Query for getting previous date in oracle in specific scenario - sql

I have the below data in a table A which I need to insert into table B along with one computed column.
TABLE A:
Account_No | Balance | As_on_date
1001 |-100 | 1-Jan-2013
1001 |-150 | 2-Jan-2013
1001 | 200 | 3-Jan-2013
1001 |-250 | 4-Jan-2013
1001 |-300 | 5-Jan-2013
1001 |-310 | 6-Jan-2013
Table B:
In table B, there should be no of days to be shown when balance is negative and
the date one which it has gone into negative.
So, for 6-Jan-2013, this table should show below data:
Account_No | Balance | As_on_date | Days_passed | Start_date
1001 | -310 | 6-Jan-2013 | 3 | 4-Jan-2013
Here, no of days should be the days when the balance has gone negative in recent time and
not from the old entry.
I need to write a SQL query to get the no of days passed and the start date from when the
balance has gone negative.
I tried to formulate a query using Lag analytical function, but I am not succeeding.
How should I check the first instance of negative balance by traversing back using LAG function?
Even the first_value function was given a try but not getting how to partition in it based on negative value.
Any help or direction on this will be really helpful.

Here's a way to achive this using analytical functions.
INSERT INTO tableb
WITH tablea_grouped1
AS (SELECT account_no,
balance,
as_on_date,
SUM (CASE WHEN balance >= 0 THEN 1 ELSE 0 END)
OVER (PARTITION BY account_no ORDER BY as_on_date)
grp
FROM tablea),
tablea_grouped2
AS (SELECT account_no,
balance,
as_on_date,
grp,
LAST_VALUE (
balance)
OVER (
PARTITION BY account_no, grp
ORDER BY as_on_date
ROWS BETWEEN UNBOUNDED PRECEDING
AND UNBOUNDED FOLLOWING)
closing_balance
FROM tablea_grouped1
WHERE balance < 0
AND grp != 0 --keep this, if starting negative balance is to be ignored
)
SELECT account_no,
closing_balance,
MAX (as_on_date),
MAX (as_on_date) - MIN (as_on_date) + 1,
MIN (as_on_date)
FROM tablea_grouped2
GROUP BY account_no, grp, closing_balance
ORDER BY account_no, MIN (as_on_date);
First, SUM is used as analytical function to assign group number to consecutive balances less than 0.
LAST_VALUE function is then used to find the last -ve balance in each group
Finally, the result is aggregated based on each group. MAX(date) gives the last date, MIN(date) gives the starting date, and the difference of the two gives number of days.
Demo at sqlfiddle.

Try this and use gone_negative to computing specified column value for insert into another table:
select temp.account_no,
temp.balance,
temp.prev_balance,
temp.on_date,
temp.prev_on_date,
case
WHEN (temp.balance < 0 and temp.prev_balance >= 0) THEN
1
else
0
end as gone_negative
from (select account_no,
balance,
on_date,
lag(balance, 1, 0) OVER(partition by account_no ORDER BY account_no) prev_balance,
lag(on_date, 1) OVER(partition by account_no ORDER BY account_no) prev_on_date
from tblA
order by account_no) temp;
Hope this helps pal.

