PL SQL function that includes multiple tables - oracle

I'm new to PL SQL and have to write a function, which has customer_id as an input and has to output a product_name of the best selling product for that customer_id.
The schema looks like this:
I found a lot of simple examples where it includes two tables, but I can't seem to find one where you have to do multiple joins and use a function, while selecting only the best selling product.
I could paste a lot of very bad code here and how I tried to approach this, but this seems to be a bit over my head for current knowledge, since I've been learning PL SQL for less than 3 days now and got this task.

With some sample data (minimal column set):
SQL> select * from products order by product_id;
PRODUCT_ID PRODUCT_NAME
---------- ----------------
1 BMW
2 Audi
SQL> select * From order_items;
PRODUCT_ID CUSTOM QUANTITY UNIT_PRICE
---------- ------ ---------- ----------
1 Little 100 1
1 Little 200 2
2 Foot 300 3
If we check some totals:
SQL> select o.product_id,
2 o.customer_id,
3 sum(o.quantity * o.unit_price) total
4 from order_items o
5 group by o.product_id, o.customer_id;
PRODUCT_ID CUSTOM TOTAL
---------- ------ ----------
2 Little 400
1 Little 100
2 Foot 900
SQL>
It says that
for customer Little, product 2 was sold with total = 400 - that's our choice for Little
for customer Little, product 1 was sold with total = 100
for customer Foot, product 2 was sold with total = 900 - that's our choice for Foot
Query might then look like this:
temp CTE calculates totals per each customer
rank_them CTE ranks them in descending order per each customer; row_number so that you get only one product, even if there are ties
finally, select the one that ranks as the highest
SQL> with
2 temp as
3 (select o.product_id,
4 o.customer_id,
5 sum(o.quantity * o.unit_price) total
6 from order_items o
7 group by o.product_id, o.customer_id
8 ),
9 rank_them as
10 (select t.customer_id,
11 t.product_id,
12 row_number() over (partition by t.customer_id order by t.total desc) rn
13 from temp t
14 )
15 select * From rank_them;
CUSTOM PRODUCT_ID RN
------ ---------- ----------
Foot 2 1 --> for Foot, product 2 ranks as the highest
Little 2 1 --> for Little, product 1 ranks as the highest
Little 1 2
SQL>
Moved to a function:
SQL> create or replace function f_product (par_customer_id in order_items.customer_id%type)
2 return products.product_name%type
3 is
4 retval products.product_name%type;
5 begin
6 with
7 temp as
8 (select o.product_id,
9 o.customer_id,
10 sum(o.quantity * o.unit_price) total
11 from order_items o
12 group by o.product_id, o.customer_id
13 ),
14 rank_them as
15 (select t.customer_id,
16 t.product_id,
17 row_number() over (partition by t.customer_id order by t.total desc) rn
18 from temp t
19 )
20 select p.product_name
21 into retval
22 from rank_them r join products p on p.product_id = r.product_id
23 where r.customer_id = par_customer_id
24 and r.rn = 1;
25
26 return retval;
27 end;
28 /
Function created.
SQL>
Testing:
SQL> select f_product ('Little') result from dual;
RESULT
--------------------------------------------------------------------------------
Audi
SQL> select f_product ('Foot') result from dual;
RESULT
--------------------------------------------------------------------------------
Audi
SQL>
Now, you can improve it so that you'd care about no data found issue (when customer didn't buy anything), ties (but you'd then return a collection or a refcursor instead of a scalar value) etc.
[EDIT] I'm sorry, ORDERS table has to be included into the temp CTE; your data model is correct, you don't have to do anything about it - my query was wrong (small screen + late hours issue; not a real excuse, just saying).
So:
with
temp as
(select i.product_id,
o.customer_id,
sum(i.quantity * i.unit_price) total
from order_items i join orders o on o.order_id = i.order_id
group by i.product_id, o.customer_id
),
The rest of my code is - otherwise - unmodified.

