imagine i have a table like this
Bikename Username NumOfUsages LastTimeOfUsage
Bike1 Haldi 5 2018-12-13 12:00:00
Bike1 Torte 1 2018-08-15 12:00:00
Bike2 Haldi 3 2018-12-15 12:00:00
Bike3 Manne 2 2018-09-16 12:00:00
Bike3 Torte 5 2018-09-16 12:00:00
now i wants a Result like this
Bikename Username NumOfUsages LastTimeOfUsage
Bike1 Haldi 5 2018-12-13 12:00:00
Bike2 Haldi 3 2018-12-15 12:00:00
Bike3 Torte 5 2018-09-16 12:00:00
as you can see i wants to have the Entry with the MAX NumofUsages only
Thanks for your Help so much...
You want correlated subquery :
select b.*
from Bikes b
where b.NumOfUsages = (select max(b1.NumOfUsages) from Bikes b1 where b1.bikename = b.bikename);
By this way, it will return entire row for maximum NumOfUsages for each bikename since it has referenced from outer query with subquery (i.e. b1.bikename = b.bikename).
you can use row_number() maximum dbms support
select * from
(
select *, row_number() over(partition by bikename order by NumOfUsages desc) rn
from table_name
)t where t.rn=1
You didn't specify the SQL flavor in your question, but in most cases simply group by the bikename and take max(numOfUsages) (and order by that as well)
Related
I have a requirement in which I need to retrieve rows in a select query in which I have to get value of END_DATE as EFFECTIVE_DATE -1 DAY for the records with same key (CARD_NBR in this case)
I have tried using it by GROUP by but I am not able to get the desired output. Could someone please help in guiding me ? The record with most recent effective date should have END_DATE as 9999-12-31 only.
Table:
CARD_NBR
SERIEL_NO
EFFECTIVE_DATE
END_DATE
12345
1
2021-01-01
9999-12-31
12345
2
2021-01-25
9999-12-31
12345
3
2021-02-15
9999-12-31
67899
1
2021-03-01
9999-12-31
67899
2
2021-04-02
9999-12-31
67899
3
2021-05-24
9999-12-31
Output:
CARD_NBR
SERIEL_NO
EFFECTIVE_DATE
END_DATE
12345
1
2021-01-01
2021-01-24
12345
2
2021-01-25
2021-02-14
12345
3
2021-02-15
9999-12-31
67899
1
2021-03-01
2021-04-01
67899
2
2021-04-02
2021-05-24
67899
3
2021-05-24
9999-12-31
You can use lead():
select t.*,
lead(effective_date - interval '1 day', 1, effective_date) over (partition by card_nbr order by effective_date) as imputed_end_date
from t;
Date manipulations are highly database-dependent so this uses Standard SQL syntax. You can incorporate this into an update, but the best approach also depends on the database.
SQLite v.3.25 now supports windows function and you can use below code to get your result.
SELECT A.CARD_NBR,
A.SRL_NO,
A.START_DT,
COALESCE(B.START_DT,A.END_DT) AS END_DT
FROM
(
SELECT A.CARD_NBR,
A.SRL_NO,
A.START_DT,
A.END_DT,
ROW_NUMBER() OVER(PARTITION BY A.CARD_NBR ORDER BY A.SRL_NO ASC) RNUM1
FROM T1 A
)A
LEFT JOIN
(
SELECT B.CARD_NBR,
B.SRL_NO,
B.START_DT,
B.END_DT,
ROW_NUMBER() OVER(PARTITION BY B.CARD_NBR ORDER BY B.SRL_NO ASC) RNUM1
FROM T1 B
)B
ON A.CARD_NBR=B.CARD_NBR
AND A.RNUM1+1=B.RNUM1
I am trying to get the time between two dates in two different rows.
dataset:
user
id
time
status
user3
4
2021-02-01 14:00:00
OUT
user1
2
2021-02-01 12:00:00
OUT
user2
1
2021-02-01 10:00:00
OUT
user1
2
2021-02-01 09:00:00
IN
user2
1
2021-02-01 08:00:00
IN
user3
4
2021-02-01 08:30:00
IN
What I am trying to obtain :
user
id
time
user3
4
07:30:00
user1
2
03:00:00
user2
1
02:00:00
The rows can be all mixed up since I dont know whenever a user will open (IN) or close (OUT) the job.
