I have a table:
id | parent_id | create_time
---|------------|--------------------
1 | 1 | 2016-08-26 00:00:00
2 | 2 | 2016-08-24 00:00:00
3 | 2 | 2016-08-22 00:00:00
4 | 4 | 2016-07-26 00:00:00
5 | 5 | 2016-07-24 00:00:00
I need to count unique 'parent_id' for each month(week,day).
Output something like this:
---------------------|----
2016-08-01 00:00:00 | 2
2016-07-01 00:00:00 | 2
But I could only do so:
SELECT date_trunc('month', create_time),count(parent_id) FROM test GROUP BY
date_trunc('month', create_time),parent_id
Result:
--------------------|---
2016-07-01 00:00:00 | 1
2016-07-01 00:00:00 | 1
2016-08-01 00:00:00 | 1
2016-08-01 00:00:00 | 2
I tried a lot of options, but do not have enough knowledge.
You need to remove the parent_id from the group by clause. And you probably want to use count(distinct parent_id) as well:
SELECT date_trunc('month', create_time),
count(distinct parent_id)
FROM test
GROUP BY date_trunc('month', create_time)
Try to use
SELECT date_trunc('month', create_time),count(distinct parent_id) FROM test GROUP BY
date_trunc('month', create_time)
Related
This question already has answers here:
Generate series of months for every row in Oracle
(1 answer)
Create all months list from a date column in ORACLE SQL
(3 answers)
Closed 1 year ago.
I want to make my table here in Oracle
+----+------------+------------+
| N | Start | End |
+----+------------+------------+
| 1 | 2018-01-01 | 2018-05-31 |
| 1 | 2018-01-01 | 2018-06-31 |
+----+------------+------------+
Into, as silly as it looks I need to insert one row for each month in the range for each in the first table
+----+------------+
| N | month| |
+----+------------+
| 1 | 2018-01-01 |
| 1 | 2018-01-01 |
| 1 | 2018-02-01 |
| 1 | 2018-02-01 |
| 1 | 2018-03-01 |
| 1 | 2018-03-01 |
| 1 | 2018-04-01 |
| 1 | 2018-04-01 |
| 1 | 2018-05-01 |
| 1 | 2018-05-01 |
| 1 | 2018-06-01 |
+----+------------+
I been trying to follow SQL: Generate Record Per Month In Date Range but I haven't had any luck figuring out the result I want.
Thanks for helping
My best guess is that you want to show all begining of months that are in the interval start to end in your table.
create table t1 as
select date'2018-01-01' start_d, date'2018-05-31' end_d from dual union all
select date'2018-01-01' start_d, date'2018-06-30' end_d from dual;
with cal as
(select add_months(date'2018-01-01', rownum-1) month_d
from dual connect by level <= 12)
select cal.month_d from cal
join t1 on cal.month_d between t1.start_d and t1.end_d
order by 1;
MONTH_D
-------------------
01.01.2018 00:00:00
01.01.2018 00:00:00
01.02.2018 00:00:00
01.02.2018 00:00:00
01.03.2018 00:00:00
01.03.2018 00:00:00
01.04.2018 00:00:00
01.04.2018 00:00:00
01.05.2018 00:00:00
01.05.2018 00:00:00
01.06.2018 00:00:00
So probaly there is a cut & paste error in your expectation for January.
Some other points
do not use reserved word as start for column names
Use DATE format to store dates to aviod invalid entries such as 2018-06-31
You can use a recursive CTE. For example:
with
n (s, e, cur) as (
select s, e, s from t
union all
select s, e, add_months(cur, 1)
from n
where add_months(cur, 1) < e
)
select cur from n;
Result:
CUR
---------
01-JAN-18
01-JAN-18
01-FEB-18
01-FEB-18
01-MAR-18
01-MAR-18
01-APR-18
01-APR-18
01-MAY-18
01-MAY-18
01-JUN-18
See running example at db<>fiddle.