Related

Query to find before and after values of a given value

If we have table Employees
EMP_ID ENAME SALARY DEPT_ID
1 abc 1000 10
2 bca 1050 10
3 dsa 2000 20
4 zxc 3000 30
5 bnm 5000 30
6 rty 5050 30
I want to get the rank of the salary with before 2 values and after 2 values including the given rank
Like if I give rank 4 it should give ranks 2,3,4,5,6 details.
output should be
5 bnm 5000 30
4 zxc 3000 30
3 dsa 2000 20
3 dsa 2000 20
2 bca 1050 10
1 abc 1000 10
I have a query
WITH dept_count AS (
SELECT
e.*,
dense_rank() over( ORDER BY salary DESC) AS rk
FROM employees e
)
SELECT
*
FROM dept_count dc
WHERE dc.rk BETWEEN (
SELECT
c.rk-2
FROM dept_count c
WHERE c.rk =4
)
AND (
SELECT
c.rk + 2
FROM dept_count c
WHERE c.rk = 4
)
but I need a query which can be simplified.
Could someone help me with this query?
You just need to use ROW_NUMBER() along with a substitution parameter :
WITH dept_count AS (
SELECT
e.*,
ROW_NUMBER() OVER( ORDER BY salary DESC) AS rk
FROM employees e
)
SELECT *
FROM dept_count
WHERE rk BETWEEN &prm - 2 AND &prm + 2

oracle: Count occurrence of a status in consecutive rows

I have a table which stores status of a customer reply in oracle. I have to count last consecutive declines from the customer.
For ex:
Id Status
-----------------------------
1. Declined
2. Accepted
3. Declined
4. Declined
This will have count = 2.
As last two were declined.
NOTE: Original Poster clarified in a "comment" that she needed a different requirement - addressing it in a separate Answer. Keeping this since it shows one possible solution for a more complicated problem that the OP's.
Assuming you want the count of the most recent consecutive "Declines" (even if followed by "Accepted" - and you want to allow for more than one customer - here is one possible solution.
Input table (I called it "t"):
SQL> select * from t;
CUSTOMER_ID DECISION_ID STATUS
----------- ----------- --------------------
10 1 Accepted
10 2 Declined
10 3 Declined
10 4 Accepted
10 5 Accepted
10 6 Declined
10 7 Declined
30 1 Declined
30 2 Accepted
30 3 Declined
30 4 Accepted
30 5 Declined
30 6 Declined
30 7 Declined
30 8 Accepted
30 9 Accepted
Query:
with t1 as
(
select customer_id,
decision_id - row_number() over
(partition by customer_id order by decision_id) as idx
from t
where status = 'Declined'
),
t2 as (select customer_id, max(idx) as max_idx from t1 group by customer_id)
select t1.customer_id, count(1) as ct
from t1 join t2 on t1.customer_id = t2.customer_id
where t1.idx = t2.max_idx
group by t1.customer_id
order by t1.customer_id
/
Query output:
CUSTOMER_ID CT
----------- ----------
10 2
30 3
#Gurmeet - Then the problem is much easier. Here is one way to solve it. If you need the result ordered by customer_id, add order by customer_id right at the end. nvl in the definition of d_A in CTE t2 is needed in case that customer never had a transaction with "Accepted" status.
Input: Same as in my other Answer.
Query: (MODIFIED to meet OP additional requirement):
with t0 as (select customer_id, status, row_number() over
(partition by customer_id order by decision_id) rn from t),
t1 as (select customer_id, max(rn) as d_all from t0 group by customer_id),
t2 as (select customer_id, nvl(max(rn), 0) as d_A from t0
where status = 'Accept' group by customer_id)
select customer_id, d_all - d_A as ct from t1 natural join t2
/
Result:
CUSTOMER_ID CT
----------- ----------
30 0
10 2
2 rows selected.

