I have table in Oracle SQL presents ID of clients and date with time of their login to application:
ID | LOGGED
----------------
11 | 2021-07-10 12:55:13.278
11 | 2021-08-10 13:58:13.211
11 | 2021-02-11 12:22:13.364
22 | 2021-01-10 08:34:13.211
33 | 2021-04-02 14:21:13.272
I need to select only these clients (ID) who has logged minimum 1 time in last month (August) and minimum 1 time in one month preceding August (June or July)
Currently we have September, so...
I need clients who has logged min 1 time in August
and min 1 time in July or Jun,
if logged in June -> not logg in July
if logged in July -> not logged in June
As a result I need like below:
ID
----
11
How can do that in Oracle SQL ? be aware that column "LOGGED" has Timestamp like: 2021-01-10 08:34:13.211
May be you consider this:
select id
from yourtable
group by id
having count(case
months_between(trunc(sysdate,'MM'),
trunc(logged,'MM')
) when 1 then 1 end
) >= 1
and count
(case when
months_between(trunc(sysdate,'MM') ,
trunc(logged,'MM')
) in (2,3) then 1 end
) = 1
I don't understand one thing:
You wrote :
minimum 1 time in one month preceding August (June or July)
and after then:
if logged in June -> not logg in July
if logged in July -> not logged in June
If you need EXACTLY one month- June or July
just consider my query above.
If you need minimum one logon in June and July, then:
select id
from yourtable
group by id
having count(case
months_between(trunc(sysdate,'MM'),
trunc(logged,'MM')
) when 1 then 1 end
) >= 1
and count
(case when
months_between(trunc(sysdate,'MM') ,
trunc(logged,'MM')
) in (2,3) then 1 end
) >= 1
Your question needs some clarification, but based on what you were describing I am seeing a couple of options.
The simplest one is probably using a combo of data densification (for generating a row for every month for each id) plus an analytical function (for enabling inter-row calculations. Here's a simple example of this:
rem create a dummy table with some more data (you do not seem to worry about the exact timestamp)
drop table logs purge;
create table logs (ID number, LOGGED timestamp);
insert into logs values (11, to_timestamp('2021-07-10 12:55:13.278','yyyy-mm-dd HH24:MI:SS.FF'));
insert into logs values (11, to_timestamp('2021-07-11 12:55:13.278','yyyy-mm-dd HH24:MI:SS.FF'));
insert into logs values (11, to_timestamp('2021-08-10 13:58:13.211','yyyy-mm-dd HH24:MI:SS.FF'));
insert into logs values (11, to_timestamp('2021-02-11 12:22:13.364','yyyy-mm-dd HH24:MI:SS.FF'));
insert into logs values (11, to_timestamp('2021-04-11 12:22:13.364','yyyy-mm-dd HH24:MI:SS.FF'));
insert into logs values (22, to_timestamp('2021-01-10 08:34:13.211','yyyy-mm-dd HH24:MI:SS.FF'));
insert into logs values (33, to_timestamp('2021-04-02 14:21:13.272','yyyy-mm-dd HH24:MI:SS.FF'));
commit;
The following SQL gets your data densified and lists the total count of logins for a month and the previous month on the same row so that you could do a comparative calculation. I have not done then, but I am hoping you get the idea.
with t as
(-- dummy artificial table just to create a time dimension for densification
select distinct to_char(sysdate - rownum,'yyyy-mm') mon
from dual connect by level < 300),
l_sparse as
(-- aggregating your login info per month
select id, to_char(logged,'yyyy-mm') mon, count(*) cnt
from logs group by id, to_char(logged,'yyyy-mm') ),
l_dense as
(-- densification with partition outer join
select t.mon, l.id, cnt from l_sparse l partition by (id)
right outer join t on (l.mon = t.mon)
)
-- final analytical function to list current and previous row info in same record
select mon, id
, cnt
, lag(cnt) over (partition by id order by mon asc) prev_cnt
from l_dense
order by id, mon;
parts of the result:
MON ID CNT PREV_CNT
------- ---------- ---------- ----------
2020-12 11
2021-01 11
2021-02 11 2
2021-03 11 2
2021-04 11 1
2021-05 11 1
2021-06 11
2021-07 11 3
2021-08 11 2 3
2021-09 11 2
2020-12 22
2021-01 22 2
2021-02 22 2
2021-03 22
2021-04 22
...