So far i tried to make a first query :
SELECT *
FROM table
WHERE DATE(date) BETWEEN '2021-02-01' AND '2021-02-02'
ORDER BY date DESC;
And then I was thinking looping on the result array, and make the math from there, but I was wondering if there is anyway to do this in SQL only.
You can use comditional aggregation as follows:
SELECT user, id,
Timediff( Max(case when status='OUT' then time end),
Min(case when status='IN' then time end) ) as diff
FROM table
WHERE DATE(date) BETWEEN '2021-02-01' AND '2021-02-02'
Group by user, id;
You can sort the result as out first then get the previous time from next row using Lead().
Though I am no expert in mariaDB but below query should work:
with userresult as(
SELECT *,lead(TIME,1)OVER(PARTITION BY id ORDER BY STATUS DESC) previoustime,ROW_NUMBER()over(partition by id order by status desc) rownum FROM USERS
where DATE(date) BETWEEN '2021-02-01' AND '2021-02-02'
)
select user,id,timediff(time,previoustime)time from userresult where rownum=1
If your mariaDB version is lower than 10.2 you can try simple sql like below:
select username,id,timediff(time,(select max(time) from users u where status='in'and u.id=us.id ))
from users us where status='out'
Try this simple solution:
SELECT T1.id, T1.time AS INTime, T1_1.time AS OUTTime
FROM T1
INNER JOIN T1 AS T1_1 ON (T1.id = T1_1.id AND T1_1.Status="OUT")
WHERE T1.Status="IN";
I have one table called prices that have a reference from table products through product_id column. I want a query that selects prices grouped by product_id with the max final date and get the value of start_date through one select with id of price grouped.
I try with the following query but I am getting a wrong value of start date. Is weird because of the result subquery return more than one row even though I use the price id on where clause. Because that I put the limit on the query but it is wrong.
select prices.produto_id, prices.id,
MAX(CASE WHEN prices.finish_date IS NULL THEN COALESCE(prices.finish_date,'9999-12-31') ELSE prices.finish_date END) as finish_date,
(select start_date from prices where prices.id = prices.id limit 1)
as start_date from prices group by prices.product_id, prices.id
How I can get the relative start date of the price id in my grouped row? I am using postgresql.
A example to view what I want with my query:
DataSet:
ID | PRODUCT_ID | START_DATE | FINISH_DATE
1 1689 2018-01-19 02:00:00 2019-11-19 23:59:59
2 1689 2019-10-11 03:00:00 2019-10-15 23:59:59
3 1689 2019-01-11 03:00:00 2019-05-15 23:59:59
4 1690 2019-11-11 03:00:00 2019-12-15 23:59:59
5 1690 2019-05-11 03:00:00 2025-12-15 23:59:59
6 1691 2019-05-11 03:00:00 null
I want this result:
ID | PRODUCT_ID | START_DATE | FINISH_DATE
1 1689 2018-01-19 02:00:00 2019-11-19 23:59:59
5 1690 2019-05-11 03:00:00 2025-12-15 23:59:59
6 1691 2019-05-11 03:00:00 9999-12-31 23:59:59
The start date should be the same value of the row before the group by.
I would recommend DISTINCT ON in Postgres:
select distinct on (p.product_id) p.*
from prices p
order by p.product_id,
p.finish_date desc nulls first;
NULL values are treated as larger than any other value, so a descending sort puts them first. However, I've included nulls first just to be explicit.
DISTINCT ON is a very handy Postgres extension, which you can learn more about in the documentation.
