I've a table data as below, now I need to fetch the record with in same code, where (Value2-Value1)*2 of one row >= (Value2-Value1) of consequtive date row. (all dates are uniform with in all codes)
---------------------------------------
code Date Value1 Value2
---------------------------------------
1 1-1-2018 13 14
1 2-1-2018 14 16
1 4-1-2018 15 18
2 1-1-2019 1 3
2 2-1-2018 2 3
2 4-1-2018 3 7
ex: output needs to be
1 1-1-2018 13 14
as I am begginer to SQL coding, tried my best, but cannot get through with compare only on consequtive dates.
Use a self join.
You can specify all the conditions you've listed in the ON clause:
SELECT T0.code, T0.Date, T0.Value1, T0.Value2
FROM Table As T0
JOIN Table As T1
ON T0.code = T1.code
AND T0.Date = DateAdd(Day, 1, T1.Date)
AND (T0.Value2 - T0.Value1) * 2 >= T1.Value2 - T1.Value1
Related
I have table1 as below.
num
value
1
10
2
15
3
20
table2
ver
value
1.0
5
2.0
15
3.0
18
Output should be as below. I need to select all rows from table1 such that table1.value <= table2.value.
num
value
1
10
2
15
I tried this, it's not working.
select from table1 where value <= (exec value from table2)
From a logical point of view what you're asking kdb to compare is:
10 15 20<=5 15 18
Because these are equal lengths, kdb assumes you mean pairwise comparison, aka
10<=5
15<=15
20<=18
to which it would return
q)10 15 20<=5 15 18
010b
What you actually seem to mean (based on your expected output) is 10 15 20<=max(5 15 18). So in that case you would want:
q)t1:([]num:1 2 3;val:10 15 20)
q)t2:([]ver:1 2 3.;val:5 15 18)
q)select from t1 where val<=exec max val from t2
num val
-------
1 10
2 15
As an aside, you can't/shouldn't have a column called value as it clashes with a keyword
value is a keyword so don't assign to it.
Assuming you want all values from table1 with value less than the max value in table2 you could do:
q)table1:([]num:til 3;val:10 15 20)
q)table2:([]ver:`float$til 3;val:5 15 18)
q)select from table1 where val<=max table2`val
num val
-------
0 10
1 15
I have a complex situation in PostgreSQL 11 where i need to generate a numbering based on a single figure which i get it from a CTE.
Below is the CTE
WITH pending_orders_to_be_processed_details
AS
(
SELECT ROW_NUMBER() OVER(ORDER BY so.create_date ) as queue_no
, name,so.create_date ::TIMESTAMP
FROM picking sp
LEFT JOIN order so ON so.name=sp.origin
WHERE sp.state IN('assigned','confirmed')
)
,orders_which_can_be_processed_today AS
(
-- This CTE will give me a count of orders
and its hourly average, Lets say count is 400 and hourly avg is 3
)
Now i need to number the details according to the hourly average, Means the first 3 orders need to be ranked as 1, next 3 to be ranked as 2 and so on, so that i can able to identify that these can be processed based on this ranking.
Input will be
name queu_number. create_date
so1 1 2021-03-11 12:00:00
so2 2 2021-03-11 13:00:00
so3 3 2021-03-11 14:00:00
so4 4 2021-03-11 15:00:00
so5 5 2021-03-11 16:00:00
so6 6 2021-03-11 17:00:00
so7 7 2021-03-11 18:00:00
so8 8 2021-03-11 19:00:00
so9 9 2021-03-11 20:00:00
The expected output will be
name rank
so1 1
so2 1
so3 1
so4 2
so5 2
so6 2
so7 3
so8 3
so9 3
Any help/suggestions.
Edit: I recently learned about a function, which fits well here:
demo:db<>fiddle
You can use the ntile() window function for that:
SELECT
*,
ntile(3) OVER (ORDER BY create_date)
FROM mytable
demo:db<>fiddle
Since you already created a cumulative row count, you can use this to create your expected rank:
SELECT
*,
floor((queue_no - 1) / 3) + 1 as rank
FROM my_cte
queue_no - 1 (so, 1 to 3 will be shifted to 0 to 2)
Diff by 3: so, 0 to 2 will be 0.x and 3 to 5 will be 1.x, ...
