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)
Related
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 | 2021-07-10 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.
I have a table which contain _id, underSubheadId, wefDate, price.
Whenever a product is created or price is edited an entry is made in this table also.
What I want is if I enter a date, I get the latest price of all distinct UnderSubheadIds before the date (or on that date if no entry found)
_id underHeadId wefDate price
1 1 2016-11-01 5
2 2 2016-11-01 50
3 1 2016-11-25 500
4 3 2016-11-01 20
5 4 2016-11-11 30
6 5 2016-11-01 40
7 3 2016-11-20 25
8 5 2016-11-15 52
If I enter 2016-11-20 as date I should get
1 5
2 50
3 25
4 30
5 52
I have achieved the result using ROW NUMBER function in SQL SERVER, but I want this result in Sqlite which don't have such function.
Also if a date like 2016-10-25(which have no entries) is entered I want the price of the date which is first.
Like for 1 we will get price as 5 as the nearest and the 1st entry is 2016-11-01.
This is the query for SQL SERVER which is working fine. But I want it for Sqlite which don't have ROW_NUMBER function.
select underSubHeadId,price from(
select underSubHeadId,price, ROW_NUMBER() OVER (Partition By underSubHeadId order by wefDate desc) rn from rates
where wefDate<='2016-11-19') newTable
where newTable.rn=1
Thank You
This is a little tricky, but here is one way:
select t.*
from t
where t.wefDate = (select max(t2.wefDate)
from t t2
where t2.underSubHeadId = t.underSubHeadId and
t2.wefdate <= '2016-11-20'
);
select underHeadId, max(price)
from t
where wefDate <= "2016-11-20"
group by underHead;
I have a raw table recording customer ids coming to a store over a particular time period. Using Impala, I would like to calculate the number of distinct customer IDs coming to the store until each day. (e.g., on day 3, 5 distinct customers visited so far)
Here is a simple example of the raw table I have:
Day ID
1 1234
1 5631
1 1234
2 1234
2 4456
2 5631
3 3482
3 3452
3 1234
3 5631
3 1234
Here is what I would like to get:
Day Count(distinct ID) until that day
1 2
2 3
3 5
Is there way to easily do this in a single query?
Not 100% sure if will work on impala
But if you have a table days. Or if you have a way of create a derivated table on the fly on impala.
CREATE TABLE days ("DayC" int);
INSERT INTO days
("DayC")
VALUES (1), (2), (3);
OR
CREATE TABLE days AS
SELECT DISTINCT "Day"
FROM sales
You can use this query
SqlFiddleDemo in Postgresql
SELECT "DayC", COUNT(DISTINCT "ID")
FROM sales
cross JOIN days
WHERE "Day" <= "DayC"
GROUP BY "DayC"
OUTPUT
| DayC | count |
|------|-------|
| 1 | 2 |
| 2 | 3 |
| 3 | 5 |
UPDATE VERSION
SELECT T."DayC", COUNT(DISTINCT "ID")
FROM sales
cross JOIN (SELECT DISTINCT "Day" as "DayC" FROM sales) T
WHERE "Day" <= T."DayC"
GROUP BY T."DayC"
try this one:
select day, count(distinct(id)) from yourtable group by day
i have table and it has following data:
USERID NAME DATEFROM DATETO
1 xxx 2014-05-10 2014-05-15
1 xxx 2014-05-20 2014-05-25
4 yyy 2014-04-20 2014-04-21
now i have sql query like :
select * from leave where datefrom>='2014-05-01' and dateto<='2014-05-31'
so now i want output :
userid name total_leave_days
1 xxx 12
4 yyy 2
(2014-05-10 - 2014-05-15 )=6 days
(2014-05-20 - 2014-05-25 )=6 days
total = 12 days for useid 1
(2014-04-20 - 2014-04-21)= 2 days for userid 4
how can i calculate this total days .?
Please try:
select
USERID,
NAME,
SUM(DATEDIFF(day, DATEFROM, DATETO)+1) total_leave_days
From leave
group by USERID, NAME
SQL Fiddle Demo
It's important to note that you need "+1" to emulate the expected calculations because there is an inherent assumption of ""start of day" for the Start date and "end of day" for end date - but dbms's don't think that way. a date is always stored as "start of day".
select
USERID
, name
, sum( datediff(day,DATEFROM,DATETO) + 1 ) as leave_days
from leavetable
group by
USERID
, name
produces this:
| USERID | NAME | LEAVE_DAYS |
|--------|------|------------|
| 1 | xxx | 12 |
| 4 | yyy | 2 |
see: http://sqlfiddle.com/#!3/ebe5d/1
You can use DateDiff.
SELECT UserID, Name, SUM(DATEDIFF(DAY, DateFrom, DateTo) + 1) AS total_leave_days
FROM leave
WHERE datefrom >= '2014-05-01' AND dateto <= '2014-05-31'
GROUP BY UserID, Name
The + 1 ,of course, is because DATEDIFF will return the exclusive count, where it sounds like you want the inclusive number of days.
Try this:
select userid, name, sum (1 + datediff(day,datefrom,dateto)) as total_leave_days
from leaves
where datefrom>='2014-05-01' and dateto<='2014-05-31'
group by userid, name
This will sum the total leaves per userid. Note that datediff will give you 5 days difference for the range 2014-05-10 to 2014-05-15, so we need to add 1 to the result to get 6 days i.e. range inclusive of both ends.
Demo