I want to group orders from the same customer if they happen within 10 minutes of the first order, then find the next first order and group them and so on.
Ex:
Customer group orders
6 1 3
2 4,5
3 8
7 1 9,10
2 11,12
3 13
id customer time
3 6 2021-05-12 12:14:22.000000
4 6 2021-05-12 12:24:24.000000
5 6 2021-05-12 12:29:16.000000
8 6 2021-05-12 13:01:40.000000
9 7 2021-05-14 12:13:11.000000
10 7 2021-05-14 12:20:01.000000
11 7 2021-05-14 12:45:00.000000
12 7 2021-05-14 12:48:41.000000
13 7 2021-05-14 12:58:16.000000
18 9 2021-05-18 12:22:13.000000
25 15 2021-05-18 13:44:02.000000
26 16 2021-05-17 09:39:02.000000
27 16 2021-05-18 19:38:43.000000
28 17 2021-05-18 15:40:02.000000
29 18 2021-05-19 15:32:53.000000
30 18 2021-05-19 15:45:56.000000
31 18 2021-05-19 16:29:09.000000
34 15 2021-05-24 15:45:14.000000
35 15 2021-05-24 15:45:14.000000
36 19 2021-05-24 17:14:53.000000
Here is what I have currently, I think that it is currently not grouping by customer when case when d.StartTime > dateadd(minute, 10, c.first_time) so it compares StartTime of all orders for all customers.
with
data as (select Customer,StartTime,Id, row_number() over(partition by Customer order by StartTime) rn from orders t),
cte as (
select d.*, StartTime as first_time
from data d
where rn = 1
union all
select d.*,
case when d.StartTime > dateadd(minute, 10, c.first_time)
then d.StartTime
else c.first_time
end
from cte c
inner join data d on d.rn = c.rn + 1
)
select c.*, dense_rank() over(partition by Customer order by first_time) grp
from cte c;'
I have two databases (MySQL & SQL Server) having similar schema so either would work for me.
Try the following on SQL Server:
SELECT customer,
ROW_NUMBER() OVER (PARTITION BY customer ORDER BY grp) AS group_no,
STRING_AGG(id, ',') AS orders
FROM
(
SELECT id,customer, [time],
(DATEDIFF(SECOND, MIN([time]) OVER (PARTITION BY CUSTOMER), [time])/60)/10 grp
FROM orders
) T
GROUP BY customer, grp
ORDER BY customer
See a demo.
According to your posted requirement, you are trying to divide the period between the first order date and the last order date into groups (or let's say time frames) each one is 10 minutes long.
What I did in this query: for each customer order, find the difference between the order date and the minimum date (first customer order date) in seconds and then divide it by 10 to get it's time frame number. i.e. for a difference = 599s the frame number = 599/60 =9m /10 = 0. for a difference = 620s the frame number = 620/60 =10m /10 = 1.
After defining the correct groups/time frames for each order you can simply use the STRING_AGG function to get the desired output. Noting that the STRING_AGG function applies to SQL Server 2017 (14.x) and later.
Related
I want to count the distinct amount of users over the last 60 days, and then, count the distinct amount of users over the last 59 days, and so on and so forth.
Ideally, the output would look like this (TARGET OUTPUT)
Day Distinct Users
60 200
59 200
58 188
57 185
56 180
[...] [...]
where 60 days is the max total possible distinct users, and then 59 would have a little less and so on and so forth.
my query looks like this.
select
count(distinct (case when datediff(day,DATE,current_date) <= 60 then USER_ID end)) as day_60,
count(distinct (case when datediff(day,DATE,current_date) <= 59 then USER_ID end)) as day_59,
count(distinct (case when datediff(day,DATE,current_date) <= 58 then USER_ID end)) as day_58
FROM Table
The issue with my query is that This outputs the data by column instead of by rows (like shown below) AND, most importantly, I have to write out this logic 60x for each of the 60 days.
Current Output:
Day_60 Day_59 Day_58
209 207 207
Is it possible to write the SQL in a way that creates the target as shown initially above?
