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
I want to first group data as per "Roll" and then further group it as per "Name" with different "Marks".
I have grouped data using Group by and having but I am not sure how to further group it.
Group by two columns :
SELECT Roll,Name FROM table GROUP BY Roll,Name;
If you want to see marks also you have aggegrate(sum) it like :
SELECT Roll,Name,sum(Marks)as 'Total Marks' FROM table GROUP BY Roll,Name;
If you want to see marks also you have aggegrate(average) it like :
SELECT Roll,Name,avg(Marks)as 'Average Marks' FROM table GROUP BY Roll,Name;
Refer this image
If I understand correctly, you want roll/name pairs that have multiple distinct values. You can do this using window functions:
select distinct roll, name, marks
from (select t.*, count(distinct marks) over (partition by roll, name) as cnt
from t
) t
where cnt > 1;
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 have table as below:
I want write a sql query to get output as below:
the query should select all the records from the table but, when multiple records have same Id column value then it should take only one record having latest Date.
E.g., Here Rudolf id 1211 is present three times in input---in output only one Rudolf record having date 06-12-2010 is selected. same thing with James.
I tried to write a query but it was not succssful. So, please help me to form a query string in sql.
Thanks in advance
You can partition your data over Date Desc and get the first row of each partition
SELECT A.Id, A.Name, A.Place, A.Date FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY Id ORDER BY Date DESC) AS rn
FROM [Table]
) A WHERE A.rn = 1
you can use WITH TIES
select top 1 PERCENT WITH TIES * from t
order by (row_number() over(partition by id order by date desc))
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=280b7412b5c0c04c208f2914b44c7ce3
As i can see from your example, duplicate rows differ only in Date. If it's a case, then simple GROUP BY with MAX aggregate function will do the job for you.
SELECT Id, Name, Place, MAX(Date)
FROM [TABLE_NAME]
GROUP BY Id, Name, Place
Here is working example: http://sqlfiddle.com/#!18/7025e/2
I have a simple query in a tabloid control that gets all the leads in one month. I then use the tabloid control to group them into lead source. And then I have an associated count column. I want to sort my report on the count descending, without doing it in the query. I keep getting an error saying you cannot sort on an aggregate.
Thanks.
you can do one more thing..
just write your query in subquery part and write order by clause in outer query.
(suppose you have group by query as follow-
select lead_source, count(*) cnt
from your_table
group by lead_source
)
so you can do as follow -
select lead_source, cnt from (
select lead_source, count(*) cnt
from your_table
group by lead_source
)
order by cnt
this your_table and group by column list you have to edit accordingly your table structure ..