SQL Server datetime ranges between records - sql

What would be the best way to get datetime ranges between records in SQL Server? I think it would be easiest to explain with an example.
I have the following data - these records start and end datetime ranges would never overlap:
ID
Start
End
1
1/27/2021 06:00:00
1/27/2021 09:00:00
2
1/27/2021 10:00:00
1/27/2021 14:00:00
3
1/27/2021 21:00:00
1/28/2021 04:00:00
4
1/28/2021 06:00:00
1/28/2021 09:00:00
I need to get the date time range between records. So the resulting SQL query would return the following result set (ID doesn't matter):
ID
Start
End
1
1/27/2021 09:00:00
1/27/2021 10:00:00
2
1/27/2021 14:00:00
1/27/2021 21:00:00
3
1/28/2021 04:00:00
1/28/2021 06:00:00
Thanks for any help in advance.


Use lead():
select t.*
from (select id, end as start, lead(start) over (order by start) as end
from t
) t
where end is not null;
Note: end is a lousy name for a column, given that it is a SQL keyword. I assume it is for illustrative purposes only.
Here is a SQL Fiddle.

Related

How can i create a new column count in SQL table where count=1 if hours column >=6 else count=0

I aim to first achieve this
id
employee
Datelog
TimeIn
TimeOut
Hours
Count
5
Two
2022-08-10
09:00:00
16:00:00
07:00:00
1
4
Two
2022-08-09
09:00:00
16:00:00
07:00:00
1
3
Two
2022-08-08
09:00:00
16:00:00
07:00:00
1
2
One
2022-08-05
09:00:00
16:00:00
07:00:00
1
1
Two
2022-08-04
09:00:00
10:00:00
01:00:00
0
and now my main objective here is to give a bonus of 2k to employees whose Totalcount per month >=3.
employee
Month
TotalCount
Bonus
Two
August
3
2000
One
August
1
0

Here's the answer using Postgres. It's pretty much generic other than extracting the month out of datelog that might have a slightly different syntax.
select employee
,max(date_part('month', datelog ))
,count(*)
,case when count(*) >= 3 then 2000 else 0 end as bonus
from t
where hours >= time '06:00:00'
group by employee
employee
max
count
bonus
Two
8
3
2000
One
8
1
0
Fiddle

SQL SELECT Difference between two days greater than 1 day

I have table T1
ID SCHEDULESTART SCHEDULEFINISH
1 2018-05-12 14:00:00 2018-05-14 11:00:00
2 2018-05-30 14:00:00 2018-06-01 11:00:00
3 2018-02-28 14:00:00 2018-03-02 11:00:00
4 2018-02-28 14:00:00 2018-03-01 11:00:00
5 2018-05-30 14:00:00 2018-05-31 11:00:00
I want to select all rows where difference in days (it's not important difference in hours) is greater than 1 day.
If SCHEDULESTART or SCHEDULEFINISH are on the same day or SCHEDULEFINISH is on next day then these rows should NOT be selected.
So the result should return rows with IDs: 1 2 3
because first row have difference in two days, second row (1st June is 2 days after 30th May ) and 3rd row (2nd March is 2 days after 28 February).
Is this possible somehow?
I know the function DAY but this will return only day number in that one month!!!
I must beging my query with
SELECT ID FROM T1 WHERE ...
Thanks in advance

In DB2, this should work:
select t1.*
from t1
where date(schedulestart) < date(schedulefinish) - 1 day;

Total time calculation in a sql query for a day where time in 24 hour format as hhmm

I have a table with date(date), left time(varchar2(4)) and arrival time(varchar2(4)). Time taken is in 24 hour format as hhmm. If a person travel 3 times a day, what will be the query to calculate total travel time in a day?
I am using oracle 11g. Kindly help. Thank you.

Convert the value to a number and report in minutes:
select to_number(substring(time, 1, 2))*60 + to_number(substring(time, 3, 2)) as minutes
Your query would look something like:
select person, sum(to_number(substring(time, 1, 2))*60 + to_number(substring(time, 3, 2))) as minutes
from t
group by person;
I see no reason to convert this back to a string -- or to even store the value as a string instead of as a number. But if you need to, you can reverse the process to get a string.

