROM (SELECT e_id, status, o_count FROM A)
GROUP BY e_id, status, o_count
Yes, just wrap it with MAX() :
SELECT E_ID,status
MAX(case when o_count = 1 then 1 ELSE 0 END) as Ocount
FROM A
GROUP BY e_id,status
Also , the sub query was unnecessary , you are not doing any logic in there.
First of all you group by e_id, status, o_count. That means you aggregate your data such that you get one row for each such combination. Don't you rather want to get one result row per e_id alone or maybe e_id plus status?
Then you have that case construct not containing any aggregate function, but only the o_count which is part of your group by clause. So you are looking at one row where you want o_count = any(1) which is exactly the same as o_count = 1 of course, because there is only one value in the specified set. You can replace the complete case expression with a mere o_count.
Then you apply distinct. But there can be no duplicates, as you are grouping by all columns used. So distinct doesn't do anything here.
Selecting from a subquery without any where clause or aggregation is also superfluous and you can select from table a directly.
Your query can be re-written as
select distinct e_id, status, o_count
from a;
I suppose you want something like this instead:
select e_id, status, max(o_count)
from a
group by e_id, status;
Or this:
select e_id, max(status), max(o_count)
from a
group by e_id;
i want to get a unique count of the of multiple columns containing the similar or different data...i am using sql server 2005...for one column i am able to take the unique count... but to take a count of multiple columns at a time, what's the query ?
You can run the following selected, getting the data from a derived table:
select count(*) from (select distinct c1, c2, from t1) dt
To get the count of combined unique column values, use
SELECT COUNT(*) FROM TableName GROUP BY UniqueColumn1, UniqueColumn2
To get the unique counts of multiple individual columns, use
SELECT COUNT(DISTINCT Column1), COUNT(DISTINCT Column2)
FROM TableName
Your question is not clear what exactly you want to achieve.
I think what you're getting at is individual SUMS from two unique columns in one query. I was able to accomplish this be using
SELECT FiscalYear, SUM(Col1) AS Col1Total, SUM(Col2) AS Col2Total
FROM TableName
GROUP BY FiscalYear
If your data is not numerical in nature, you can use CASE statements
SELECT FiscalYear, SUM(CASE WHEN ColA = 'abc' THEN 1 ELSE 0 END) AS ColATotal,
SUM(CASE WHEN ColB = 'xyz' THEN 1 ELSE 0 END) AS ColBTotal
FROM TableName
GROUP BY FiscalYear
Hope this helps!