Now round these result to 0, 1, 2, ...
If you want to start with 1 instead of 0, add 1
Suppose I have data formatted in the following way (FYI, total row count is over 30K):
customer_id order_date order_rank
A 2017-02-19 1
A 2017-02-24 2
A 2017-03-31 3
A 2017-07-03 4
A 2017-08-10 5
B 2016-04-24 1
B 2016-04-30 2
C 2016-07-18 1
C 2016-09-01 2
C 2016-09-13 3
I need a 4th column, let's call it days_since_last_order which, in the case where order_rank = 1 then 0 else calculate the number of days since the previous order (with rank n-1).
So, the above would return:
customer_id order_date order_rank days_since_last_order
A 2017-02-19 1 0
A 2017-02-24 2 5
A 2017-03-31 3 35
A 2017-07-03 4 94
A 2017-08-10 5 38
B 2016-04-24 1 0
B 2016-04-30 2 6
C 2016-07-18 1 79
C 2016-09-01 2 45
C 2016-09-13 3 12
Is there an easier way to calculate the above with a window function (or similar) rather than join the entire dataset against itself (eg. on A.order_rank = B.order_rank - 1) and doing the calc?
Thanks!
use the lag window function
SELECT
customer_id
, order_date
, order_rank
, COALESCE(
DATE(order_date)
- DATE(LAG(order_date) OVER (PARTITION BY customer_id ORDER BY order_date))
, 0)
FROM <table_name>
I need to find users who have posted three times or more, three months in a row. I wrote this query:
select count(id), owneruserid, extract(month from creationdate) as postmonth from posts
group by owneruserid, postmonth
having count(id) >=3
order by owneruserid, postmonth
And I get this:
count owneruserid postmonth
36 -1 1
23 -1 2
45 -1 3
41 -1 4
18 -1 5
24 -1 6
31 -1 7
78 -1 8
83 -1 9
17 -1 10
88 -1 11
127 -1 12
3 6 11
3 7 12
4 8 1
8 8 12
4 12 4
3 12 5
3 22 2
4 22 4
(truncated)
Which is great. How can I query for users who posted three times or more, three months or more in a row? Thanks.
This is called the Islands and Gaps problem, specifically it's an Island problem with a date range. You should,
Fix this question up.
Flag it to be sent to dba.stackexchange.com
To solve this,
Create a pseudo column with a window that has 1 if the row preceding it does not correspond to the preceding mont
Create groups out of that with COUNT()
Check to make sure the count(*) for the group is greater than or equal to three.
Query,
SELECT l.id, creationdaterange, count(*)
FROM (
SELECT t.id,
t.creationdate,
count(range_reset) OVER (PARTITION BY t.id ORDER BY creationdate) AS creationdaterange
FROM (
SELECT id,
creationdate,
CASE
WHEN date_trunc('month',creationdate::date)::date - interval '1 month' = date_trunc('month',lag(creationdate))::date OVER (PARTITION BY id ORDER BY creationdate)
THEN 1
END AS range_reset
FROM post
ORDER BY id, creationdate
) AS t;
) AS l
GROUP BY t.id, creationdaterange
HAVING count(*) >= 3;
I want to get a table that constructs a column that tracks how many times an id appears in a given week. If the id appears once it is given a 1, if it appears twice it is given a 2, but if it appears more than two times it is given a 0.
id date
a 2015-11-10
a 2015-11-25
a 2015-11-09
b 2015-11-10
b 2015-11-09
a 2015-11-05
b 2015-11-23
b 2015-11-28
b 2015-12-04
a 2015-11-10
b 2015-12-04
a 2015-12-07
a 2015-12-09
c 2015-11-30
a 2015-12-06
c 2015-10-31
c 2015-11-04
b 2015-12-01
a 2015-10-30
a 2015-12-14
the one week intervals are given as follows
1 - 2015-10-30 to 2015-11-05
2 - 2015-11-06 to 2015-11-12
3 - 2015-11-13 to 2015-11-19
4 - 2015-11-20 to 2015-11-26
5 - 2015-11-27 to 2015-12-03
6 - 2015-12-04 to 2015-12-10
7 - 2015-12-11 to 2015-12-17
The table should look like this.