Try this
with data as (
SELECT id, product_id,
max(COALESCE(finish_date,'9999-12-31')) as finish_date from prices group by 1,2)
select d.*, p.start_date from data d join prices p on p.id = d.id;
It surely isnt' the most elegant solution, but it should work.
I have the following table, I am using SQL Server 2008
BayNo FixDateTime FixType
1 04/05/2015 16:15:00 tyre change
1 12/05/2015 00:15:00 oil change
1 12/05/2015 08:15:00 engine tuning
1 04/05/2016 08:11:00 car tuning
2 13/05/2015 19:30:00 puncture
2 14/05/2015 08:00:00 light repair
2 15/05/2015 10:30:00 super op
2 20/05/2015 12:30:00 wiper change
2 12/05/2016 09:30:00 denting
2 12/05/2016 10:30:00 wiper repair
2 12/06/2016 10:30:00 exhaust repair
4 12/05/2016 05:30:00 stereo unlock
4 17/05/2016 15:05:00 door handle repair
on any given day need do find the highest number of fixes made on a given bay number, and if that calculated number is repeated then it should also appear in the resultset
so would like to see the result set as follows
BayNo FixDateTime noOfFixes
1 12/05/2015 00:15:00 2
2 12/05/2016 09:30:00 2
4 12/05/2016 05:30:00 1
4 17/05/2016 15:05:00 1
I manage to get the counts of each but struggling to get the max and keep the highest calculated repeated value. can someone help please
Use window functions.
Get the count for each day by bayno and also find the min fixdatetime for each day per bayno.
Then use dense_rank to compute the highest ranked row for each bayno based on the number of fixes.
Finally get the highest ranked rows.
select distinct bayno,minfixdatetime,no_of_fixes
from (
select bayno,minfixdatetime,no_of_fixes
,dense_rank() over(partition by bayno order by no_of_fixes desc) rnk
from (
select t.*,
count(*) over(partition by bayno,cast(fixdatetime as date)) no_of_fixes,
min(fixdatetime) over(partition by bayno,cast(fixdatetime as date)) minfixdatetime
from tablename t
) x
) y
where rnk = 1
Sample Demo
You are looking for rank() or dense_rank(). I would right the query like this:
select bayno, thedate, numFixes
from (select bayno, cast(fixdatetime) as date) as thedate,
count(*) as numFixes,
rank() over (partition by cast(fixdatetime as date) order by count(*) desc) as seqnum
from t
group by bayno, cast(fixdatetime as date)
) b
where seqnum = 1;
Note that this returns the date in question. The date does not have a time component.
I have a table
TIMESTAMP ID Name
5/30/2016 11:45 1 Ben
5/30/2016 11:45 2 Ben
5/30/2016 23:15 2 Ben
5/30/2016 7:30 1 Peter
5/30/2016 6:05 1 Peter
5/30/2016 14:40 2 May
5/30/2016 1:05 1 May
Now, I need to get the MIN timestamp for each distinct Name.
Then if there are more than one MIN entry, choose the one with the MAX ID.
So the result should be
TIMESTAMP ID Name
5/30/2016 11:45 2 Ben
5/30/2016 6:05 1 Peter
5/30/2016 1:05 1 May
I tried using the query below:
SELECT MIN(TIMESTAMP),NAME FROM TBLSAMPLE WHERE TIMESTAMP BETWEEN TO_DATE('5/30/2016', 'MM/DD/YYYY' ) AND TO_DATE('5/30/2016', 'MM/DD/YYYY' ) + 1
GROUP BY NAME
and I could get the minimum time. But once I add in MAX(ID) the result return an entry that does not match any of the rows.
Your help are really appreciated.
You can do this with row_number():
select t.*
from (select t.*,
row_number() over (partition by name order by timestamp asc, id desc) as seqnum
from tblsample t
) t
where seqnum = 1;
Your question doesn't specify a condition on the dates. But if you want to add a where clause, then add it to the subquery.