I have pretty simple table which has 2 column. First one show time (timestamp), the second one show speed of car at that time (float8).
| DATE_TIME | SPEED |
|---------------------|-------|
| 2018-11-09 00:00:00 | 256 |
| 2018-11-09 01:00:00 | 659 |
| 2018-11-09 02:00:00 | 256 |
| other dates | xxx |
| 2018-11-21 21:00:00 | 651 |
| 2018-11-21 22:00:00 | 515 |
| 2018-11-21 23:00:00 | 849 |
Lets say we have period from 9 november to 21 november. How to group that period by week. In fact I want such result:
| DATE_TIME | AVG_SPEED |
|---------------------|-----------|
| 9-11 November | XXX |
| 12-18 November | YYY |
| 19-21 November | ZZZ |
I use PostgreSQL 10.4.
I use such SQL Statement to know the number of the week of the certain date:
SELECT EXTRACT(WEEK FROM TIMESTAMP '2018-11-09 00:00:00');
EDIT:
#tim-biegeleisen when I set period from '2018-11-01' to '2018-11-13' your sql statement return 2 result:
In fact I need such result:
2018-11-01 00:00:00 | 2018-11-04 23:00:00
2018-11-05 00:00:00 | 2018-11-11 23:00:00
2018-11-12 00:00:00 | 2018-11-13 05:00:00
As you can see in the calendar there are 3 week in that period.
We can do this using a calendar table. This answer assumes that a week begins with the first date in your data set. You could also do this assuming something else, e.g. a standard week according to something else.
WITH dates AS (
SELECT date_trunc('day', dd)::date AS dt
FROM generate_series
( '2018-11-09'::timestamp
, '2018-11-21'::timestamp
, '1 day'::interval) dd
),
cte AS (
SELECT t1.dt, t2.DATE_TIME, t2.SPEED,
EXTRACT(week from t1.dt) week
FROM dates t1
LEFT JOIN yourTable t2
ON t1.dt = t2.DATE_TIME::date
)
SELECT
MIN(dt)::text || '-' || MAX(dt) AS DATE_TIME,
AVG(SPEED) AS AVG_SPEED
FROM cte
GROUP BY
week
ORDER BY
MIN(dt);
Demo
I am trying to write a query that will allow to me to count the number of active subscriptions by day in Redshift.
I have the following table:
sub_id | start_date | end_date
---------------------------------------
20001 | 2017-09-01 | NULL
20002 | 2017-08-01 | 2017-08-29
20003 | 2016-01-01 | 2017-04-25
20004 | 2016-07-01 | 2017-09-03
I would like to be able to state, for each date between two dates how many subscriptions are active, such that:
date | active_subs
------------------------
2016-06-30 | 1
2016-07-01 | 2
... |
2017-04-24 | 2
2017-04-25 | 1
... |
2017-07-31 | 1
2017-08-01 | 2
... |
2017-08-28 | 2
2017-08-29 | 1
2017-08-30 | 1
2017-08-31 | 1
2017-09-01 | 2
2017-09-02 | 2
2017-09-03 | 1
I have a reference table from which a query can draw 1 row per day with the table name of date and the relevant column being date.ref_date (in the YYYY-MM-DD format)
Do i write this query using window functions or is there a better way?
Thanks
If I understood you correctly, you don't need nor window functions, joins(except to the date table) or cumulative count. You can do this:
SELECT t.date,
COUNT(s.sub_id) as active_subs
FROM dateTable t
LEFT JOIN YourTable s
ON(t.dateCol between s.start_date
AND COALESCE(s.end_date,<Put A late date here>))
GROUP BY t.date
I would do this as:
with cte as (
select start_date as dte, 1 as inc
from t
union all
select coalesce(end_date, current_date), -1 as inc
from t
)
select dte,
sum(sum(inc)) over (order by dte)
from cte
group by dte
order by dte;
There may be off-by-one errors, depending on whether you count stops on the date given or on the next day.
I have following situation where i need to get several values between two invoices date.
So query is giving data based on invoices now what i need to do is for some values fetch data between this invoice date and last invoice date
already tried ways
1) sub query will easily solve this but as i have to do this for 4-5 column and its a 15 gb database so that's not possible.
2) if i go like this
left join (select inv.date ,inv,actno from invoice inv) as invo on invo.actno=act.id and invo.date < inv.date
then it will give all the data less then that date but i need only one data that will be less than main invoice date.
3) we can not get second max value in subquery of from clause because outer invoice is not grouped so it might be max or midlle or least .
