i WanT a query to find the number of days between the longest and least tenured employee still working for the company. The output should include the number of employees with the longest-tenure, the number of employees with the least-tenure, and the number of days between both the longest-tenured and least-tenured hiring dates.
this is what i have so far,i am having trouble getting the difference to appear in output
select a.count, date_tenure
from
(
select distinct hire_date ,current_date - hire_date as date_tenure, count(id) over (partition by hire_date) as count, rank() over (order by current_date-hire_date) as rank_asc,
rank() over (order by current_date-hire_date desc) as rank_desc
from employees
where termination_date is null
order by 2 desc) a
where rank_desc = 1
or rank_asc = 1
table : employees
id
hire_date
termination_date
Related
I am attempting to pull 10 random records from each month of this year using this query here but I get an error "ERROR: relation "c1" does not exist
"
Not sure where I'm going wrong - I think it may be I'm using Mysql syntax instead, but how do I resolve this?
My desired output is like this
Month
Another header
2021-01
random email 1
2021-01
random email 2
total of ten random emails from January, then ten more for each month this year (til November of course as Dec yet to happen)..
With CTE AS
(
Select month,
email,
Row_Number() Over (Partition By month Order By FLOOR(RANDOM()*(1-1000000+1))) AS RN
From (
SELECT
DISTINCT(TO_CHAR(DATE_TRUNC('month', timestamp ), 'YYYY-MM')) AS month
,CASE
WHEN
JSON_EXTRACT_PATH_TEXT(json_extract_array_element_text (form_data,0),'name') = 'email'
THEN
JSON_EXTRACT_PATH_TEXT(json_extract_array_element_text (form_data,0),'value')
END AS email
FROM form_submits_y2 fs
WHERE fs.website_id IN (791)
AND month LIKE '2021%'
GROUP BY 1,2
ORDER BY 1 ASC
)
)
SELECT *
FROM CTE C1
LEFT JOIN
(SELECT RN
,month
,email
FROM CTE C2
WHERE C2.month = C1.month
ORDER BY RANDOM() LIMIT 10) C3
ON C1.RN = C3.RN
ORDER By month ASC```
You can't reference an outer table inside a derived table with a regular join. You need to use left join lateral to make that work
I did end up finding a more elegant solution to my query here via this source from github :
SELECT
month
,email
FROM
(
Select month,
email,
Row_Number() Over (Partition By month Order By FLOOR(RANDOM()*(1-1000000+1))) AS RN
From (
SELECT
TO_CHAR(DATE_TRUNC('month', timestamp ), 'YYYY-MM') AS month
,CASE
WHEN JSON_EXTRACT_PATH_TEXT(json_extract_array_element_text (form_data,0),'name') = 'email'
THEN JSON_EXTRACT_PATH_TEXT(json_extract_array_element_text (form_data,0),'value')
END AS email
FROM form_submits_y2 fs
WHERE fs.website_id IN (791)
AND month LIKE '2021%'
GROUP BY 1,2
ORDER BY 1 ASC
)
) q
WHERE
RN <=10
ORDER BY month ASC
I want to get the number of consecutive days from the current date using Postgres SQL.
enter image description here
Above is the scenario in which I have highlighted consecutive days count should be like this.
Below is the SQL query which I have created but it's not returning the expected result
with grouped_dates as (
select user_id, created_at::timestamp::date,
(created_at::timestamp::date - (row_number() over (partition by user_id order by created_at::timestamp::date) || ' days')::interval)::date as grouping_date
from watch_history
)
select * , dense_rank() over (partition by grouping_date order by created_at::timestamp::date) as in_streak
from grouped_dates where user_id = 702
order by created_at::timestamp::date
Can anyone please help me to resolve this issue?
If anyhow we can able to apply distinct for created_at field to below query then I will get solutions for my issue.
WITH list AS
(
SELECT user_id,
(created_at::timestamp::date - (row_number() over (partition by user_id order by created_at::timestamp::date) || ' days')::interval)::date as next_day
FROM watch_history
)
SELECT user_id, count(*) AS number_of_consecutive_days
FROM list
WHERE next_day IS NOT NULL
GROUP BY user_id
Does anyone have an idea how to apply distinct to created_at for the above mentioned query ?
To get the "number of consecutive days" for the same user_id :
WITH list AS
(
SELECT user_id
, array_agg(created_at) OVER (PARTITION BY user_id ORDER BY created_at RANGE BETWEEN CURRENT ROW AND '1 day' FOLLOWING) AS consecutive_days
FROM watch_history
)
SELECT user_id, count(DISTINCT d.day) AS number_of_consecutive_days
FROM list
CROSS JOIN LATERAL unnest(consecutive_days) AS d(day)
WHERE array_length(consecutive_days, 1) > 1
GROUP BY user_id
To get the list of "consecutive days" for the same user_id :
WITH list AS
(
SELECT user_id
, array_agg(created_at) OVER (PARTITION BY user_id ORDER BY created_at RANGE BETWEEN CURRENT ROW AND '1 day' FOLLOWING) AS consecutive_days
FROM watch_history
)
SELECT user_id
, array_agg(DISTINCT d.day ORDER BY d.day) AS list_of_consecutive_days
FROM list
CROSS JOIN LATERAL unnest(consecutive_days) AS d(day)
WHERE array_length(consecutive_days, 1) > 1
GROUP BY user_id
full example & result in dbfiddle
I have a table with:
ID (id client), date_start (subscription of SaaS), date_end (could be a date value or be NULL).