You can see for ID 11 that for 2021-08 you have logins for the current and previous month, so you can math on it. (Would require another subselect/with branch).
Alternatives to this would be:
interrow calculation plus time math between two logged timestamps
pattern matching
Did not drill into those, not enough info about your real requirement.
Related
I work with crashes and mileage for the same year which is Year in table. Crashes are are there for every record, but annual mileage is not. NULLs for mileage could be at the beginning or at the end of the time period for certain customer. Also, couple of annual mileage records can be missing as well. I do not know how to overcome this. I try to do it in CASE statement but then I do not know how to code it properly. Issue needs to be resolved in SQL and use SQL Server.
This is how the output looks like and I need to have mileage for every single year for each customer.
The info I am pulling from is proprietary database and the records themselves should be untouched as is. I just need code in query which will modify my current output to output where I have mileage for every year. I appreciate any input!
Year
Customer
Crashes
Annual_Mileage
2009
123
5
3453453
2010
123
1
NULL
2011
123
0
54545
2012
123
14
376457435
2013
123
3
63453453
2014
123
4
NULL
2015
123
15
6346747
2016
123
0
NULL
2017
123
2
534534
2018
123
7
NULL
2019
123
11
NULL
2020
123
15
565435
2021
123
12
474567546
2022
123
7
NULL
Desired Results
Year
Customer
Crashes
Annual_Mileage
2009
123
5
3453453
2010
123
1
175399 (prior value is taken)
2011
123
0
54545
2012
123
14
376457435
2013
123
3
63453453
2014
123
4
34900100 (avg of 2 adjacent values)
2015
123
15
6346747
2016
123
0
3440641 (avg of 2 adjacent values)
2017
123
2
534534
2018
123
7
534534 ( prior value is taken)
2019
123
11
549985 (avg of 2 adjacent values)
2020
123
15
565435
2021
123
12
474567546
2022
123
7
474567546 (prior value is taken)
SELECT Year,
Customer,
Crashes,
CASE
WHEN Annual_Mlg IS NOT NULL THEN Annual_Mlg
WHEN Annual_Mlg IS NULL THEN
CASE
WHEN PREV.Annual_Mlg IS NOT NULL
AND NEXT.Annual_Mlg IS NOT NULL
THEN ( PREV.Annual_Mlg + NEXT.Annual_Mlg ) / 2
ELSE 0
END
END AS Annual_Mlg
FROM #table
The above code doesn't work, but I just need to start somehow and that what I have currently.
I understand what I need to do I just do not know how to code it in SQL.
After i applied row_number () function i got this output for first 2 clients and for the rest of the 4 clients row_number() function gave correct output. i have no idea why is that. I thought may be because i used "full join" before to combine mileage and crashes table?
enter image description here
Your use of #table tells me that you're using MS SQL Server (a temporary table, probably in a stored procedure).
You want to:
select all the rows in #table
joined with the matching row (if any) for the previous year, and
joined with the matching row (if any) for the next year
Then it's easy. Assuming the primary key on your #table is composed of the year and customer columns, something like this ought to do you:
select t.year ,
t.customer ,
t.crashes ,
annual_milage = coalesce(
t.annual_milage ,
( coalesce( p.annual_mileage, 0 ) +
coalesce( n.annual_mileage, 0 )
) / 2
)
from #table t -- take all the rows
left join #table p on p.year = t.year - 1 -- with the matching row for
and p.customer = t.customer -- the previous year (if any)
left join #table n on n.year = t.year + 1 -- and the matching row for
and n.customer = t.customer -- the next year (if any)
Notes:
What value you default to if the previous or next year doesn't exist is up to you (zero? some arbitrary value?)
Is the previous/next year guaranteed to be the current year +/- 1?
If not, you may have to use derived tables as the source for the
prev/next data, selecting the closest previous/next year (that sort
of thing rather complicates the query significantly).
Edited To Note:
If you have discontiguous years for each customer such that the "previous" and "next" years for a given customer are not necessarily the current year +/- 1, then something like this is probably the most straightforward way to find the previous/next year.