Here's on way to do it.
Select all records from my_table where the balance is positive.
Do a self-join and get all the records that have a as_on_date is greater than the current row, but the amounts are in negative
Once we get these, we cut-off the rows WHERE the date difference between the current and the previous row for as_on_date is > 1. We then filter the results a outer sub query
The Final select just groups the rows and gets the min, max values for the filtered rows which are grouped.
Query:
SELECT
account_no,
min(case when row_number = 1 then balance end) as balance,
max(mt2_date) as As_on_date,
max(mt2_date) - mt1_date as Days_passed,
min(mt2_date) as Start_date
FROM
(
SELECT
*,
MIN(break_date) OVER( PARTITION BY mt1_date ) AS min_break_date,
ROW_NUMBER() OVER( PARTITION BY mt1_date ORDER BY mt2_date desc ) AS row_number
FROM
(
SELECT
mt1.account_no,
mt2.balance,
mt1.as_on_date as mt1_date,
mt2.as_on_date as mt2_date,
case when mt2.as_on_date - lag(mt2.as_on_date,1) over () > 1 then mt2.as_on_date end as break_date
FROM
my_table mt1
JOIN my_table mt2 ON ( mt2.balance < mt1.balance AND mt2.as_on_date > mt1.as_on_date )
WHERE
MT1.balance > 0
order by
mt1.as_on_date,
mt2.as_on_date ) sub_query
) T
WHERE
min_break_date is null
OR mt2_date < min_break_date
GROUP BY
mt1_date,
account_no
SQLFIDDLE
I have a added a few more rows in the FIDDLE, just to test it out

Related

SQL Divide previous row balance by current row balance and insert that value into current rows column "Growth"

I have a table where like this.
Year
ProcessDate
Month
Balance
RowNum
Calculation
2022
20220430
4
22855547
1
2022
20220330
3
22644455
2
2022
20220230
2
22588666
3
2022
20220130
1
33545444
4
2022
20221230
12
22466666
5
I need to take the previous row of each column and divide that amount by the current row.
Ex: Row 1 calculation should = Row 2 Balance / Row 1 Balance (22644455/22855547 = .99% )
Row 2 calculation should = Row 3 Balance / Row 2 Balance etc....
Table is just a Temporary table I created titled #MonthlyLoanBalance2.
Now I just need to take it a step further.
Let me know what and how you would go about doing this.
Thank you in advance!
Insert into #MonthlytLoanBalance2 (
Year
,ProcessDate
,Month
,Balance
,RowNum
)
select
--CloseYearMonth,
left(ProcessDate,4) as 'Year',
ProcessDate,
--x.LOANTypeKey,
SUBSTRING(CAST(x.ProcessDate as varchar(38)),5,2) as 'Month',
sum(x.currentBalance) as Balance
,ROW_NUMBER()over (order by ProcessDate desc) as RowNum
from
(
select
distinct LoanServiceKey,
LoanTypeKey,
AccountNumber,
CurrentBalance,
OpenDateKey,
CloseDateKey,
ProcessDate
from
cu.LAFactLoanSnapShot
where LoanStatus = 'Open'
and LoanTypeKey = 0
and ProcessDate in (select DateKey from dimDate
where IsLastDayOfMonth = 'Y'
and DateKey > convert(varchar, getdate()-4000, 112)
)
) x
group by ProcessDate
order by ProcessDate desc;``
I am assuming your data is already prepared as shown in the table. Now you can try Lead() function to resolve your issue. Remember format() function is used for taking only two precision.
SELECT *,
FORMAT((ISNULL(LEAD(Balance,1) OVER (ORDER BY RowNum), 1)/Balance),'N2') Calculation
FROM #MonthlytLoanBalance2

How to select only decreasing delta in values with increasing delta also present in data?

