#dbhouseparty - Real World Problem Solving with SQL

Copyright © 2020, Oracle and/or its affiliates
Real World Problem Solving
with SQL
Kim Berg Hansen
Oracle ACE Director
Chris Saxon
Developer Advocate
(sql-problem-solving)

2 Copyright © 2020, Oracle and/or its affiliates | Confidential: Internal/Restricted/Highly Restricted
[Date]
Chris Saxon
Developer Advocate
Chris is part of Steven Feuerstein's Oracle
Developer Advocate team and the answer
team on Ask TOM. It's his job to help you
get the best out of Oracle Database and
have fun while doing so!
About your presenters:
Kim Berg Hansen
Oracle ACE Director
Kim is an Oracle ACE Director and SQL
Evangelist. He's the author of the recent
book "Practical Oracle SQL" and SQL quiz-
master supreme on Oracle Dev Gym. When
not writing SQL he likes to drink craft beer
and read sci-fi.

It's PARTYTIME!
…who bought the fireworks?

Forecast zero stock?
How can I

Copyright © 2020, Oracle and/or its affiliates. All rights reserved. |
0
100
200
300
400
500
600
700
800
0
5
10
15
20
25
30
35
40
45
9:00 10:00 11:00 12:00 13:00 14:00 15:00 16:00 17:00
Actual sales Predicted sales
Remaining stock
Total sales

Copyright © 2020, Oracle and/or its affiliates. All rights reserved. |Copyright © 2020 Oracle and/or its affiliates. blogs.oracle.com/sql www.youtube.com/c/TheMagicOfSQL @ChrisRSaxon
for shops loop
for hours loop
… complex fn …
end loop
end loop;
5 stores
2 hours
=> not scalable!

Calculate Total Sales
, sum(nem.qtynem) over (
partition by nem.shopid
order by nem.hour
rows between unbounded preceding
and current row
) sumnem

Calculate Remaining Stock
, greatest(nem.startnem - nvl(
sum(nem.qtynem) over (
order by nem.hour
and 1 preceding
)
,0),0) stocknem

select …
, max ( hour ) + numtodsinterval(
max (stocknem) keep (
dense_rank last order by hour
) / max (qtynem) keep (
dense_rank last order by hour
), 'hour'
) zerohour
where stocknem > 0
Find Zero Stock Time

Full SQL
with shop as (
select s.shopid
, s.containers * 250 startnem
from fw_store s
), budget as (
select db.shopid
, db.budgetdate + numtodsinterval(hb.hour,'hour') budgethour
, db.budgetnem * hb.percent / 100 budgetnem
from fw_daybudget db
cross join fw_hourbudget hb
), nem as (
select budget.shopid
, shop.startnem
, budget.budgethour hour
, case
when budget.budgethour < sysdate
then 'S'
else 'B'
end salesbudget
, case
when budget.budgethour < sysdate
then nvl(sales.salesnem,0)
else budget.budgetnem
end qtynem
from shop
join budget
on budget.shopid = shop.shopid
left outer join fw_sales sales
on sales.shopid = budget.shopid
and sales.saleshour = budget.budgethour
)
…
…
select shopid
, max(hour)
+ numtodsinterval(
max(stocknem) keep (dense_rank last order by hour)
/ max(qtynem) keep (dense_rank last order by hour)
,'hour'
) zerohour
from (
select nem.shopid
, nem.hour
, nem.salesbudget
, nem.qtynem
, sum(nem.qtynem) over (
order by nem.hour
rows between unbounded preceding and current row
) sumnem
, greatest(nem.startnem - nvl(
sum(nem.qtynem) over (
order by nem.hour
rows between unbounded preceding and 1 preceding
)
,0),0) stocknem
from nem
)
where stocknem > 0
group by shopid
order by shopid;

Time to get the drinks!
Copyright © 2020 Oracle and/or its affiliates.

