How can i use sql query for the following - sql-server

My data table sampletime in one column and sample value in another column contain data like follow
sampletime value
----------------------------
2016-03-02 08:31:14 1
2016-03-02 09:31:14 2
2016-03-02 12:31:14 3
2016-03-04 08:31:14 4
2016-03-04 09:31:14 5
2016-03-05 08:31:14 3
I need two minimum sample time in each day. How can I group?
Query
SELECT rn.sampletime AS stime
FROM rn_qos_data_0007 rn
INNER JOIN s_qos_data qos
ON qos.table_id = rn.table_id
AND qos.qos = 'QOS_CPU_USAGE'
AND Substring(qos.origin, 1, 4) = 'A0C3'
AND qos.host = '10.98.48.100'
WHERE rn.sampletime BETWEEN '2016/01/01' AND '2016/06/22'
GROUP BY rn.sampletime

You need ROW_NUMBER window function
Select * From
(
select row_number()over(partition by cast(sampletime as date) order by sampletime) RN,*
From ..
) A
Where RN <=2

Related

How to Sum (MAX values) from different value groups in same column SQL Server

I have a table like this:
Date
Consec_Days
2015-01-01
1
2015-01-03
1
2015-01-06
1
2015-01-07
2
2015-01-09
1
2015-01-12
1
2015-01-13
2
2015-01-14
3
2015-01-17
1
I need to Sum the max value (days) for each of the consecutive groupings where Consec_Days are > 1. So the correct result would be 5 days.
This is a type of gaps-and-islands problem.
There are many solutions, here is one simple one
Get the start points of each group using LAG
Calculate a grouping ID using a windowed conditional count
Group by that ID and take the highest sum
WITH StartPoints AS (
SELECT *,
IsStart = CASE WHEN LAG(Consec_Days) OVER (ORDER BY Date) = 1 THEN 1 END
FROM YourTable t
),
Groupings AS (
SELECT *,
GroupId = COUNT(IsStart) OVER (ORDER BY Date)
FROM StartPoints
WHERE Consec_Days > 1
)
SELECT TOP (1)
SUM(Consec_Days)
FROM Groupings
GROUP BY
GroupId
ORDER BY
SUM(Consec_Days) DESC;
db<>fiddle
with cte as (
select Consec_Days,
coalesce(lead(Consec_Days) over (order by Date), 1) as next
from YourTable
)
select sum(Consec_Days)
from cte
where Consec_Days <> 1 and next = 1
db<>fiddle

Joined tables - how to aggregate problem (sum)?

I have two tables joined like this:
SELECT count(DISTINCT T1.ContractNumber) AS nr_of_contracts,
count(T3.DateofInstallmentPayment) AS nr_of_paid_installents,
count(T3.DateofDueInstallment) AS nr_of_installments,
sum(T1.DisbursementAmount) AS disbursed_amount
FROM q.T1
LEFT JOIN q.T3
ON T1.ContractNumber=T3.ContractNumber
WHERE DateOfDisbursement BETWEEN '2019-12-01' AND '2019-12-31'
AND T3.DateofDueInstallment < GETDATE()
where T1 table contains data about clients (per contract number) and T3 about their payment schedules (per every instalment).
What I want is to have paid off amount (disbursement amount) of contracts from table T1 (aggregated by contract number) and not by every instalment. When I tried to select just sum(T1.Disbursement amount) then I receive sum but for all instalment which is incorrect.
T1:
Contract Number
DisbursementDate
Disbursement Amount
1
2019-12-01
1000
2
2019-12-01
2000
3
2019-12-01
3000
T3:
Contract Number
DateofDueInstallment
DateofInstallmentPayment
1
2020-01-01
2020-01-01
1
2020-02-01
2020-02-06
1
2020-03-01
2020-04-01
What I get after joining two tables for Contract Number = 1 is sum(DisbursementAmount) = 3000.
Contract Number
sum(DisbursementAmount)
1
3000
What I want after joining two tables for Contract Number = 1 is sum(DisbursementAmount) = 1000.
Contract Number
sum(DisbursementAmount)
1
1000
Something like this,not tested - a subquery with a different aggregation column
SELECT T1.product, T1.NrOfInstallment, count(DISTINCT T1.ContractNumber),
SELECT paid_amound FROM(SELECT ContractNumber, sum(tt.DisbursementAmount + (tt.ContractNumber*0.01))
- sum(tt.ContractNumber*0.01) as paid_amount
FROM abc.T1 as tt WHERE tt.ContractNumber = T1.ContractNumber GROUP BY ContractNumber) AS t)
FROM abc.T1
LEFT JOIN bde.T3
ON T1.ContractNumber=T3.ContractNumber
GROUP BY T1.product, T1.NrOfInstallment

SQL Server - select column using in having count()