I have a warehouse which returns stock. Stock diminishes (orders) and increases (supplies). I have only stock values, nothing else. Say I have this data sorted by stock quantity ascending:
time
stock
00:11
7144 <--- current
00:10
7280
00:09
7416 <--- note increase, means new supply arrived
00:08
2259
00:07
2333
00:06
2538
00:05
2999
00:04
3074
00:03
3104 <--- start
I need to derive orders excluding supply, so max(qty)-min(qty) do not work, function has to mind the sudden increase and return only diminishing delta. So I expect in given values formula to be: orders = (3104-2259)+(7416-7144)
How would you approach this task?
Thanks.
You can join two "adjacent" rows and check that the stock of the earlier row is more than the stock of the later row to find your orders. Then sum over the difference of these stocks.
Assuming that the column time is unique, you can use following query:
SELECT SUM(t1.stock - t2.stock) AS sum_of_orders
FROM my_table t1
INNER JOIN my_table t2 ON t2.time > t1.time
AND NOT EXISTS(
SELECT 1
FROM my_table t3
WHERE t3.time > t1.time AND t3.time < t2.time)
WHERE t1.stock > t2.stock;
This is a type of gaps-and-islands problem. You can identify the "islands" by when the stock increases. Then within each island you want the max and min. You can summarize the islands using:
select min(stock), max(stock)
from (select t.*,
sum(case when prev_stock > stock then 0 else 1 end) over (order by time) as grp
from (select t.*,
lag(stock) over (order by time) as prev_stock
from t
) t
) t
group by grp;
Then one more summary gives the total you want:
select sum(max_stock - min_stock)
from (select select min(stock) as min_stock, max(stock) as max_stock
from (select t.*,
sum(case when prev_stock > stock then 0 else 1 end) over (order by time) as grp
from (select t.*,
lag(stock) over (order by time) as prev_stock
from t
) t
) t
group by grp
) t;
Here is a db<>fiddle.

Sum for a rolling total

I have the following query:
select b.month_date,total_signups,active_users from
(
SELECT date_trunc('month',confirmed_at) as month_date
, count(distinct id) as total_signups
FROM follower.users
WHERE confirmed_at::date >= dateadd(day,-90,getdate())::date
and (deleted_at is null or deleted_at > date_trunc('month',confirmed_at))
group by 1
) a ,
(
SELECT date_trunc('month', inv.created_at) AS month_date
,COUNT(DISTINCT em.user_id) AS active_users
FROM follower.invitees inv
INNER JOIN follower.events
ON inv.event_id = em.event_id
where inv.created_at::date >= dateadd(day,-90,getdate())::date
GROUP BY 1
) b
where a.month_date=b.month_date
This returns three columns month date, total signups and active users, what I need is a rolling total for all users in the fourth column (rolling total of signups). I've tried over and partition functions with no luck. Could someone help? Appreciate it very much.
Try adding this column definition to your first Select:
SUM(total_signups)
OVER (ORDER BY b.month_date ASC rows between unbounded preceding and current row)
AS running_total
Here's a mini-demo

Find the true start end dates for customers that have multiple accounts in SQL Server 2014

