So I have a table like this:
Taxi Client Time
Tom A 1
Tom A 2
Tom B 3
Tom A 4
Tom A 5
Tom A 6
Tom B 7
Tom B 8
Bob A 1
Bob A 2
Bob A 3
and the expected result will be like this:
Tom 3
Bob 1
I have used the partition function to count the consecutive value but the result become this:
Tom A 2
Tom A 3
Tom B 2
Bob A 2
Please help, I am not good in English, thanks!
This is a variation of a gaps-and-islands problem. You can solve it using window functions:
select taxi, count(*)
from (select t.taxi, t.client, count(*) as num_times
from (select t.*,
row_number() over (partition by taxi order by time) as seqnum,
row_number() over (partition by taxi, client order by time) as seqnum_c
from t
) t
group by t.taxi, t.client, (seqnum - seqnum_c)
having count(*) >= 2
)
group by taxi;
use distinct count
select taxi ,count( distinct cient)
from table_name
group by taxi
It seems your expected output is wrong
I don't see where you get the number 3 from. If you're trying to do what your question says and group by client in consecutive order only and then get the number of different groups, I can help you out with the following query. Bob has 1 group and Tom has 4.
Partition by taxi, ORDER BY taxi, time and check if this client matches the previous client for this taxi. If yes, do not count this row. If no, count this row, this is a new group.
SELECT FEE.taxi,
SUM(FEE.clientNotSameAsPreviousInSequence)
FROM
(
SELECT taxi,
CASE
WHEN PreviousClient IS NULL THEN
1
WHEN PreviousClient <> client THEN
1
ELSE
0
END AS clientNotSameAsPreviousInSequence
FROM
(
SELECT *,
LAG(client) OVER (PARTITION BY taxi ORDER BY taxi, time) AS PreviousClient
FROM table
) taxisWithPreviousClient
) FEE
GROUP BY FEE.taxi;
Related
LoginHistory table
Date Name Login
----------------------------------------
03/20/2021 Amy 1
03/20/2021 Lily 1
03/20/2021 Nancy 1
03/21/2021 Amy 1
03/21/2021 Lily 1
03/21/2021 Leo 1
03/22/2021 Amy 1
03/22/2021 Lisa 1
03/22/2021 Leo 1
03/23/2021 Lily 1
03/23/2021 Lisa 1
03/23/2021 Leo 1
I want to find the people and their login date who was login instance 3 times in consecutive dates. For example, my output should has Amy, because she was login 3/20,3/21 and 3/22. For Lily, she shouldn't be in my output, because even she login 3 times, the date(3/20,3/21 and 3/23) is not in consecutive order.
output should be:
Date Name Login
----------------------------------------
03/20/2021 Amy 1
03/21/2021 Amy 1
03/21/2021 Leo 1
03/22/2021 Amy 1
03/22/2021 Leo 1
03/23/2021 Leo 1
Thanks.
Based on the specific sample data provided, you could use analytic min and max to get the first and last date for each name, count the difference in days and the number of logins which must be 3 with 2 days between first and last date.
You haven't specific a RDBMS so the date functions may need amending as appropriate, however all RDBMS support the same functionality.
select date, name
from (
select *,
DateDiff(day,Min(date) over(partition by name),
Max(date) over(partition by name))diff,
Count(*) over(partition by name) qty
from t
)t
where diff=2 and qty=3
order by date;
To produce a table of the consecutive logins, you can first anchor your search on the action that is the last in the sequence. Then, you can join all the preceding dates to that original result:
with vals(v) as (
select 1
union all
select 2
)
select c2.* from (
select c.* from loginhistory c where
(select count(*) from loginhistory c1 cross join vals v
where c1.name = c.name and c.dt = c1.dt + interval '1' day * v.v) = 2
) t1
join loginhistory c2 on t1.name = c2.name and c2.dt <= t1.dt and (c2.dt + interval '2' day) >= t1.dt
order by c2.dt
select * from LoginHistory where name in (
select name
from LoginHistory
where date between <start> and <end> -- must be exactly three dates in the range
group by name
having count(distinct date) = 3
)
select manager, count(*) over (partition by manager) cnt
from dbtable
group by manager
This will provide me the count of manager but if I need a count of senior_manager how will I get it?
|--------------------|------------------|
| Manager |Senior_Manager |
|--------------------|------------------|
| John |Arpit |
| John |govind |
| John |olive |
| Domnic |kelvin |
| Domnic |paul |
|--------------------|------------------|
Result
John 3
Domnic 2
Your code returns "1" for all managers -- because it counts the number of rows after the group by.
If you want to count the number of rows in the table for a given manager, then you want aggregation, not analytic functions:
Select manager, count(*) as cnt
from dbtable
group by manager;
I'm not sure if this answers your question, but it at least addresses the issue that the your query does not do much that is useful.
EDIT:
For the revised question, it simply seems:
Select senior_manager, count(*) as cnt
from dbtable
group by senior_manager;
The result you wanted can be retrieved by
select manager, count(*) over (partition by manager) cnt
from dbtable
This means each manager will be associated with the count of rows in the partition where {manager} value equals that exact manager. According to the table above this is what you expect to get.