So I need a cumulative count of active clients month by month.
any idea on how to write that in Postgres and achieve this result?
Starting from this, but I don't know how to proceed
select
date_trunc('month', c.date_start)::date,
count(*)
from customer
Please check next solution:
select
subscrubed_date,
subscrubed_customers,
unsubscrubed_customers,
coalesce(subscrubed_customers, 0) - coalesce(unsubscrubed_customers, 0) cumulative
from (
select distinct
date_trunc('month', c.date_start)::date subscrubed_date,
sum(1) over (order by date_trunc('month', c.date_start)) subscrubed_customers
from customer c
order by subscrubed_date
) subscribed
left join (
select distinct
date_trunc('month', c.date_end)::date unsubscrubed_date,
sum(1) over (order by date_trunc('month', c.date_end)) unsubscrubed_customers
from customer c
where date_end is not null
order by unsubscrubed_date
) unsubscribed on subscribed.subscrubed_date = unsubscribed.unsubscrubed_date;
share SQL query
You have a table of customers. With a start date and sometimes an end date. As you want to group by date, but there are two dates in the table, you need to split these first.
Then, you may have months where only customers came and others where only customers left. So, you'll want a full outer join of the two sets.
For a cumulative sum (also called a running total), use SUM OVER.
with came as
(
select date_trunc('month', date_start) as month, count(*) as cnt
from customer
group by date_trunc('month', date_start)
)
, went as
(
select date_trunc('month', date_end) as month, count(*) as cnt
from customer
where date_end is not null
group by date_trunc('month', date_end)
)
select
month,
came.cnt as cust_new,
went.cnt as cust_gone,
sum(came.cnt - went.cnt) over (order by month) as cust_active
from came full outer join went using (month)
order by month;
I have a denormalized table with the columns:
buyer_id
order_id
item_id
item_price
item_category
I would like to return something that returns 1 row per buyer_id
buyer_id, sum(item_price), item_category
-- but ONLY for the category with the highest rank of sales along that specific buyer_id.
I can't get row_number() or partition to work because I need to order by the sum of item_price relative to item_category relative to buyer. Am I overlooking anything obvious?
You need a few layers of fudging here:
SELECT buyer_id, item_sum, item_category
FROM (
SELECT buyer_id,
rank() OVER (PARTITION BY buyer_id ORDER BY item_sum DESC) AS rnk,
item_sum, item_category
FROM (
SELECT buyer_id, sum(item_price) AS item_sum, item_category
FROM my_table
GROUP BY 1, 3) AS sub2) AS sub
WHERE rnk = 1;
In sub2 you calculate the sum of 'item_price' for each 'item_category' for each 'buyer_id'. In sub you rank these with a window function by 'buyer_id', ordering by 'item_sum' in descending order (so the highest 'item_sum' comes first). In the main query you select those rows where rnk = 1.
I have following data in the table as below and I am looking for a way to group the continuous time intervals for each id to return:
CREATE TABLE DUMMY
(
ID VARCHAR2(10 BYTE),
TIME_STAMP VARCHAR2(8 BYTE),
NAME VARCHAR2(255 BYTE)
);
SELECT ID, min(TIME_STAMP) "startDate", max(TIME_STAMP) "endDate", NAME
GROUP BY ID , NAME
something like
100 20011128 20011203 David
100 20011204 20011207 Unknown
100 20011208 20011215 David
100 20011216 20011220 Sara
and so on ...
ps. I have a sample script, but i don't know how to attach my file.
Hi every one here is more input:
There is only one record with time_stamp for a specific ID.
Users can be different, for example for day 1 David, day 2 unknown, day 3 David and so on.
So there is one row for every day of year for each ID but with different users.
Now, i want to see the break point, differences base on time_stamp intervals from day one
until last day for a specific ID in day order from begin day until last day.
Query Result should be :
ID NAME MIN_DATE MAX_DATE
100 David 20011128 20050407
100 Sara 20050408 20050417
100 David 20050418 20080416
100 Unknown 20080417 20080507
100 David 20080508 20080508
100 Unknown 20080509 20080607
100 David 20080608 20080608
100 Unknown 20080609 20080921
100 David 20080922 20080922
100 Unknown 20080923 20081231
100 David 20090101 20090405
thanks
Hi again, many thanks to everyone, i have solved the problem, here is the solution:
select id, min(time_stamp), max(time_stamp), name
from ( select id, time_stamp, name,
max(rn) over (order by time_stamp) grp
from ( select id, time_stamp, name,
case
when lag(name) over (order by time_stamp) <> name or
row_number() over (order by time_stamp) = 1
then row_number() over (order by time_stamp)
end rn
from dummy
)
)
group by id, grp, name
order by 1
Select
ID,
Name,
min(time_stamp) min_date,
max(time_stamp) max_date
from
Dummy
group by
Id,
Name
That should work.
IF you want the date range for each Id, but all the names you can do:
Select
d.Id,
d.Name,
dr.min_date,
dr.max_date
from
Dummy d
JOIN
(Select
Id,
min(time_stamp) min_date,
max(time_stamp) max_date
from
Dummy
group by
Id
) dr
on ( dr.Id = d.Id)