We use a derived table in our from clause, and assign a sequential number in lieu of year for each customer, using the ranking function row_number() function. This query, then
select row_nbr = row_number() over (
partition by x.customer
order by x.year
) ,
x.*
from #table x
would produce results along these lines:
row_nbr
customer
year
...
1
123
1992
...
2
123
1993
...
3
123
1995
...
4
123
2020
...
1
456
2001
...
2
456
2005
...
3
456
2020
...
And that leads us to this:
select year = t.year ,
customer = t.customer ,
crashes = t.crashes ,
annual_mileage = coalesce(
t.mileage,
coalesce(
t.annual_mileage,
(
coalesce(p.annual_mileage,0) +
coalesce(n.annual_mileage,0)
) / 2
),
)
from (
select row_nbr = row_number() over (
partition by x.customer
order by x.year
) ,
x.*
from #table x
) t
left join #table p on p.customer = t.customer and p.row_nbr = t.row_nbr-1
left join #table n on n.customer = t.customer and n.row_nbr = t.row_nbr+1
Essentially date field is updated every month along with other fields, however one field is only updated ~6 times throughout the year. For months where that field is not updated, looking to show the most recent previous data
Date
Emp_no
Sales
Group
Jan
1234
100
Med
Feb
1234
200
---
Mar
1234
170
---
Apr
1234
150
Low
May
1234
180
---
Jun
1234
90
High
Jul
1234
100
---
Need it to show:
Date
Emp_no
Sales
Group
Jan
1234
100
Med
Feb
1234
200
Med
Mar
1234
170
Med
Apr
1234
150
Low
May
1234
180
Low
Jun
1234
90
High
Jul
1234
100
High
This field is not updated at set intervals, could be 1-4 months of Nulls in a row
Tried something like this to get the second most recent date but unsure how to deal with the fact that i could need between 1-4 months prior
LAG(Group)
OVER(PARTITION BY emp_no
ORDER BY date)
Thanks!
This is the traditional "gaps and islands" problem.
There are various ways to solve it, a simple version will work for you.
First, create a new identifier that splits the rows in to "groups", where only the first row in each group is NOT NULL.
SUM(CASE WHEN "group" IS NOT NULL THEN 1 ELSE 0 END) OVER (PARTION BY emp_no ORDER BY "date") AS emp_group_id
Then you can use MAX() in another window function, as all "groups" will only have one NOT NULL value.
WITH
gaps
AS
(
SELECT
t.*,
SUM(
CASE WHEN "group" IS NOT NULL
THEN 1
ELSE 0
END
)
OVER (
PARTITION BY emp_no
ORDER BY "date"
)
AS emp_group_id
FROM
your_table t
)
SELECT
"date",
emp_no,
sales,
MAX("group")
OVER (
PARTITION BY emp_no, emp_group_id
)
AS "group"
FROM
gaps
Edit
Ignore all that.
Oracle has IGNORE NULLS.
LAST_VALUE("group" IGNORE NULLS)
OVER (
PARTITION BY emp_no
ORDER BY "date"
ROWS BETWEEN UNBOUNDED PRECEDING
AND CURRENT ROW
)
AS "group"
I have a table of EMPLOYEES that contains information about the DATE and WORKTIME per that day. Fx:
ID | DATE | WORKTIME |
----------------------------------------
1 | 1-Sep-2014 | 4 |
2 | 2-Sep-2014 | 6 |
1 | 3-Sep-2014 | 5.5 |
1 | 4-Sep-2014 | 7 |
2 | 4-Sep-2014 | 4 |
1 | 9-Sep-2014 | 8 |
and so on.
Question: How can I create a query that would allow me to calculate amount of time worked per week (HOURS_PERWEEK). I understand that I need a summation of WORKTIME together with grouping considering both, ID and week, but so far my trials as well as googling didnt yield any results. Any ideas on this? Thank you in advance!
edit:
Got a solution of
select id, sum (worktime), trunc(date, 'IW') week
from employees
group by id, TRUNC(date, 'IW');
But will need somehow to connect that particular output with DATE table by updating a newly created column such as WEEKLY_TIME. Any hints on that?