id interval count
a 1 2
b 1 0
c 1 2
a 2 0
b 2 2
c 2 0
a 3 0
b 3 0
c 3 0
a 4 1
b 4 1
c 4 0
a 5 0
b 5 2
c 5 1
a 6 0
b 6 2
c 6 0
a 7 1
b 7 0
c 7 0
The interval column doesn't have to be there, I simply added it for clarity.
I am new to sql and am unsure how to break the dates into intervals. The only thing I have is grouping by date and counting.
Select id ,date, count (*) as frequency
from data_1
group by id, date having frequency <= 2;
Looking at just the data you provided, this does the trick:
SELECT v.id,
i.interval,
coalesce((CASE WHEN sub.cnt < 3 THEN sub.cnt ELSE 0 END), 0) AS count
FROM (VALUES('a'), ('b'), ('c')) v(id)
CROSS JOIN generate_series(1, 7) i(interval)
LEFT JOIN (
SELECT id, ((date - '2015-10-30')/7 + 1)::int AS interval, count(*) AS cnt
FROM my_table
GROUP BY 1, 2) sub USING (id, interval)
ORDER BY 2, 1;
A few words of explanation:
You have three id values which are here recreated with a VALUES clause. If you have many more or don't know beforehand which id's to enumerate, you can always replace the VALUES clause with a sub-query.
You provide a specific date range over 7 weeks. Since you might have weeks where a certain id is not present you need to generate a series of the interval values and CROSS JOIN that to the id values above. This yields the 21 rows you are looking for.
Then you calculate the occurrences of ids in intervals. You can subtract a date from another date which will give you the number of days in between. So subtract the date of the row from the earliest date, divide that by 7 to get the interval period, add 1 to make the interval 1-based and convert to integer. You can then convert counts of > 2 to 0 and NULL to 0 with a combination of CASE and coalesce().
The query outputs the interval too, otherwise you will have no clue what the data refers to. Optionally, you can turn this into a column which shows the date range of the interval.
More flexible solution
If you have more ids and a larger date range, you can use the below version which first determines the distinct ids and the date range. Note that the interval is now 0-based to make calculations easier. Not that it matters much because instead of the interval number, the corresponding date range is displayed.
WITH mi AS (
SELECT min(date) AS min, ((max(date) - min(date))/7)::int AS intv FROM my_table)
SELECT v.id,
to_char((mi.min + i.intv * 7)::timestamp, 'YYYY-mm-dd') || ' - ' ||
to_char((mi.min + i.intv * 7 + 6)::timestamp, 'YYYY-mm-dd') AS period,
coalesce((CASE WHEN sub.cnt < 3 THEN sub.cnt ELSE 0 END), 0) AS count
FROM mi,
(SELECT DISTINCT id FROM my_table) v
CROSS JOIN LATERAL generate_series(0, mi.intv) i(intv)
LEFT JOIN LATERAL (
SELECT id, ((date - mi.min)/7)::int AS intv, count(*) AS cnt
FROM my_table
GROUP BY 1, 2) sub USING (id, intv)
ORDER BY 2, 1;
SQLFiddle with both solutions.
Assuming you have a table of all users, this will do the trick.
select
users.id,
interval_table.id,
CASE
WHEN count(log_table.user_id)>2 THEN 0
ELSE count(log_table.user_id)
END
from users
cross join interval_table
left outer join log_table
on users.id = log_table.user_id
and log_table.event_date >= interval_table.start_interval
and log_table.event_date < interval_table.stop_interval
group by users.id, interval_table.id
order by interval_table.id, users.id
Check it out: http://sqlfiddle.com/#!15/1a822/21