I need help in business days calculation.
I've two tables
1) One table ACTUAL_TABLE containing order date and contact date with timestamp datatypes.
2) The second table BUSINESS_DATES has each of the calendar dates listed and has a flag to indicate weekend days.
using these two tables, I need to ensure business days and not calendar days (which is the current logic) is calculated between these two fields.
My thought process was to first get a range of dates by comparing ORDER_DATE with TABLE_DATE field and then do a similar comparison of CONTACT_DATE to TABLE_DATE field. This would get me a range from the BUSINESS_DATES table which I can then use to calculate count of days, sum(Holiday_WKND_Flag) fields making the result look like:
Order# | Count(*) As DAYS | SUM(WEEKEND DATES)
100 | 25 | 8
However this only works when I use a specific order number and cant' bring all order numbers in a sub query.
My Query:
SELECT SUM(Holiday_WKND_Flag), COUNT(*) FROM
(
SELECT
* FROM
BUSINESS_DATES
WHERE BUSINESS.Business BETWEEN (SELECT ORDER_DATE FROM ACTUAL_TABLE
WHERE ORDER# = '100'
)
AND
(SELECT CONTACT_DATE FROM ACTUAL_TABLE
WHERE ORDER# = '100'
)
TEMP
Uploading the table structure for your reference.
SELECT ORDER#, SUM(Holiday_WKND_Flag), COUNT(*)
FROM business_dates bd
INNER JOIN actual_table at ON bd.table_date BETWEEN at.order_date AND at.contact_date
GROUP BY ORDER#
Instead of joining on a BETWEEN (which always results in a bad Product Join) followed by a COUNT you better assign a bussines day number to each date (in best case this is calculated only once and added as a column to your calendar table). Then it's two Equi-Joins and no aggregation needed:
WITH cte AS
(
SELECT
Cast(table_date AS DATE) AS table_date,
-- assign a consecutive number to each busines day, i.e. not increased during weekends, etc.
Sum(CASE WHEN Holiday_WKND_Flag = 1 THEN 0 ELSE 1 end)
Over (ORDER BY table_date
ROWS Unbounded Preceding) AS business_day_nbr
FROM business_dates
)
SELECT ORDER#,
Cast(t.contact_date AS DATE) - Cast(t.order_date AS DATE) AS #_of_days
b2.business_day_nbr - b1.business_day_nbr AS #_of_business_days
FROM actual_table AS t
JOIN cte AS b1
ON Cast(t.order_date AS DATE) = b1.table_date
JOIN cte AS b2
ON Cast(t.contact_date AS DATE) = b2.table_date
Btw, why are table_date and order_date timestamp instead of a date?
Porting from Oracle?
You can use this query. Hope it helps
select order#,
order_date,
contact_date,
(select count(1)
from business_dates_table
where table_date between a.order_date and a.contact_date
and holiday_wknd_flag = 0
) business_days
from actual_table a
Related
I can't get my head around whether this is even possible, but I feel like I might have done it before and lost that bit of code. I am trying to craft a select statement that contains an inner join on a subquery to show the number of days between two dates from the same table.
A simple example of the data structure would look like:
Name ID Date Day Hours
Bill 1 3/3/20 Thursday 8
Fred 2 4/3/20 Monday 6
Bill 1 8/3/20 Tuesday 2
Based on this data, I want to select each row plus an extra column which is the number of days between the date from each row for each ID. Something like:
Select * from tblData
Inner join (datediff(Select Top(1) Date from tblData where Date < Date), Date) And ID = ID)
or for simplicity:
Select * from tblData
Inner join (datediff(Select Top(1) Date from tblData where Date < 8/3/20), 8/3/20) And ID = 1)
The resulting dataset would look like:
Name ID Date Day Hours DaysBtwn
Bill 1 3/3/20 Thursday 8 4 (Assuming there was an earlier row in the table)
Fred 2 4/3/20 Monday 6 5 (Assuming there was an earlier row in the table)
Bill 1 8/3/20 Tuesday 2 5 (Based on the previous row date being 3/3/20 for Bill)
Does this make sense and am I trying to do this the wrong way? I want to do this for about 600000 rows in table and therefore efficiency is the key, so if there is a better way to do this, i'm open to suggestions.
You can use lag():
select t.*, datediff(day, lag(date) over(partition by id order by date), date) diff
from mytable t
I think you just want lag():
select t.*,
datediff(day,
lag(date) over (partition by name order by date),
date
) as diff
from tblData t;
Note: If you want to filter the data so rows in the result set are used for the lag() but not in the result set, then use a subquery:
select t.*
from (select t.*,
datediff(day,
lag(date) over (partition by name order by date),
date
) as diff
from tblData t
) t
where date < '2020-08-03';
Also note the use of the date constant as a string in YYYY-MM-DD format.