Using below data in CTE format -
with data_cte(dates,userid) as
(select * from values
('2022-05-01'::date,'UID1'),
('2022-05-01'::date,'UID2'),
('2022-05-02'::date,'UID1'),
('2022-05-02'::date,'UID2'),
('2022-05-03'::date,'UID1'),
('2022-05-03'::date,'UID2'),
('2022-05-03'::date,'UID3'),
('2022-05-04'::date,'UID1'),
('2022-05-04'::date,'UID1'),
('2022-05-04'::date,'UID2'),
('2022-05-04'::date,'UID3'),
('2022-05-04'::date,'UID4'),
('2022-05-05'::date,'UID1'),
('2022-05-06'::date,'UID1'),
('2022-05-07'::date,'UID1'),
('2022-05-07'::date,'UID2'),
('2022-05-08'::date,'UID1')
)
Query to get all dates and count and distinct counts -
select dates,count(userid) cnt, count(distinct userid) cnt_d
from data_cte
group by dates;
DATES
CNT
CNT_D
2022-05-01
2
2
2022-05-02
2
2
2022-05-03
3
3
2022-05-04
5
4
2022-05-05
1
1
2022-05-06
1
1
2022-05-08
1
1
2022-05-07
2
2
Query to get difference of date from current date
select dates,datediff(day,dates,current_date()) ddiff,
count(userid) cnt,
count(distinct userid) cnt_d
from data_cte
group by dates;
DATES
DDIFF
CNT
CNT_D
2022-05-01
45
2
2
2022-05-02
44
2
2
2022-05-03
43
3
3
2022-05-04
42
5
4
2022-05-05
41
1
1
2022-05-06
40
1
1
2022-05-08
38
1
1
2022-05-07
39
2
2
Get records with date difference beyond a certain range only -
include clause having
select datediff(day,dates,current_date()) ddiff,
count(userid) cnt,
count(distinct userid) cnt_d
from data_cte
group by dates
having ddiff<=43;
DDIFF
CNT
CNT_D
43
3
3
42
5
4
41
1
1
39
2
2
38
1
1
40
1
1
If you need to prefix 'day' to each date diff count, you can
add and outer query to previously fetched data-set and add the needed prefix to the date diff column as following -
I am using CTE syntax, but you may use sub-query given you will select from table -
,cte_1 as (
select datediff(day,dates,current_date()) ddiff,
count(userid) cnt,
count(distinct userid) cnt_d
from data_cte
group by dates
having ddiff<=43)
select 'day_'||to_char(ddiff) days,
cnt,
cnt_d
from cte_1;
DAYS
CNT
CNT_D
day_43
3
3
day_42
5
4
day_41
1
1
day_39
2
2
day_38
1
1
day_40
1
1
Updated the answer to get distinct user count for number of days range.
A clause can be included in the final query to limit to number of days needed.
with data_cte(dates,userid) as
(select * from values
('2022-05-01'::date,'UID1'),
('2022-05-01'::date,'UID2'),
('2022-05-02'::date,'UID1'),
('2022-05-02'::date,'UID2'),
('2022-05-03'::date,'UID5'),
('2022-05-03'::date,'UID2'),
('2022-05-03'::date,'UID3'),
('2022-05-04'::date,'UID1'),
('2022-05-04'::date,'UID6'),
('2022-05-04'::date,'UID2'),
('2022-05-04'::date,'UID3'),
('2022-05-04'::date,'UID4'),
('2022-05-05'::date,'UID7'),
('2022-05-06'::date,'UID1'),
('2022-05-07'::date,'UID8'),
('2022-05-07'::date,'UID2'),
('2022-05-08'::date,'UID9')
),cte_1 as
(select datediff(day,dates,current_date()) ddiff,userid
from data_cte), cte_2 as
(select distinct ddiff from cte_1 )
select cte_2.ddiff,
(select count(distinct userid)
from cte_1 where cte_1.ddiff <= cte_2.ddiff) cnt
from cte_2
order by cte_2.ddiff desc
DDIFF
CNT
47
9
46
9
45
9
44
8
43
5
42
4
41
3
40
1
You can do unpivot after getting your current output.