Your example:
select manager, count(*) over (partition by manager) cnt
from dbtable
group by manager
Yields the following results:
MANAGER CNT
Domnic 1
John 1
If you drop the group by, you get:
MANAGER CNT
Domnic 2
Domnic 2
John 3
John 3
John 3
Are those the counts you're looking for? If so, then you can eliminate the duplicate rows with distinct:
select distinct manager, count(*) over (partition by manager) cnt
from dbtable
Which gives:
MANAGER CNT
John 3
Domnic 2
I have the following table:
NAME | SCORE
ALICE | 100
BOB | 90
CHARLES| 90
DUKE | 80
EVE | 70
...
My question is the following:
How can I extract with one query the name of the three best players? In my example the query should return four rows (ALICE, BOB, CHARLES and DUKE) because there are two silver medalists (they both have 90 points).
Thank You in advance.
Oracle has the DENSE_RANK analytical function for that exact purpose:
select name, score from (
select name, score, dense_rank() over(order by score desc nulls last) rank
-- ^^^^^^^^^^
-- reject NULL score at the end
from t
) V
where rank < 4
order by rank, name
See http://sqlfiddle.com/#!4/88445/5
How about the following
select *
from table1
where score >=
(select score from (
select score, rownum r from (
select distinct score from table1 order by score desc
) where rownum <= 3
) where r = 3)
order by score desc
See also this SQLFiddle: http://sqlfiddle.com/#!4/23e68/1
I have the following table . I need to get the last but one event associate for each event
event_id event_date event_associate
1 2/14/2014 ben
1 2/15/2014 ben
1 2/16/2014 steve
1 2/17/2014 steve // this associate is the last but one for event 1
1 2/18/2014 paul
2 2/19/2014 paul
2 2/20/2014 paul // this associate is the last but one for event 2
2 2/21/2014 ben
3 2/22/2014 paul
3 2/23/2014 paul
3 2/24/2014 ben
3 2/25/2014 steve // this associate is the last but one for event 3
3 2/26/2014 ben
I need to find out who was the last but one event_associate for each event . The result should be
event_id event_associate
1 steve
2 paul
3 steve
I know in order to do this I need to maximize event_date and exclude the last event_associate
So I tried
SELECT event_id , event_associate
WHERE NOT EXISTS (
SELECT *
FROM mytable
WHERE event_date = MAX(event_date)
)
QUALIFY ROW_NUMBER() OVER ( PARTITION BY event_id ORDER BY event_date DESC) = 1
But I do not know how to use EXISTS in this case .
You are quite close, you just need the 2nd row based on ROW_NUMBER:
select t.*,
row_number()
over (partition by event_id
order by event_date desc)
from tab as t
qualify
row_number()
over (partition by event_id
order by event_date desc) = 2
-- or simply
-- qualify rn = 2
I am facing a very common issue regarding "Selecting top N rows for each group in a table".
Consider a table with id, name, hair_colour, score columns.
I want a resultset such that, for each hair colour, get me top 3 scorer names.
To solve this i got exactly what i need on Rick Osborne's blogpost "sql-getting-top-n-rows-for-a-grouped-query"
That solution doesn't work as expected when my scores are equal.
In above example the result as follow.
id name hair score ranknum
---------------------------------
12 Kit Blonde 10 1
9 Becca Blonde 9 2
8 Katie Blonde 8 3
3 Sarah Brunette 10 1
4 Deborah Brunette 9 2 - ------- - - > if
1 Kim Brunette 8 3
Consider the row 4 Deborah Brunette 9 2. If this also has same score (10) same as Sarah, then ranknum will be 2,2,3 for "Brunette" type of hair.
What's the solution to this?
If you're using SQL Server 2005 or newer, you can use the ranking functions and a CTE to achieve this:
;WITH HairColors AS
(SELECT id, name, hair, score,
ROW_NUMBER() OVER(PARTITION BY hair ORDER BY score DESC) as 'RowNum'
)
SELECT id, name, hair, score
FROM HairColors
WHERE RowNum <= 3
This CTE will "partition" your data by the value of the hair column, and each partition is then order by score (descending) and gets a row number; the highest score for each partition is 1, then 2 etc.
So if you want to the TOP 3 of each group, select only those rows from the CTE that have a RowNum of 3 or less (1, 2, 3) --> there you go!
The way the algorithm comes up with the rank, is to count the number of rows in the cross-product with a score equal to or greater than the girl in question, in order to generate rank. Hence in the problem case you're talking about, Sarah's grid would look like
a.name | a.score | b.name | b.score
-------+---------+---------+--------
Sarah | 9 | Sarah | 9
Sarah | 9 | Deborah | 9
and similarly for Deborah, which is why both girls get a rank of 2 here.
The problem is that when there's a tie, all girls take the lowest value in the tied range due to this count, when you'd want them to take the highest value instead. I think a simple change can fix this:
Instead of a greater-than-or-equal comparison, use a strict greater-than comparison to count the number of girls who are strictly better. Then, add one to that and you have your rank (which will deal with ties as appropriate). So the inner select would be:
SELECT a.id, COUNT(*) + 1 AS ranknum
FROM girl AS a
INNER JOIN girl AS b ON (a.hair = b.hair) AND (a.score < b.score)
GROUP BY a.id
HAVING COUNT(*) <= 3
Can anyone see any problems with this approach that have escaped my notice?
Use this compound select which handles OP problem properly
SELECT g.* FROM girls as g
WHERE g.score > IFNULL( (SELECT g2.score FROM girls as g2
WHERE g.hair=g2.hair ORDER BY g2.score DESC LIMIT 3,1), 0)
Note that you need to use IFNULL here to handle case when table girls has less rows for some type of hair then we want to see in sql answer (in OP case it is 3 items).