4) we can not send values of other table in subquery of join table.
ex
create table inv (id serial ,date timestamp without time zone);
insert into inv (date) values('2017-01-31 00:00:00'),('2017-01-30 00:00:00'),('2017-01-29 00:00:00'),('2017-01-28 00:00:00'),('2017-01-27 00:00:00');
select date as d1 from inv;
id | date
----+---------------------
1 | 2017-01-31 00:00:00
2 | 2017-01-30 00:00:00
3 | 2017-01-29 00:00:00
4 | 2017-01-28 00:00:00
5 | 2017-01-27 00:00:00
(5 rows)
I need this
id |date |date | id
1 | 2017-01-31 00:00:00 | 2017-01-30 00:00:00 | 2
2 | 2017-01-30 00:00:00 | 2017-01-29 00:00:00 | 3
3 | 2017-01-29 00:00:00 | 2017-01-28 00:00:00 | 4
4 | 2017-01-28 00:00:00 | 2017-01-27 00:00:00 | 5
5 | 2017-01-27 00:00:00 |
I can't do subquery in select as database is big and need to do this for 4-5 column
UPDATE 1
I need this from same table but using it twice in FROM clause as my requirement is that I need several data joined from invoice table and then there is 4-5 column in which I need things like sum of amount paid between last and this invoice.
So I can take both invoice date in subquery and get the data between them
UPDATE 2
lag will not solve this
select i.id,i.date, lag(date) over (order by date) from inv i order by id ;
id | date | lag
----+---------------------+---------------------
1 | 2017-01-31 00:00:00 | 2017-01-30 00:00:00
2 | 2017-01-30 00:00:00 | 2017-01-29 00:00:00
3 | 2017-01-29 00:00:00 | 2017-01-28 00:00:00
4 | 2017-01-28 00:00:00 | 2017-01-27 00:00:00
5 | 2017-01-27 00:00:00 |
(5 rows)
Time: 0.480 ms
test=# select i.id,i.date, lag(date) over (order by date) from inv i where id=2 order by id ;
id | date | lag
----+---------------------+-----
2 | 2017-01-30 00:00:00 |
(1 row)
Time: 0.525 ms
test=# select i.id,i.date, lag(date) over (order by date) from inv i where id in (2,3) order by id ;
id | date | lag
----+---------------------+---------------------
2 | 2017-01-30 00:00:00 | 2017-01-29 00:00:00
3 | 2017-01-29 00:00:00 |
it will calculate on the data it will get from the table in that query it is bounded in that query see here 3 has a lag but could not get it cause query is not allowing it to have it ....something in left join needs to be done so the lag date can be taken from same table but calling it again in from clause Thanks Again buddy
Like here?:
t=# select date as d1,
lag(date) over (order by date)
from inv
order by 1 desc;
d1 | lag
---------------------+---------------------
2017-01-31 00:00:00 | 2017-01-30 00:00:00
2017-01-30 00:00:00 | 2017-01-29 00:00:00
2017-01-29 00:00:00 | 2017-01-28 00:00:00
2017-01-28 00:00:00 | 2017-01-27 00:00:00
2017-01-27 00:00:00 |
(5 rows)
Time: 1.416 ms
I have a table as below in the database, how can I write a SQL to show the expected result?
first_date: the first order_date on the table ORDER BY order_date ASC
days_to_date: (order_date - first_date) in number of days
My table:
id | order_date | order_ref
---+------------------------
1 | 2015-03-01 | BC101
2 | 2015-03-01 | BC102
3 | 2015-03-02 | BC103
4 | 2015-03-03 | BC104
Expected result:
id | order_date | first_date | days_to_date
---+------------+------------+-------------
1 | 2015-03-01 | 2015-03-01 | 0
2 | 2015-03-01 | 2015-03-01 | 0
3 | 2015-03-02 | 2015-03-01 | 1
4 | 2015-03-03 | 2015-03-01 | 2
Other notes:
I'm using HSQLDB 2.0.0, but prefer solving the problem of getting the first_date displayed on every row in general cases if that's possible
Thanks in advance
Try
select id, order_date,
(select min(order_date) from your_table) as first_date,
datediff('day', (select min(order_date) from your_table), order_date) as days_to_date
from your_table
order by order_date