This is my first question (and sorry for my English)
I have this table in SQL Server:
id_patient | date | id_drug
----------------------------------------------------
1 20200101 A
1 20200102 A
1 20200103 A
1 20200104 A
1 20200105 A
1 20200110 A
2 20200101 A
2 20200105 B
2 20200106 C
2 20200107 D
2 20200108 E
2 20200110 L
3 20200101 A
3 20200102 A
3 20200103 A
3 20200104 A
3 20200105 C
3 20200106 C
4 20200105 A
4 20200106 D
4 20200107 D
5 20200105 A
5 20200106 A
5 20200107 C
5 20200108 D
I would like to extract patient and drug for all patients who have taken at least 3 different drugs in a given period
I have tried:
select id_patient, count(distinct ID_drug)
from table
where date between XXX and YYY
group by id_patient
having count(Distinct ID_drug) > 3
but in this way -YES- I get all patients with 3 or more different id_drug in this date range but I can't get the ID_drug because in the count()
For example, I'd like to obtain:
Who help me ?
Thanks
You can use string_agg() in the most recent versions of SQL Server:
select id_patient, count(distinct ID_drug),
string_agg(id_drug, ',')
from table
where date between XXX and YYY
group by id_patient
having count(Distinct ID_drug) > 3;
If you want the original rows, you can use window functions. Unfortunately, SQL Server does not support count(distinct) as a window function, but there is an easy work-around using dense_rank():
select t.*
from (select t.*,
(dense_rank() over (partition by id_patient order by id_drug) +
dense_rank() over (partition by id_patient order by id_drug desc)
) as num_drugs
from t
where . . .
) t
where num_drugs >= 3;
SELECT id_patient,
ID_drug
FROM table
WHERE id_patient IN (
SELECT id_patient
FROM table
WHERE date
BETWEEN XXX
AND YYY
GROUP BY id_patient
HAVING COUNT(DISTINCT ID_drug) >= 3
)
GROUP BY id_patient,
ID_drug;

Finding max date difference on a single column

in the below table example - Table A, we have entries for four different ID's 1,2,3,4 with the respective status and its time. I wanted to find the "ID" which took the maximum amount of time to change the "Status" from Started to Completed. In the below example it is ID = 4. I wanted to run a query and find the results, where we currently has approximately million records in a table. It would be really great, if someone provide an effective way to retrieve this data.
Table A
ID Status Date(YYYY-DD-MM HH:MM:SS)
1. Started 2017-01-01 01:00:00
1. Completed 2017-01-01 02:00:00
2. Started 2017-10-02 03:00:00
2. Completed 2017-10-02 05:00:00
3. Started 2017-15-03 06:00:00
3. Completed 2017-15-03 09:00:00
4. Started 2017-22-04 10:00:00
4. Completed 2017-22-04 15:00:00
Thanks!
Bruce
You can query as below:
Select top 1 with ties Id from #yourDate y1
join #yourDate y2
On y1.Id = y2.Id
and y1.[STatus] = 'Started'
and y2.[STatus] = 'Completed'
order by Row_number() over(order by datediff(mi,y1.[Date], y2.[date]) desc)
SELECT
started.ID, timediff(completed.date, started.date) as elapsed_time
FROM TABLE_A as started
INNER JOIN TABLE_A as completed ON (completed.ID=started.ID AND completed.status='Completed')
WHERE started.status='Started'
ORDER BY elapsed_time desc
be sure there's a index on TABLE_A for the columns ID, date
I haven't run this sql but it may solve your problem.
select a.id, max(DATEDIFF(SECOND, a.date, b.date + 1)) from TableA as a
join TableA as b on a.id = b.id
where a.status="started" and b.status="completed"
Here's a way with a correlated sub-query. Just uncomment the TOP 1 to get ID 4 in this case. This is based off your comments that there is only 1 "started" record, but could be multiple "completed" records for each ID.
declare #TableA table (ID int, Status varchar(64), Date datetime)
insert into #TableA
values
(1,'Started','2017-01-01 01:00:00'),
(1,'Completed','2017-01-01 02:00:00'),
(2,'Started','2017-02-10 03:00:00'),
(2,'Completed','2017-02-10 05:00:00'),
(3,'Started','2017-03-15 06:00:00'),
(3,'Completed','2017-03-15 09:00:00'),
(4,'Started','2017-04-22 10:00:00'),
(4,'Completed','2017-04-22 15:00:00')
select --top 1
s.ID
,datediff(minute,s.Date,e.EndDate) as TimeDifference
from #TableA s
inner join(
select
ID
,max(Date) as EndDate
from #TableA
where Status = 'Completed'
group by ID) e on e.ID = s.ID
where
s.Status = 'Started'
order by
datediff(minute,s.Date,e.EndDate) desc
RETURNS
+----+----------------+
| ID | TimeDifference |
+----+----------------+
| 4 | 300 |
| 3 | 180 |
| 2 | 120 |
| 1 | 60 |
+----+----------------+
If you know that 'started' will always be the earliest point in time for each ID and the last 'completed' record you are considering will always be the latest point in time for each ID, the following should have good performance for a large number of records:
SELECT TOP 1
id
, DATEDIFF(s, MIN([Date]), MAX([date])) AS Elapsed
FROM #TableA
GROUP BY ID
ORDER BY DATEDIFF(s, MIN([Date]), MAX([date])) DESC

How to get non matching records in sql server in same table?

I have a table say StudentBillDetails and in this table data is saved annually and yrid is referenced to some other table. Now I am stuck with a problem. I want to retrieve non matching records as described below.
Stid BillNo Yrid
1 525 3
1 525 1
1 525 4
2 443 4
2 442 1
2 443 3
In above given table structure as you can see for three years StId 1 has same value but StId 2 has a confliction in Yrid 1. So I want to get these type of records.
If you just want to flag Stid values which have conflicts then the following simple query should work:
SELECT Stid
FROM yourTable
GROUP BY Stid
HAVING COUNT(DISTINCT BillNo) > 1
If you want the entire records you could try joining your table to the above query:
SELECT t1.*
FROM yourTable t1
INNER JOIN
( SELECT Stid FROM yourTable GROUP BY Stid HAVING COUNT(DISTINCT BillNo) > 1 ) t2
ON t1.Stid = t2.Stid

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