sample one.
select
*
from (
select
209 Day_60,
207 Day_59,
207 Day_58
)unpivot ( cnt for days in (Day_60,Day_59,Day_58));
I have a table as following (using bigquery):
id
year
month
day
rating
111
2020
11
30
4
111
2020
12
01
4
112
2020
11
30
5
113
2020
11
30
5
Is there a way in which I can select ids that have ratings that are consecutively (two or more consecutive records) low (low as in both records' ratings less than 4.5)?
For example, my desired output is:
id
year
month
day
rating
111
2020
11
30
4
111
2020
12
01
4
If you want all rows, then you need to look at both the previous rating and the next rating:
SELECT t.*
FROM (SELECT t.*,
LAG(rating) OVER (PARTITION BY id ORDER BY year, month, day ASC) AS prev_rating,
LEAD(rating) OVER (PARTITION BY id ORDER BY year, month, day ASC) AS next_rating,
FROM dataset.table t
) t
WHERE (rating < 4.5 and prev_rating < 4.5) OR
(rating < 4.5 and next_rating < 4.5)
Below is for BigQuery Standard SQL
select * except(grp, seq_len)
from (
select *, sum(1) over(partition by grp) seq_len
from (
select *,
countif(rating >= 4.5) over(partition by id order by year, month, day) grp
from `project.dataset.table`
)
where rating < 4.5
)
where seq_len > 1
I have a query that return this result. How can i limit the occurrence of a value from the 4th column.
19 1 _BOURC01 1
20 1 _BOURC01 3 2019-11-18
20 1 _BOURC01 3 2017-01-02
21 1 _BOURC01 6
22 1 _BOURC01 10
23 1 _BOURC01 13 2016-06-06
24 1 _BOURC01 21 2016-09-19
My Query:
SELECT "_44_SpeakerSpeech"."id" AS "id", "_44_SpeakerSpeech"."active" AS "active", "_44_SpeakerSpeech"."id_speaker" AS "id_speaker", "_44_SpeakerSpeech"."Speech" AS "Speech", "34 Program Weekend"."date" AS "date"
FROM "_44_SpeakerSpeech"
LEFT JOIN "_34_programWeekend" "34 Program Weekend" ON "_44_SpeakerSpeech"."Speech" = "34 Program Weekend"."theme_id"
WHERE "id_speaker" = "_BOURC01"
ORDER BY id_speaker, Speech, date DESC
Thanks
I think this is what you want here:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY s.id, s.active, s.id_speaker, s.Speech
ORDER BY p.date DESC) rn
FROM "_44_SpeakerSpeech" s
LEFT JOIN "_34_programWeekend" p ON s.Speech = p.theme_id
WHERE s.id_speaker = '_BOURC01'
)
SELECT id, active, id_speaker, Speech, date
FROM cte
WHERE rn = 1;
This logic assumes that when two or more records all have the same columns values (excluding the date), you want to retain only the latest record.
I have a dataset with different clients, and their sales count. Over time, some clients get added and deleted from the data. How do I make sure that when I look at the sales counts, that I am only using a selection of the clients that were in the data set all the time? Ie if I have a client that doesn't have a record for 2018-03, then I don't want that client to be part of the entire query. If a clients does not have a record in 2020-03, then I also do not want this client to be part of the entire query.
For example, the following query:
select DATE_PART (y, sold_date)as year, DATE_PART (mm, sold_date) as month, count(distinct(client))
from sales_data
where sold_date > '2018-01-01'
group by year, month
order by year,month
Yields
year month count
2018 1 78
2018 2 83
2018 3 80
2018 4 83
2018 5 84
2018 6 81
2018 7 83
2018 8 90
2018 9 89
2018 10 95
2018 11 94
2018 12 97
2019 1 102
2019 2 103
2019 3 102
2019 4 105
2019 5 103
2019 6 104
2019 7 104
2019 8 106
2019 9 106
2019 10 108
2019 11 109
2019 12 104
2020 1 104
2020 2 102
2020 3 103
2020 4 98
2020 5 97
2020 6 79
So I want to only use the clients that are in all months, they should not be more than 78, because there can not be more users than the minimal month (2018-1).