I have a checking account table that contains columns Cust_id (customer id), Open_Date (start date), and Closed_Date (end date). There is one row for each account. A customer can open multiple accounts at any given point. I would like to know how long the person has been a customer.
eg 1:
CREATE TABLE [Cust]
(
[Cust_id] [varchar](10) NULL,
[Open_Date] [date] NULL,
[Closed_Date] [date] NULL
)
insert into [Cust] values ('a123', '10/01/2019', '10/15/2019')
insert into [Cust] values ('a123', '10/12/2019', '11/01/2019')
Ideally I would like to insert this into a table with just one row, that says this person has been a customer from 10/01/2019 to 11/01/2019. (as he opened his second account before he closed his previous one.
Similarly eg 2:
insert into [Cust] values ('b245', '07/01/2019', '09/15/2019')
insert into [Cust] values ('b245', '10/12/2019', '12/01/2019')
I would like to see 2 rows in this case- one that shows he was a customer from 07/01 to 09/15 and then again from 10/12 to 12/01.
Can you point me to the best way to get this?
I would approach this as a gaps and islands problem. You want to group together groups of adjacents rows whose periods overlap.
Here is one way to solve it using lag() and a cumulative sum(). Everytime the open date is greater than the closed date of the previous record, a new group starts.
select
cust_id,
min(open_date) open_date,
max(closed_date) closed_date
from (
select
t.*,
sum(case when not open_date <= lag_closed_date then 1 else 0 end)
over(partition by cust_id order by open_date) grp
from (
select
t.*,
lag(closed_date) over (partition by cust_id order by open_date) lag_closed_date
from cust t
) t
) t
group by cust_id, grp
In this db fiddle with your sample data, the query produces:
cust_id | open_date | closed_date
:------ | :--------- | :----------
a123 | 2019-10-01 | 2019-11-01
b245 | 2019-07-01 | 2019-09-15
b245 | 2019-10-12 | 2019-12-01
I would solve this with recursion. While this is certainly very heavy, it should accommodate even the most complex account timings (assuming your data has such). However, if the sample data provided is as complex as you need to solve for, I highly recommend sticking with the solution provided above. It is much more concise and clear.
WITH x (cust_id, open_date, closed_date, lvl, grp) AS (
SELECT cust_id, open_date, closed_date, 1, 1
FROM (
SELECT cust_id
, open_date
, closed_date
, row_number()
OVER (PARTITION BY cust_id ORDER BY closed_date DESC, open_date) AS rn
FROM cust
) AS t
WHERE rn = 1
UNION ALL
SELECT cust_id, open_date, closed_date, lvl, grp
FROM (
SELECT c.cust_id
, c.open_date
, c.closed_date
, x.lvl + 1 AS lvl
, x.grp + CASE WHEN c.closed_date < x.open_date THEN 1 ELSE 0 END AS grp
, row_number() OVER (PARTITION BY c.cust_id ORDER BY c.closed_date DESC) AS rn
FROM cust c
JOIN x
ON x.cust_id = c.cust_id
AND c.open_date < x.open_date
) AS t
WHERE t.rn = 1
)
SELECT cust_id, min(open_date) AS first_open_date, max(closed_date) AS last_closed_date
FROM x
GROUP BY cust_id, grp
ORDER BY cust_id, grp
I would also add the caveat that I don't run on SQL Server, so there could be syntax differences that I didn't account for. Hopefully they are minor, if present.
you can try something like that:
select distinct
cust_id,
(select min(Open_Date)
from Cust as b
where b.cust_id = a.cust_id and
a.Open_Date <= b.Closed_Date and
a.Closed_Date >= b.Open_Date
),
(select max(Closed_Date)
from Cust as b
where b.cust_id = a.cust_id and
a.Open_Date <= b.Closed_Date and
a.Closed_Date >= b.Open_Date
)
from Cust as a
so, for every row - you're selecting minimal and maximal dates from all overlapping ranges, later distinct filters out duplicates

Sum Until Value Reached - Teradata

In Teradata, I need a query to first identify all members in the MEM TABLE that currently have a negative balance, let's call that CUR_BAL. Then, for all of those members only, sum all transactions from the TRAN TABLE in order by date until the sum of those transactions is equal to the CUR_BAL.
Editing to add a third ADJ table that contains MEM_NBR, ADJ_DT and ADJ_AMT that need to be included in the running total in order to capture all of the records.
I would like the outcome to include the MEM.MEM_NBR, MEM.CUR_BAL, TRAN.TRAN_DATE OR ADJ.ADJ_DT (date associated with the transaction that resulted in the running total to equal CUR_BAL), MEM.LST_UPD_DT. I don't need to know if the balance is negative as a result of a transaction or adjustment, just the date that it went negative.
Thank you!
select
mem_nbr,
cur_bal,
tran_date,
tran_type
from (
select
a.mem_nbr,
a.cur_bal,
b.tran_date,
b.tran_type,
a.lst_upd_dt,
sum(b.tran_amt) over (partition by b.mem_nbr order by b.tran_date rows between unbounded preceding and current row) as cumulative_bal
from mem a
inner join (
select
mem_nbr,
tran_date,
tran_amt,
'Tran' as tran_type
from tran
union all
select
mem_nbr,
adj_date,
adj_amt,
'Adj' as tran_type
from adj
) b
on a.mem_nbr = b.mem_nbr
where a.cur_bal < 0
qualify cumulative_bal < 0
) z
qualify rank() over (partition by mem_nbr order by tran_date) = 1
The subquery picks up all instances where the cumulative balance is negative, then the outer query picks up the earliest instance of it. If you want the latest, add desc after tran_date in the final qualify line.