I am sorry if the question is silly because i'm new to SQL Server. I want to select top 5 records for each day of specified month.
e.g.
top 5 records for day 1 in month september
top 5 records for day 2 in month september
top 5 records for day 3 in month september
.
.
top 5 records for day 31 in month september
and show these all records as a one result.
Let's say you're checking speeding records for the month June 2012, and you wanted the top 5 speeds (by speed desc).
SELECT *
FROM (
SELECT *, RowNum = Row_number() over (partition by Cast(EventTime as Date)
order by Speed desc)
FROM Events
WHERE EventTime >= '20120601'
AND EventTime < '20120701'
) X
WHERE RowNum <= 5
Try this one,
WITH TopFiveRecords
AS
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY dayColumn ORDER BY colName DESC) RN
FROM tableName
)
SELECT *
FROM TopFiveRecords
WHERE RN <= 5
-- AND date condition here ....
dayColumn the column that contains the date of the month
colName the column to be sorted
Related
I have below audit table. Need to find Min Row where year is Year Getdate()) compared to min Row where Year is Year(getdate())-1
You can try to use ROW_NUMBER window function PARTITION BY ID & year of the date column.
SELECT *
FROM (
SELECT *,ROW_NUMBER() OVER(PARTITION BY ID,YEAR(Date) ORDER BY Date) rn
FROM T
) t1
WHERE rn = 1
I can't get my head around whether this is even possible, but I feel like I might have done it before and lost that bit of code. I am trying to craft a select statement that contains an inner join on a subquery to show the number of days between two dates from the same table.
A simple example of the data structure would look like:
Name ID Date Day Hours
Bill 1 3/3/20 Thursday 8
Fred 2 4/3/20 Monday 6
Bill 1 8/3/20 Tuesday 2
Based on this data, I want to select each row plus an extra column which is the number of days between the date from each row for each ID. Something like:
Select * from tblData
Inner join (datediff(Select Top(1) Date from tblData where Date < Date), Date) And ID = ID)
or for simplicity:
Select * from tblData
Inner join (datediff(Select Top(1) Date from tblData where Date < 8/3/20), 8/3/20) And ID = 1)
The resulting dataset would look like:
Name ID Date Day Hours DaysBtwn
Bill 1 3/3/20 Thursday 8 4 (Assuming there was an earlier row in the table)
Fred 2 4/3/20 Monday 6 5 (Assuming there was an earlier row in the table)
Bill 1 8/3/20 Tuesday 2 5 (Based on the previous row date being 3/3/20 for Bill)
Does this make sense and am I trying to do this the wrong way? I want to do this for about 600000 rows in table and therefore efficiency is the key, so if there is a better way to do this, i'm open to suggestions.
You can use lag():
select t.*, datediff(day, lag(date) over(partition by id order by date), date) diff
from mytable t
I think you just want lag():
select t.*,
datediff(day,
lag(date) over (partition by name order by date),
date
) as diff
from tblData t;
Note: If you want to filter the data so rows in the result set are used for the lag() but not in the result set, then use a subquery:
select t.*
from (select t.*,
datediff(day,
lag(date) over (partition by name order by date),
date
) as diff
from tblData t
) t
where date < '2020-08-03';
Also note the use of the date constant as a string in YYYY-MM-DD format.
Dear Stack Overflow community,
I am looking for the patient id where the two consecutive dates after the very first one are less than 7 days.
So differences between 2nd and 1st date <= 7 days
and differences between 3rd and 2nd date <= 7 days
Example:
ID Date
1 9/8/2014
1 9/9/2014
1 9/10/2014
2 5/31/2014
2 7/20/2014
2 9/8/2014
For patient 1, the two dates following it are less than 7 days apart.
For patient 2 however, the following date are more than 7 days apart (50 days).
I am trying to write an SQL query that just output the patient id "1".