You can find the start of the ISO week, which will always be a Monday, using TRUNC("DATE", 'IW').
So if, in the query, you GROUP BY the id and the start of the week TRUNC("DATE", 'IW') then you can SELECT the id and aggregate to find the SUM the WORKTIME column for each id.
Since this appears to be a homework question and you haven't attempted a query, I'll leave it at this to point you in the correct direction and you can complete the query.
Update
Now I need to create another column (lets call it WEEKLY_TIME) and populate it with values from the current output, so that Sep 1,3,4 (for ID=1) would all contain value 16.5, specifying that on that day (that is within the certain week) that person worked 16.5 in total. And for ID=2 it would then be a value of 10 for both Sep 2 and 4.
For this, if I understand correctly, you appear to not want to use aggregation functions and want to use the analytic version of the function:
select id,
"DATE",
trunc("DATE", 'IW') week,
worktime,
sum (worktime) OVER (PARTITION BY id, trunc("DATE", 'IW'))
AS weekly_time
from employees;
Which, for the sample data:
CREATE TABLE employees (ID, "DATE", WORKTIME) AS
SELECT 1, DATE '2014-09-01', 4 FROM DUAL UNION ALL
SELECT 2, DATE '2014-09-02', 6 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-03', 5.5 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-04', 7 FROM DUAL UNION ALL
SELECT 2, DATE '2014-09-04', 4 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-09', 8 FROM DUAL;
Outputs:
ID
DATE
WEEK
WORKTIME
WEEKLY_TIME
1
2014-09-01 00:00:00
2014-09-01 00:00:00
4
16.5
1
2014-09-03 00:00:00
2014-09-01 00:00:00
5.5
16.5
1
2014-09-04 00:00:00
2014-09-01 00:00:00
7
16.5
1
2014-09-09 00:00:00
2014-09-08 00:00:00
8
8
2
2014-09-04 00:00:00
2014-09-01 00:00:00
4
10
2
2014-09-02 00:00:00
2014-09-01 00:00:00
6
10
db<>fiddle here
edit: answer submitted without noticing "Oracle" tag. Otherwise, question answered here: Oracle SQL - Sum and group data by week
Select employee_Id,
DATEPART(week, workday) as [Week],
sum (worktime) as [Weekly Hours]
from WORK
group by employee_id, DATEPART(week, workday)
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=238b229156a383fa3c466b6c3c2dee1e
I have table in Oracle SQL presents ID of clients and date with time of their login to application:
ID | LOGGED
----------------
11 | 2021-09-10 12:55:13.278
11 | 2021-08-10 13:58:13.211
11 | 2021-02-11 12:22:13.364
22 | 2021-09-15 08:34:13.211
33 | 2021-04-02 14:21:13.272
How can I select only these IDs, which logged the first time during last 30 days ?
So as a result I need something like below:
ID
---
22
Because only ID 22 logged first time during last 30 days -> 2021-09-15 08:34:13.211
How can I do that in Oracle SQL ?
Use this
Select id from table where trunc(logged)
>=
Trunc(sysdate-30) group by id having count(*) =1
Or better condition is using min, max
Select id from table where trunc(logged)
>=
Trunc(sysdate-30) group by id having min(logged)
=max(logged)
Trying to build a query to have sum on sales column by month, 3 months, 6 months, 11 months and prior year 11 months. What are possible options to go about it.
I tried datediff and date related functions didn't get intended results.
Would like some suggestions on how to go about it?
This should get you started.
CREATE TABLE sales (
product varchar (1),
month int,
amount int
)
insert into sales values ('a',1,5); insert into sales values ('a',1,33);
insert into sales values ('a',2,32); insert into sales values ('b',1,12);
insert into sales values ('b',2,4); insert into sales values ('c',1,5);
insert into sales values ('c',2,11); insert into sales values ('c',2,13);
SELECT
product,
SUM(CASE WHEN month = 1 THEN amount END) AS Month1,
SUM(CASE WHEN month = 2 THEN amount END) AS Month2
FROM
sales
GROUP BY
product
output:
product | Month1 | Month2
--------------------------------
a | 38 | 32
b | 12 | 4
c | 5 | 24