FYI, I am using Amazon Redshift here but I am OK with a query that's rdbms agnostic or works for SQL-Server/Oracle/MySQL/PostgreSQL, I am just interested in a pattern on how to solve this issue effectively.
If I'm understanding what you want correctly, and if this is just a one-off query, you could use a correlated subquery in the where clause:
SELECT
DATE_PART(y, s.sold_date) AS year,
DATE_PART(mm, s.sold_date) AS month,
COUNT(DISTINCT s.client)
FROM
sales_data AS s
WHERE
EXISTS (
SELECT sd.client FROM sales_data AS sd WHERE DATE_PART(y,
sd.sold_date) = 2018 AND DATE_PART(mm, sd.sold_date) = 1 AND
sd.client = s.client
) AND
s.sold_date > '2018-01-01'
GROUP BY
year,
month
ORDER
DATE_PART(y, s.sold_date),
DATE_PART(mm, s.sold_date)
presence in all months can be done with 2-step aggregation:
group sales data by customer ID having all months
group sales data joined to (1) by year, month
like this (=12 can be a dynamic expression, depending on the amount of history you have)
with
stable_customers as (
select customer_id
from sales_data
group by 1
having count(distinct date_trunc('month' from sold_date)=12
)
select
DATE_PART (y, sold_date) as year
,DATE_PART (mm, sold_date) as month,
,count(1)
from sales_date
join stable_customers
using (customer_id)
where sold_date > '2018-01-01'
group by year, month
order by year,month
Use window functions. Unfortunately, SQL Server does not support count(distinct) as a window function. Fortunately, there is a simple work-around using dense_rank():
select year, month, count(distinct client)
from (select sd.*, year, month,
(dense_rank() over (order by year, month) +
dense_rank() over (order by year desc, month desc)
) as num_months,
(dense_rank() over (partition by client order by year, month) +
dense_rank() over (partition by client order by year desc, month desc)
) as num_months_client
from sales_data sd cross apply
(values (year(sold_date), month(sold_date))) v(year, month)
where sd.sold_date > '2018-01-01'
) sd
where num_months_client = num_months
group by year, month
order by year, month;
Note: This looks at all months that are in the data. If all clients are missing 2019-03, then that months is not considered at all.
I have a table with items, dates, and prices and I am trying to find a way to write a SELECT query in PostgreSQL which will skip rows with duplicate prices so that, only the first and last occurrence of the same price in a row would stay. After the price change, it can go back to the previous value and it should be preserved as well.
id date price item
1 20.10.2018 10 a
2 21.10.2018 10 a
3 22.10.2018 10 a
4 23.10.2018 15 a
5 24.10.2018 15 a
6 25.10.2018 15 a
7 26.10.2018 10 a
8 27.10.2018 10 a
9 28.10.2018 10 a
10 29.10.2018 10 a
11 26.10.2018 3 b
12 27.10.2018 3 b
13 28.10.2018 3 b
14 29.10.2018 3 c
Result:
id date price item
1 20.10.2018 10 a
3 22.10.2018 10 a
4 23.10.2018 15 a
6 25.10.2018 15 a
7 26.10.2018 10 a
10 29.10.2018 10 a
11 26.10.2018 3 b
13 28.10.2018 3 b
14 29.10.2018 3 c
You can use lag() and lead():
select id, date, price, item
from (select t.*,
lag(price) over (partition by item order by date) as prev_price,
lead(price) over (partition by item order by date) as next_price
from t
) t
where prev_price is null or prev_price <> price or
next_price is null or next_price <> price