I have a table with a foreign_key_id column and a date column.
For each row that has the same foreign key, there is a different date, and if I order by foreign_key_id, date , 90% of the time all the dates are consecutive.
There are some edge cases though, where there are multiple entries with the same foreign_key that don't have consecutive dates.
Trying to come up with an easy way to identify all the foreign_key_id 's that don't have consecutive dates. Any ideas?
I was thinking of left joining on to a generated series, somehow partitioning by track id, but keep hitting a mental wall. My sql query editor keeps crashing, so that is adding some more unrelated frustration
EDIT:
I ended up doing an order by foreign_key_id, date , copying and pasting the result in excel, and then finding what I needed by doing this type of logic formula:
=IF( (B91 = B90), (F91 =(F90 + 1)) , 1 ) , where b is the foreign key column and F is the date column
but wondering if something similar could be done in sql. here's what I had when I gave up and went to excel:
select to_char(date_range.days, 'yyyy-mm-dd') as x
, data.*
from (
select generate_series('2019-04-30'::date,'2019-11-05'::date, '1 day')::date as days
) as date_range
left join(
select foreign_key_id, date
from table_a
order by foreign_key_id, date
) data on data.date = date_range.days
where foreign_key_id is null
You could do that, sure. No joins needed either. Use LAG(datecol) OVER(PARTITION BY foreignkeycol ORDER BY datecol) to get the date of the previous row for the same fk, diff it to the current date to show how many intervals (days? Minutes?) have passed since that date and then wrap it all in something that does WHERE thedifference <> 1 (Or however you define consecutive - if consecutive to you is "every 2 days" then it would be anything that doesn't have a difference of 2)
If you want both rows either side of the gap, use LEAD (same format as LAG) to get the next date and calc two diffs, then do WHERE difftoprev <> 1 or difftonext <>1 etc
It would look something like this (untested)
WITH cte AS (
SELECT foreignkeycol, datecol,
LAG(datecol) OVER(PARTITION BY foreignkeycol ORDER BY datecol) as prevdate,
LEAD(datecol) OVER(PARTITION BY foreignkeycol ORDER BY datecol) as nextdate
FROM table
)
SELECT *
FROM cte
WHERE
DATE_PART('day', datecol - prevdate) <> 1 OR
DATE_PART('day', nextdate - datecol) <> 1
I would use lead():
select t.*
from (select t.*,
lead(date) over (partition by foreign_key_id order by date) as next_date
from t
) t
where next_date <> date + interval '1 day';
This will provide each row where the next row does not have the expected date.
I'm working on a bit of PostgreSQL to grab the first 10 and last 10 invoices of every month between certain dates. I am having unexpected output in the lateral joins. Firstly the limit is not working, and each of the array_agg aggregates is returning hundreds of rows instead of limiting to 10. Secondly, the aggregates appear to be the same, even though one is ordered ASC and the other DESC.
How can I retrieve only the first 10 and last 10 invoices of each month group?
SELECT first.invoice_month,
array_agg(first.id) first_ten,
array_agg(last.id) last_ten
FROM public.invoice i
JOIN LATERAL (
SELECT id, to_char(invoice_date, 'Mon-yy') AS invoice_month
FROM public.invoice
WHERE id = i.id
ORDER BY invoice_date, id ASC
LIMIT 10
) first ON i.id = first.id
JOIN LATERAL (
SELECT id, to_char(invoice_date, 'Mon-yy') AS invoice_month
FROM public.invoice
WHERE id = i.id
ORDER BY invoice_date, id DESC
LIMIT 10
) last on i.id = last.id
WHERE i.invoice_date BETWEEN date '2017-10-01' AND date '2018-09-30'
GROUP BY first.invoice_month, last.invoice_month;
This can be done with a recursive query that will generate the interval of months for who we need to find the first and last 10 invoices.
WITH RECURSIVE all_months AS (
SELECT date_trunc('month','2018-01-01'::TIMESTAMP) as c_date, date_trunc('month', '2018-05-11'::TIMESTAMP) as end_date, to_char('2018-01-01'::timestamp, 'YYYY-MM') as current_month
UNION
SELECT c_date + interval '1 month' as c_date,
end_date,
to_char(c_date + INTERVAL '1 month', 'YYYY-MM') as current_month
FROM all_months
WHERE c_date + INTERVAL '1 month' <= end_date
),
invocies_with_month as (
SELECT *, to_char(invoice_date::TIMESTAMP, 'YYYY-MM') invoice_month FROM invoice
)
SELECT current_month, array_agg(first_10.id), 'FIRST 10' as type FROM all_months
JOIN LATERAL (
SELECT * FROM invocies_with_month
WHERE all_months.current_month = invoice_month AND invoice_date >= '2018-01-01' AND invoice_date <= '2018-05-11'
ORDER BY invoice_date ASC limit 10
) first_10 ON TRUE
GROUP BY current_month
UNION
SELECT current_month, array_agg(last_10.id), 'LAST 10' as type FROM all_months
JOIN LATERAL (
SELECT * FROM invocies_with_month
WHERE all_months.current_month = invoice_month AND invoice_date >= '2018-01-01' AND invoice_date <= '2018-05-11'
ORDER BY invoice_date DESC limit 10
) last_10 ON TRUE
GROUP BY current_month;
In the code above, '2018-01-01' and '2018-05-11' represent the dates between we want to find the invoices. Based on those dates, we generate the months (2018-01, 2018-02, 2018-03, 2018-04, 2018-05) that we need to find the invoices for.
