I have the current sql table bellow:
id user date
1 john 2021-08-20
3 john 2021-08-24
5 john 2021-08-25
8 will 2021-08-25
9 will 2021-08-20
6 will 2021-08-18
I need to return the id's who have the most recent date, and with that, return a count of how many times the user appeared. And the id isn't always numered in crescent order by date, as the example bellow.
id user count
5 john 3
8 will 3
You can use qualify to get the most recent row:
select t.*
from t
where 1=1
qualify row_number() over (partition by name order by date desc) = 1;
You can add a window function for the count:
select t.*,
count(*) over (partition by name) as cnt
from t
where 1=1
qualify row_number() over (partition by name order by date desc) = 1;
You can use the following query:
SELECT id, user, count(user) AS count
FROM table_name
Group by user
having date = max(date);
Related
Can anyone help me? I have a table result like this:
id_user
score
type
001
30
play
001
40
play
001
30
redeem
002
20
play
002
30
redeem
I want to sum column score group by id_user base on type 'play' and after that I want show ranking using find_in_set. Like this is the result of the table that I want to display:
id_user
total
rank
001
70
1
002
20
2
Previously I used the rank() function in MySQL version 10.4, but it does not work in MySQL version 15.1. This is my previous query:
SELECT id_user, SUM(score) AS total,
RANK() OVER (ORDER BY total DESC) AS rank
FROM result
WHERE type='play'
GROUP BY id_user
I have made some changes in your query. It's working now. Instead of column alias total SUM(score) needs to be used in order by clause of Rank() function's over(). And since Rank is a reserve word I used rnk instead.
DB-Fiddle:
create table result (id_user varchar(5), score int, type varchar(20));
insert into result values('001',30 ,'play');
insert into result values('001',40 ,'play');
insert into result values('001',30 ,'redeem');
insert into result values('002',20 ,'play');
insert into result values('002',30 ',redeem');
Query:
select id_user, SUM(score) AS total, RANK() OVER (ORDER BY SUM(score) DESC) AS rnk FROM result where type='play' GROUP BY id_user
Output:
id_user
total
rnk
001
70
1
002
20
2
db<>fiddle here
If your MySQL version doesn't support rank() you can use subquery to achieve same result:
Query:
select id_user, SUM(score) AS total,
coalesce((select count(distinct id_user) from result r2
where type='play'
group by id_user
having sum(r2.score)>sum(r.score) ),0)+1 AS rnk
FROM result r where type='play'
GROUP BY id_user
Output:
id_user
total
rnk
001
70
1
002
20
2
db<>fiddle here
I have a table of records I want to sort by earliest date first then by userid.
If the user associated to the date also has other records in that table I want to group those under the earliest date.
Desired output
Id UserId Date
1 2 1/1/2020
2 2 2/1/2020
3 2 3/1/2020
4 1 1/2/2020
5 1 2/2/2020
6 3 1/4/2020
7 4 1/5/2020
In this example UserId 2 has the earliest record in that table, so that record should be first followed by his additional records in date asc order
You seems want :
select t.*
from table t
order by min(date) over (partition by userid), date;
Some database product doesn't support window function with order by, so you can do instead :
select t.*, min(date) over (partition by userid) as mndate
from table t
order by mndate, date;
If I understand what you want...
You could do this (sample with DB2 syntax):
SELECT tab.UserId, tab.Date, tab.*
FROM DB2SIS.TABLE_NAME tab
ORDER BY tab.Date ASC, tab.UserId ASC
This way UserId and Date will appear repeatedly. Instead of 'tab.*' use each field you want to show, then UserId and Date will not repeat.
I have a table that looks like the following
id effective_date number_of_int_customers
123 10/01/19 0
123 02/01/20 3
456 10/01/19 6
456 02/01/20 6
789 10/01/19 5
789 02/01/20 4
999 10/01/19 0
999 02/01/20 1
I want to write a query that looks at each ID to see if the salespeople have newly started working internationally between October 1st and February 1st.
The result I am looking for is the following:
id effective_date number_of_int_customers
123 02/01/20 3
999 02/01/20 1
The result would return only the salespeople who originally had 0 international customers and now have at least 1.
I have seen similar posts here that use nested queries to pull records where the first date and last have different values. But I only want to pull records where the original value was 0. Is there a way to do this in one query in SQL?
In your case, a simple aggregation would do -- assuming that 0 is the earliest value:
select id, max(number_of_int_customers)
from t
where effective_date in ('2019-10-01', '2020-02-01')
group by id
having min(number_of_int_customers) = 0;
Obviously, this is not correct if the values can decrease to zero. But this having clause fixes that problem:
having min(case when number_of_int_customers = 0 then effective_date end) = min(effective_date)
An alternative is to use window functions, such asfirst_value():
select distinct id, last_noic
from (select t.*,
first_value(number_of_int_customers) over (partition by id order by effective_date) as first_noic,
first_value(number_of_int_customers) over (partition by id order by effective_date desc) as last_noic,
from t
where effective_date in ('2019-10-01', '2020-02-01')
) t
where first_noic = 0;
Hmmm, on second thought, I like lag() better:
select id, number_of_int_customers
from (select t.*,
lag(number_of_int_customers) over (partition by id order by effective_date) as prev_noic
from t
where effective_date in ('2019-10-01', '2020-02-01')
) t
where prev_noic = 0;
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;
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