Stock picking routes?
How can I find

select … from (
select …
, nvl(sum(i.qty) over (
partition by i.item
order by i.purch, i.loc
and 1 preceding
),0) sum_prv_qty
from order join inventory … ) s
where s.sum_prv_qty < s.ord_qty
FIFO Picking

select … from (
select …
partition by i.item
order by i.loc
and 1 preceding
),0) sum_prv_qty
Shortest Route Picking

select … from (
select …
partition by i.item
order by i.qty desc, i.loc
and 1 preceding
),0) sum_prv_qty
Least Picks

select … from (
select …
partition by i.item
order by i.qty, i.loc
and 1 preceding
),0) sum_prv_qty
Clean out smallest quantities

Copyright © 2020, Oracle and/or its affiliates. All rights reserved. |Copyright © 2020 Oracle and/or its affiliates.
1 2 3 4

select … from (
select …
, dense_rank() over (
order by warehouse, aisle
) aisle_no
from ( <pick query> ) s where …
) s2 order by s2.warehouse, s2.aisle_no
, case when mod(s2.aisle_no,2) = 1
then s2.position
else -s2.position
end;
Up odds, down evens

SLA durations
P1 => 4 hours
P2 => 8 hours
P3 => 22 hours
P4 => 44 hours
P5 => 55 hours
24 x 7
Business
hours
07:00 – 18:00,
Monday - Friday
Find ticket
breach date

Time to work
off the calories!
Ryan McGuire / Gratisography

Find consecutive rows?
How can I

RUN_DATE TIME_IN_S DISTANCE_IN_KM
01 Apr 2020 310 1
02 Apr 2020 1,600 5
03 Apr 2020 3,580 11
06 Apr 2020 1,550 5
07 Apr 2020 300 1
10 Apr 2020 280 1
13 Apr 2020 1,530 5
14 Apr 2020 295 1
15 Apr 2020 292 1
#1
#3
#2
#4

current value = previous value + 1

RUN_DATE RN RUN_DATE - RN TIME_IN_S DISTANCE_IN_KM
01 Apr 2020 1 31 Mar 2020 310 1
02 Apr 2020 2 31 Mar 2020 1,600 5
03 Apr 2020 3 31 Mar 2020 3,580 11
06 Apr 2020 4 02 Apr 2020 1,550 5
07 Apr 2020 5 02 Apr 2020 300 1
10 Apr 2020 6 04 Apr 2020 280 1
13 Apr 2020 7 06 Apr 2020 1,530 5
14 Apr 2020 8 06 Apr 2020 295 1
15 Apr 2020 9 06 Apr 2020 292 1
-
-
-
-
-
-
-
-
-

with grps as (
select run_date ,
run_date -
row_number ()
over ( order by run_date ) grp
from running_log r
)
select min ( run_date ), count (*)
from grps
group by grp
Tabibitosan Method

match_recognize (
order by run_date
measures
first ( run_date ) as start_date,
count (*) as days
pattern ( init consecutive* )
define
consecutive as
run_date = prev ( run_date ) + 1
);
Pattern Matching
Pattern variables
Regex

Count all children of a node?
How can I

select empno
, lpad(' ', (level-1)*2) || ename
, (select count(*) from emp sub
start with sub.mgr = emp.empno
connect by sub.mgr = prior sub.empno
) subs
from emp
start with mgr is null
connect by mgr = prior empno
order siblings by empno;
Connect by Subquery

11
34 1
11 0 0 0
00
0
Level 1
Level 2
Level 3
Level 4

with hierarchy as ( … )
select …
from hierarchy match_recognize (
order by rn
measures …
one row per match
after match skip to next row
pattern ( strt higher* )
define
higher as higher.lvl > strt.lvl
);
Pattern Matching

465005
55
37008
4
11.61
0.035
0
2
4
6
8
10
12
14
0
50000
100000
150000
200000
250000
300000
350000
400000
450000
500000
Consistent Gets Sorts Duration
30x slower
8455x more gets
9252x more sorts
Connect By
Match_Recognize
=>

Slide deck on analytic function use cases
http://bit.ly/kibeha_analytic_pptx
Slide deck on MATCH_RECOGNIZE
http://bit.ly/kibeha_patmatch4_pptx
Article on warehouse picking
http://bit.ly/kibeha_analytics_article
Article on hierarchical child count
http://bit.ly/kibeha_match_recognize_article
Further Reading

asktom.oracle.com
#MakeDataGreatAgain
Ryan McGuire / Gratisography

#dbhouseparty - Real World Problem Solving with SQL

More Related Content

What's hot

Similar to #dbhouseparty - Real World Problem Solving with SQL

More from Tammy Bednar

Recently uploaded

#dbhouseparty - Real World Problem Solving with SQL