We store this data in all_months.
After we get the months, we do a lateral join in order to join the invoices for every month. We need 2 lateral joins in order to get the first and last 10 invoices.
Finally, the result is represented as:
current_month - the month
array_agg - ids of all selected invoices for that month
type - type of the selected invoices ('first 10' or 'last 10').
So in the current implementation, you will have 2 rows for each month (if there is at least 1 invoice for that month). You can easily join that in one row if you need to.
LIMIT is working fine. It's your query that's broken. JOIN is just 100% the wrong tool here; it doesn't even do anything close to what you need. By joining up to 10 rows with up to another 10 rows, you get up to 100 rows back. There's also no reason to self join just to combine filters.
Consider instead window queries. In particular, we have the dense_rank function, which can number every row in the result set according to groups:
SELECT
invoice_month,
time_of_month,
ARRAY_AGG(id) invoice_ids
FROM (
SELECT
id,
invoice_month,
-- Categorize as end or beginning of month
CASE
WHEN month_rank <= 10 THEN 'beginning'
WHEN month_reverse_rank <= 10 THEN 'end'
ELSE 'bug' -- Should never happen. Just a fall back in case of a bug.
END AS time_of_month
FROM (
SELECT
id,
invoice_month,
dense_rank() OVER (PARTITION BY invoice_month ORDER BY invoice_date) month_rank,
dense_rank() OVER (PARTITION BY invoice_month ORDER BY invoice_date DESC) month_rank_reverse
FROM (
SELECT
id,
invoice_date,
to_char(invoice_date, 'Mon-yy') AS invoice_month
FROM public.invoice
WHERE invoice_date BETWEEN date '2017-10-01' AND date '2018-09-30'
) AS fiscal_year_invoices
) ranked_invoices
-- Get first and last 10
WHERE month_rank <= 10 OR month_reverse_rank <= 10
) first_and_last_by_month
GROUP BY
invoice_month,
time_of_month
Don't be intimidated by the length. This query is actually very straightforward; it just needed a few subqueries.
This is what it does logically:
Fetch the rows for the fiscal year in question
Assign a "rank" to the row within its month, both counting from the beginning and from the end
Filter out everything that doesn't rank in the 10 top for its month (counting from either direction)
Adds an indicator as to whether it was at the beginning or end of the month. (Note that if there's less than 20 rows in a month, it will categorize more of them as "beginning".)
Aggregate the IDs together
This is the tool set designed for the job you're trying to do. If really needed, you can adjust this approach slightly to get them into the same row, but you have to aggregate before joining the results together and then join on the month; you can't join and then aggregate.
SELECT oi.created_at, count(oi.id_order_item)
FROM order_item oi
The result is the follwoing:
2016-05-05 1562
2016-05-06 3865
2016-05-09 1
...etc
The problem is that I need information for all days even if there were no id_order_item for this date.
Expected result:
Date Quantity
2016-05-05 1562
2016-05-06 3865
2016-05-07 0
2016-05-08 0
2016-05-09 1
You can't count something that is not in the database. So you need to generate the missing dates in order to be able to "count" them.
SELECT d.dt, count(oi.id_order_item)
FROM (
select dt::date
from generate_series(
(select min(created_at) from order_item),
(select max(created_at) from order_item), interval '1' day) as x (dt)
) d
left join order_item oi on oi.created_at = d.dt
group by d.dt
order by d.dt;
The query gets the minimum and maximum date form the existing order items.
If you want the count for a specific date range you can remove the sub-selects:
SELECT d.dt, count(oi.id_order_item)
FROM (
select dt::date
from generate_series(date '2016-05-01', date '2016-05-31', interval '1' day) as x (dt)
) d
left join order_item oi on oi.created_at = d.dt
group by d.dt
order by d.dt;
SQLFiddle: http://sqlfiddle.com/#!15/49024/5
Friend, Postgresql Count function ignores Null values. It literally does not consider null values in the column you are searching. For this reason you need to include oi.created_at in a Group By clause
PostgreSql searches row by row sequentially. Because an integral part of your query is Count, and count basically stops the query for that row, your dates with null id_order_item are being ignored. If you group by oi.created_at this column will trump the count and return 0 values for you.
