What if a SQL code as below?
Proc SQL;
SELECT DISTINCT ID,SUM(AMOUNT) AS M,SUM(NO) AS CNT
FROM CUSTOMER_LIST
GROUP BY ID
ORDER BY CNT DESC;
QUIT;
Use DISTINCT with GROUP BY. Any possible error will occur when using this combination Or DISTINCT just a redundant word?
Thanks~
Use DISTINCT with GROUP BY. Any possible error will occur when using this combination? Or DISTINCT just a redundant word?
This won't error, but that's just unnecessary redondancy. GROUP BY ID guarantees that each ID will appear only on one row in the resulset. There is no benefit for adding DISTINCT here - and it makes the intent of the query harder to understand.
On the other hand, there are situations where you would use DISTINCT without GROUP BY: typically when you want to deduplicate a set of columns, but do not need to use aggregate functions (SUM(), COUNT()...).
SELECT ID,SUM(AMOUNT) AS M,SUM(NO) AS CNT
FROM CUSTOMER_LIST
GROUP BY ID
ORDER BY CNT DESC;
We already group by id so no need distinct id
Trying to get an overall distinct count of the employees for a range of records which has a group by on it.
I've tried using the "over()" clause but couldn't get that to work. Best to explain using an example so please see my script below and wanted result below.
EDIT:
I should mention I'm hoping for a solution that does not use a sub-query based on my "sales_detail" table below because in my real example, the "sales_detail" table is a very complex sub-query.
Here's the result I want. Column "wanted_result" should be 9:
Sample script:
CREATE TEMPORARY TABLE [sales_detail] (
[employee] varchar(100),[customer] varchar(100),[startdate] varchar(100),[enddate] varchar(100),[saleday] int,[timeframe] varchar(100),[saleqty] numeric(18,4)
);
INSERT INTO [sales_detail]
([employee],[customer],[startdate],[enddate],[saleday],[timeframe],[saleqty])
VALUES
('Wendy','Chris','8/1/2019','8/12/2019','5','Afternoon','1'),
('Wendy','Chris','8/1/2019','8/12/2019','5','Morning','5'),
('Wendy','Chris','8/1/2019','8/12/2019','6','Morning','6'),
('Dexter','Chris','8/1/2019','8/12/2019','2','Mid','2.5'),
('Jennifer','Chris','8/1/2019','8/12/2019','4','Morning','2.75'),
('Lila','Chris','8/1/2019','8/12/2019','2','Morning','3.75'),
('Rita','Chris','8/1/2019','8/12/2019','2','Mid','1'),
('Tony','Chris','8/1/2019','8/12/2019','4','Mid','2'),
('Tony','Chris','8/1/2019','8/12/2019','1','Morning','6'),
('Mike','Chris','8/1/2019','8/12/2019','4','Mid','1.5'),
('Logan','Chris','8/1/2019','8/12/2019','3','Morning','6.25'),
('Blake','Chris','8/1/2019','8/12/2019','4','Afternoon','0.5')
;
SELECT
[timeframe],
SUM([saleqty]) AS [total_qty],
COUNT(DISTINCT [s].[employee]) AS [employee_count1],
SUM(COUNT(DISTINCT [s].[employee])) OVER() AS [employee_count2],
9 AS [wanted_result]
FROM (
SELECT
[employee],[customer],[startdate],[enddate],[saleday],[timeframe],[saleqty]
FROM
[sales_detail]
) AS [s]
GROUP BY
[timeframe]
;
If I understand correctly, you are simply looking for a COUNT(DISTINCT) for all employees in the table? I believe this query will return the results you are looking for:
SELECT
[timeframe],
SUM([saleqty]) AS [total_qty],
COUNT(DISTINCT [s].[employee]) AS [employee_count1],
(SELECT COUNT(DISTINCT [employee]) FROM [sales_detail]) AS [employee_count2],
9 AS [wanted_result]
FROM #sales_detail [s]
GROUP BY
[timeframe]
You can try this below option-
SELECT
[timeframe],
SUM([saleqty]) AS [total_qty],
COUNT(DISTINCT [s].[employee]) AS [employee_count1],
SUM(COUNT(DISTINCT [s].[employee])) OVER() AS [employee_count2],
[wanted_result]
-- select count form sub query
FROM (
SELECT
[employee],[customer],[startdate],[enddate],[saleday],[timeframe],[saleqty],
(select COUNT(DISTINCT [employee]) from [sales_detail]) AS [wanted_result]
--caculate the count with first sub query
FROM [sales_detail]
) AS [s]
GROUP BY
[timeframe],[wanted_result]
Use a trick where you only count each person on the first day they are seen:
select timeframe, sum(saleqty) as total_qty),
count(distinct employee) as employee_count1,
sum( (seqnum = 1)::int ) as employee_count2
9 as wanted_result
from (select sd.*,
row_number() over (partition by employee order by startdate) as seqnum
from sales_detail sd
) sd
group by timeframe;
Note: From the perspective of performance, your complex subquery is only evaluated once.
I am trying to work on public dataset bigquery-public-data.austin_crime.crime of the BigQuery. My goal is to get the output as three column that shows the
discription(of the crime), count of them, and top district for that particular description(crime).