Thanks for your help :)
You want to use lead(), but this is complicated because you want this only for the first three rows. I think I would go for:
select t.*
from (select t.*,
lead(date, 1) over (partition by id order by date) as next_date,
lead(date, 2) over (partition by id order by date) as next_date_2,
row_number() over (partition by id order by date) as seqnum
from t
) t
where seqnum = 1 and
next_date <= date + interval '7' day and
next_date2 <= next_date + interval '7' day;
You can try using window function lag()
select * from
(
select id,date,lag(date) over(order by date) as prevdate
from tablename
)A where datediff(day,date,prevdate)<=7
I'm pretty new to SQL and have this problem:
I have a filled table with a date column and other not interesting columns.
date | name | name2
2015-03-20 | peter | pan
2015-03-20 | john | wick
2015-03-18 | harry | potter
What im doing right now is counting everything for a date
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
what i want to do now is counting the resulting lines and only returning them if there are less then 10 resulting lines.
What i tried so far is surrounding the whole query with a temp table and the counting everything which gives me the number of resulting lines (yeah)
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
)
select count(*)
from temp_count
What is still missing the check if the number is smaller then 10.
I was searching in this Forum and came across some "having" structs to use, but that forced me to use a "group by", which i can't.
I was thinking about something like this :
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
)
select *
from temp_count
having count(*) < 10
maybe im too tired to think of an easy solution, but i can't solve this so far
Edit: A picture for clarification since my english is horrible
http://imgur.com/1O6zwoh
I want to see the 2 columned results ONLY IF there are less then 10 rows overall
I think you just need to move your having clause to the inner query so that it is paired with the GROUP BY:
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
having count(*) < 10
)
select *
from temp_count
If what you want is to know whether the total # of records (after grouping), are returned, then you could do this:
with temp_count (date, counter) as
(
select date, counter=count(*)
from testtable
where date >= current date - 10 days
group by date
)
select date, counter
from (
select date, counter, rseq=row_number() over (order by date)
from temp_count
) x
group by date, counter
having max(rseq) >= 10
This will return 0 rows if there are less than 10 total, and will deliver ALL the results if there are 10 or more (you can just get the first 10 rows if needed with this also).
In your temp_count table, you can filter results with the WHERE clause:
with temp_count (date, counter) as
(
select date, count(distinct date)
from testtable
where date >= current date - 10 days
group by date
)
select *
from temp_count
where counter < 10
Something like:
with t(dt, rn, cnt) as (
select dt, row_number() over (order by dt) as rn
, count(1) as cnt
from testtable
where dt >= current date - 10 days
group by dt
)
select dt, cnt
from t where 10 >= (select max(rn) from t);
will do what you want (I think)
I have this query in sql server 2012
select sum(user_number),
sum(media_number),
month_name from (
select TOP 100
count(distinct a.answer_group_guid) as 'user_number',
count(distinct a.media_guid) as 'media_number',
datename(mm,answer_datetime) as 'month_name' ,year(answer_datetime) as 'year'
from
tb_answers as a
left outer join
tb_media as m ON m.user_guid = 'userguid' and m.media_guid=a.media_guid
where
m.user_guid = 'userguid'
group by concat(year(answer_datetime),'',month(answer_datetime)),datename(mm,answer_datetime),year(answer_datetime)
order by year(answer_datetime) desc) as aa
group by month_name,year
order by month_name desc,year desc;
it get this result
Out
user_number media_number month_name
5 1 September
2 1 October
1 1 October
1 1 August
But I need only the first occurence of octuber month
as
user_number media_number month_name
5 1 September
2 1 October
1 1 August
You simply need to use a ranking function like ROW_NUMBER(). Use it to number the records partitioning by month_name, and select only the records which are number 1 in each partition, i.e.
Add this to the select list of your query:
ROW_NUMBER() OVER(PARTITION BY month_name ORDER By XXX) as RowNumber
This will number the rows which have the same month_name with consecutive numbers, starting by 1, and in the order specified by XXX.
NOTE: specify the order in XXX to decide which of the month rows is number one and will be returned by the query
And then, do a select from the resulting query, filtering by RowNumber = 1
SELECT Q.user_number, Q.media_number, Q.month_name
FROM(
-- your query + RowNumber) Q
WHERE Q.RowNumber = 1
NOTE: if you need some ordering in your result, you'll have to move the ORDER BY out of the subselect, and write it beside the WHERE Q.RowNumber=1