SELECT oi.created_at, count(oi.id_order_item)
FROM order_item oi
Group by io.created_at
From TechontheNet (my most trusted source of information):
Because you have listed one column in your SELECT statement that is not encapsulated in the count function, you must use a GROUP BY clause. The department field must, therefore, be listed in the GROUP BY section.
Some info on Count in PostgreSql
http://www.postgresqltutorial.com/postgresql-count-function/
http://www.techonthenet.com/postgresql/functions/count.php
Solution #1 You need Date Table where you stored all date data. Then do a left join depending on period.
Solution #2
WITH DateTable AS
(
SELECT DATEADD(dd, 1, CONVERT(DATETIME, GETDATE())) AS CreateDateTime, 1 AS Cnter
UNION ALL
SELECT DATEADD(dd, -1, CreateDateTime), DateTable.Cnter + 1
FROM DateTable
WHERE DateTable.Cnter + 1 <= 5
)
Generate Temporary table based on your input and then do a left Join.
I would appreciate a little expert help please.
in an SQL SELECT statement I am trying to get the last day with data per month for the last year.
Example, I am easily able to get the last day of each month and join that to my data table, but the problem is, if the last day of the month does not have data, then there is no returned data. What I need is for the SELECT to return the last day with data for the month.
This is probably easy to do, but to be honest, my brain fart is starting to hurt.
I've attached the select below that works for returning the data for only the last day of the month for the last 12 months.
Thanks in advance for your help!
SELECT fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,fd.column_name
FROM super_table fd,
(SELECT TRUNC(daterange,'MM')-1 first_of_month
FROM (
select TRUNC(sysdate-365,'MM') + level as DateRange
from dual
connect by level<=365)
GROUP BY TRUNC(daterange,'MM')) fom
WHERE fd.cust_id = :CUST_ID
AND fd.coll_date > SYSDATE-400
AND TRUNC(fd.coll_date) = fom.first_of_month
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)
You probably need to group your data so that each month's data is in the group, and then within the group select the maximum date present. The sub-query might be:
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY YEAR(coll_date) * 100 + MONTH(coll_date);
This presumes that the functions YEAR() and MONTH() exist to extract the year and month from a date as an integer value. Clearly, this doesn't constrain the range of dates - you can do that, too. If you don't have the functions in Oracle, then you do some sort of manipulation to get the equivalent result.
Using information from Rhose (thanks):
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY TO_CHAR(coll_date, 'YYYYMM');
This achieves the same net result, putting all dates from the same calendar month into a group and then determining the maximum value present within that group.
Here's another approach, if ANSI row_number() is supported:
with RevDayRanked(itemDate,rn) as (
select
cast(coll_date as date),
row_number() over (
partition by datediff(month,coll_date,'2000-01-01') -- rewrite datediff as needed for your platform
order by coll_date desc
)
from super_table
)
select itemDate
from RevDayRanked
where rn = 1;
Rows numbered 1 will be nondeterministically chosen among rows on the last active date of the month, so you don't need distinct. If you want information out of the table for all rows on these dates, use rank() over days instead of row_number() over coll_date values, so a value of 1 appears for any row on the last active date of the month, and select the additional columns you need:
with RevDayRanked(cust_id, server_name, coll_date, rk) as (
select
cust_id, server_name, coll_date,
rank() over (
partition by datediff(month,coll_date,'2000-01-01')
order by cast(coll_date as date) desc
)
from super_table
)
select cust_id, server_name, coll_date
from RevDayRanked
where rk = 1;
If row_number() and rank() aren't supported, another approach is this (for the second query above). Select all rows from your table for which there's no row in the table from a later day in the same month.
select
cust_id, server_name, coll_date
from super_table as ST1
where not exists (
select *
from super_table as ST2
where datediff(month,ST1.coll_date,ST2.coll_date) = 0
and cast(ST2.coll_date as date) > cast(ST1.coll_date as date)
)
If you have to do this kind of thing a lot, see if you can create an index over computed columns that hold cast(coll_date as date) and a month indicator like datediff(month,'2001-01-01',coll_date). That'll make more of the predicates SARGs.
Putting the above pieces together, would something like this work for you?
SELECT fd.cust_id,
fd.server_name,
fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,
fd.column_name
FROM super_table fd,
WHERE fd.cust_id = :CUST_ID
AND TRUNC(fd.coll_date) IN (
SELECT MAX(TRUNC(coll_date))
FROM super_table
WHERE coll_date > SYSDATE - 400
AND cust_id = :CUST_ID
GROUP BY TO_CHAR(coll_date,'YYYYMM')
)
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)