There are 2 answers, If you want to sum time only on date then it can be done as:-
select curr_date,
sum(24 * (to_date(arrival_time, 'HH24:mi:ss')- to_date(left_time, 'HH24:mi:ss'))) as difference
from sql_prac group by curr_date,arrival_time,left_time;
The sample output is as follows:-
select curr_date,left_time,arrival_time from sql_prac;
CURR_DATE LEFT_TIME ARRIVAL_TIME
--------- -------------------- --------------------
30-JUN-17 00:00:00 15:00:00
30-JUL-17 03:30:00 11:30:00
30-AUG-17 03:00:00 12:30:00
30-SEP-17 04:00:00 17:00:00
30-JUN-17 00:00:00 15:00:00
30-JUL-17 03:30:00 11:30:00
30-AUG-17 03:00:00 12:30:00
30-SEP-17 04:00:00 17:00:00
30-SEP-17 04:00:00 17:00:00
9 rows selected
select curr_date,sum(24 * (to_date(arrival_time, 'HH24:mi:ss')- to_date(left_time, 'HH24:mi:ss'))) as difference
from sql_prac group by curr_date,arrival_time,left_time;
CURR_DATE DIFFERENCE
--------- ----------
30-JUN-17 30
30-JUL-17 16
30-SEP-17 39
30-AUG-17 19

If you want to sum it by person and date then it can be done as:-
select dept,curr_date,sum(24 * (to_date(arrival_time, 'HH24:mi:ss')- to_date(left_time, 'HH24:mi:ss'))) as difference
from sql_prac group by dept,curr_date,arrival_time,left_time order by Dept;
The sample output is as follows:-
Data in table is:-
select dept,curr_date,left_time,arrival_time from sql_prac;
DEPT CURR_DATE LEFT_TIME ARRIVAL_TIME
-------------------- --------- -------------------- --------------------
A 30-SEP-17 04:00:00 17:00:00
B 30-SEP-17 04:00:00 17:00:00
C 30-AUG-17 03:00:00 12:30:00
D 30-DEC-17 04:00:00 17:00:00
A 30-SEP-17 04:00:00 17:00:00
B 30-JUL-17 03:30:00 11:30:00
C 30-AUG-17 03:00:00 12:30:00
D 30-SEP-17 04:00:00 17:00:00
R 30-SEP-17 04:00:00 17:00:00
Data fetched using the query
select dept,curr_date,sum(24 * (to_date(arrival_time, 'HH24:mi:ss')- to_date(left_time, 'HH24:mi:ss'))) as difference
from sql_prac group by dept,curr_date,arrival_time,left_time order by Dept;
DEPT CURR_DATE DIFFERENCE
-------------------- --------- ----------
A 30-SEP-17 26
B 30-JUL-17 8
B 30-SEP-17 13
C 30-AUG-17 19
D 30-SEP-17 13
D 30-DEC-17 13
R 30-SEP-17 13

Select Every Nth Record From SQL

I am having a Database in which data is been logged in regular interval of time i e for 5 minutes say it is been logged for 24 hours as shown in below table.
Date and Time Value
2016-09-17 14:00:00 25.26
2016-09-17 14:05:00 24.29
2016-09-17 14:10:00 25.22
2016-09-17 14:20:00 25.10
2016-09-17 23:55:00 20.21
I want To display Every 1 hour reading using SQL query There are chances the some reading may be missing The expected Output should be.
Date and Time Value
2016-09-17 14:00:00 25.26
2016-09-17 15:00:00 27.29
2016-09-17 16:00:00 28.12
2016-09-17 17:00:00 22.11
There are chances my be that some reading may be missing. like
Date and Time Value
2016-09-17 14:35:00 25.26
This reading may be missing
Please Suggest SQL query for the same

SELECT t1.DateCol,
t1.Value
FROM yourTable t1
INNER JOIN
(
SELECT MIN(DateCol) AS firstDate
FROM yourTable
GROUP BY FORMAT(DateCol, 'dd/MM/yyyy hh')
) t2
ON t1.DateCol = t2.firstDate
If you instead wanted to group by every 15 minutes, you could try:
GROUP BY CONCAT(FORMAT(DateCol, 'dd/MM/yyyy hh'),
FLOOR(DATEPART(MINUTE, DateCol) / 15))

How to get a total time?

Using VB6 and MS Access
Table:
ID Lunch_Intime, Lunch_Outtime
001 13:00:00 14:00:00
002 12:00:00 13:00:00
003 12:00:00 15:00:00
004 14:00:00 16:00:00
So on…
Lunch_Intime, Lunch_Outtime column data type is text.
I want to get a Total_Lunch_Time for the id.
Tried Query:
Select Lunch_Intime,
Lunch_Outtime,
Lunch_Outtime - Lunch_Intime as Total_Lunch_Time
from table
...but it's showing:
Total_Lunch_Time
#error
#error
So on..,
How to make a query for total_Lunch_Time?
Expected Output.
ID Lunch_Intime, Lunch_Outtime Total_Lunch_Time
001 13:00:00 14:00:00 01:00:00
002 12:00:00 13:00:00 01:00:00
003 12:00:00 15:00:00 03:00:00
004 14:00:00 16:00:00 02:00:00

In addition to converting your "time" values from text to date/time, I think you want to apply Format() to the elapsed times.
SELECT
ID
, Lunch_Intime
, Lunch_Outtime
, Format(CDate(Lunch_Outtime) - Cdate(Lunch_Intime),
"hh:nn:ss") AS Total_Lunch_Time
FROM
table;

You must cast the hours fields into date/time using CDate() before subtracting them.