I am able to get the first two columns with this query.
select
a.description,
count(*) as district_count
from `bigquery-public-data.austin_crime.crime` a
group by description order by district_count desc
and was hoping I can get that done with one query and then I tried this in order to get the third column showing me the Top district for that particular description (crime) by adding the code below
select
a.description,
count(*) as district_count,
(
select district from
( select
district, rank() over(order by COUNT(*) desc) as rank
FROM `bigquery-public-data.austin_crime.crime`
where description = a.description
group by district
) where rank = 1
) as top_District
from `bigquery-public-data.austin_crime.crime` a
group by description
order by district_count desc
The error i am getting is this. "Correlated subqueries that reference other tables are not supported unless they can be de-correlated, such as by transforming them into an efficient JOIN."
I think i can do that by joins. Can someone has better solution possibly to do that using without join.
Below is for BigQuery Standard SQL
#standardSQL
SELECT description,
ANY_VALUE(district_count) AS district_count,
STRING_AGG(district ORDER BY cnt DESC LIMIT 1) AS top_district
FROM (
SELECT description, district,
COUNT(1) OVER(PARTITION BY description) AS district_count,
COUNT(1) OVER(PARTITION BY description, district) AS cnt
FROM `bigquery-public-data.austin_crime.crime`
)
GROUP BY description
-- ORDER BY district_count DESC
I am using SQL 2012 and trying to identify rows where the SourceDataID column has two unique entries in the PartyCode column, and I'm having difficulties.
SELECT PartyCode, SourceDataID, count (*) as CNT
FROM CustomerOrderLocation (nolock)
GROUP BY PartyCode, SourceDataID
HAVING (Count(PartyCode)>1)
ORDER BY PartyCode
Results are returning as such:
W3333 948_O 31
(party code/sourcedataid/CNT)
This is showing me the total entries where the Partycode and the SourceDataID are listed together in the table. However, I need it to show a count of any instances where W333 lists 948_O as the SourceDataID more than once.
I'm not having luck structuring the query to pull the results I am looking to get. How can I do this?
A CTE coupled with the PARTITION BY function is helpful in finding duplicates of this manner. Code below:
WITH CTE AS(
SELECT PartyCode, SourceDataID,
ROW_NUMBER()OVER(PARTITION BY SourceDataID ORDER BY SourceDataID) RN
FROM CustomerOrderLocation (NOLOCK))
SELECT * FROM CTE WHERE RN > 1
This should return every duplicate PartyCode attached to a SourceDataID.
If you want to see the entire result, change the last SELECT statement to:
SELECT * FROM CTE ORDER BY PartyCode, RN
Thanks for the help everyone. I did not do the best job of describing the issue but this is the query I ended up creating to get my result set.
;with cte1 (sourcedataid, partycode) as (select sourcedataid, partycode from customerorderparty (nolock) group by PartyCode, SourceDataID)
select count(sourcedataid), sourcedataid from cte1 group by sourcedataid having count(sourcedataid) >1
I'm looking get two things from a query, given a set of contraints:
The first match
The total number of matches
I can get the first match by:
SELECT TOP 1
ID,
ReportYear,
Name,
SignDate,
...
FROM Table
WHERE
...
ORDER BY ... //I can put in here which one I want to take
And then I can get the match count if I use
SELECT
MIN(ID),
MIN(ReportYear),
MIN(Name),
MIN(SignDate),
... ,
COUNT(*) as MatchCount
FROM Table
WHERE
...
GROUP BY
??? // I don't really want any grouping
I really want to avoid both grouping and using an aggregate function on all my results. This question SQL Server Select COUNT without using aggregate function or group by suggests the answer would be
SELECT TOP 1
ID,
ReportYear,
Name,
SignDate,
... ,
##ROWCOUNT as MatchCount
FROM Table
This works without the TOP 1, but when it's in there, ##ROWCOUNT = number of rows returned, which is 1. How can I get essentially the output of COUNT(*) (whatever's left after the where clause) without any grouping or need to aggregate all the columns?
What I don't want to do is repeat each of these twice, once for the first row and then again for the ##ROWCOUNT. I'm not finding a way I can properly use GROUP BY, because I strictly want the number of items that match my criteria, and I want columns that if I GROUPed them would throw this number off - unless I'm misunderstanding GROUP BY.
Assuming you are using a newish version of SQL Server (2008+ from memory) then you can use analytic functions.
Simplifying things somewhat, they are a way of way of doing an aggregate over a set of data instead of a group - an extension on basic aggregates.
Instead of this:
SELECT
... ,
COUNT(*) as MatchCount
FROM Table
WHERE
...
You do this:
SELECT
... ,
COUNT(*) as MatchCount OVER (PARTITION BY <group fields> ORDER BY <order fields> )
FROM Table
WHERE
...
GROUP BY
Without actually running some code, I can't recall exactly which aggregates that you can't use in this fashion. Count is fine though.
Well, you can use OVER clause, which is an window function.
SELECT TOP (1)
OrderID, CustID, EmpID,
COUNT(*) OVER() AS MatchCount
FROM Sales.Orders
WHERE OrderID % 2 = 1
ORDER BY OrderID DESC
Try next query:
select top 1
*, count(*) over () rowsCount
from
(
select
*, dense_rank() over (order by ValueForOrder) n
from
myTable
) t
where
n = 1