3rd highest salary in oracle

I had been looking for the query to find the 3rd highest salary from the database (using Oracle database). I found the below query -
SELECT *
FROM
( SELECT e.*, row_number() over (order by sal DESC) rn FROM emp e
)
WHERE rn = 3;
I do not have oracle installed in my system, so I'm not try it out. But I want to know if the below query will work or not. If not, then why ?
WITH Sal_sort AS
(SELECT DISTINCT sal FROM salary ORDER BY sal DESC
)
SELECT * FROM Salary S, Sal_sort SS WHERE S.Sal = SS.Sal AND SS.rownum = 3;
Input Data
emp_no emp_fname emp_lname salary
1 aa bb 30
2 ee yy 31
3 rr uu 32
4 tt ii 33
5 tt ii 33
6 tt ii 33
7 tt ii 33
8 tt ii 30
9 tt ii 31
Example:
select * from ee;
select emp_no,salary ,dense_rank() over (order by salary ) dr
from ee
Output
emp_no salary dr
1 30 1
8 30 1
9 31 2
2 31 2
3 32 3
4 33 4
5 33 4
6 33 4
7 33 4
So much easier in version 12 of the database and higher now.
SELECT *
FROM employees
ORDER BY salary DESC OFFSET 2 ROWS FETCH NEXT 1 ROWS ONLY
Tim talks about this feature here.
And if you take a look at the plan, you can see it's not magic, the optimizer is using analytic functions to derive the results.
JUST ONE LINE
select * from (
select salary,dense_rank() over (order by salary desc) rank from employees) where rank=3;
or
select * from (
select a.*,dense_rank() over (order by a.salary desc) rank from employees a) where rank=3;
Without Dense_rank()
SELECT salary FROM employees
ORDER BY salary DESC
OFFSET 2
FETCH 1 NEXT ONE ROWS ONLY;
With Dense_rank()
SELECT salary
FROM
(
SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) as rank from employees
)
WHERE rank = 3;

Oracle pivot with dynamic data

I am new to the oracle database and I am trying to use PIVOT to convert rows into columns. I have following tables..
table 1
solid solname
--------------
1 xxxxxx
2 yyyyyyy
table2
id name abbrv
----------------------------------
1 test db tdb
2 Prdocuiton db pdb
table3
id solId
-------------
1 1
1 2
1 3
1 4
1 5
1 7
1 8
1 9
1 22
1 23
1 24
1 25
2 26
2 27
1 28
1 29
1 32
1 33
1 34
1 35
1 36
1 37
3 38
1 39
1 40
1 43
1 44
table 3 is mapper table for table 1 and table 3.
I need to create a view with the columns in table2 and extra column for each solname's. So the view looks like
id name abbrv xxxxxxx yyyyyyy
--------------------------------------------------
So is there a way to do this using PIVOT in oracle database?
For Dynamic SQL Pivoting you need to do something similar :
create or replace view sol_view
as
select
t1.solname,
t2.name,
count(t3.abbrv),
from
table1 t1,
table2 t2,
table3 t3
where
t1.solid = t3.solid
and t2.id = t3.id
group by
t1.solname,
t3.name
select * from table( pivot('select * from sol_view') )
Caveat: I have never tried this but understood the logic from here:
http://technology.amis.nl/2006/05/24/dynamic-sql-pivoting-stealing-antons-thunder/
For Static SQL Pivoting, try something roughly along these lines. Never tried or tested though:
with pivot_data as (
select t1.solname, t2.name, t3.abbrv
from table1 t1, table2 t2, table3 t3
where t1.solid = t3.solid
and t2.id = t3.id
)
select *
from pivot_data
pivot (
count(abbrv)
for solname
in ('xxxxxx','yyyyyyy')
);
It was not really defined what you want to store in the xxxx and yyyy columns, maybe 1/blank, Y/N, ... ? However, your query might look close to something like this:
SELECT * FROM (
SELECT *
FROM table3 t3
JOIN table2 t2 USING (id)
JOIN table1 t1 USING (solid)
) PIVOT (
COUNT(*) FOR (solname) IN (
('xxx') AS "XXX",
('yyy') AS "YYY"
)
)
You can find more information and additional references on My Blog
TableName - **tblItem**
Id ItemName RecipeName Quantity
1 Sugar mutter paneer 200
2 Tea mutter paneer 100
3 Tomato mutter paneer 500
4 Onion mutter paneer 300
5 Ginger mutter paneer 300
6 Capsicum mutter paneer 300
7 Sugar mutter paneer 200
8 Tea mutter paneer 100
9 Onion mutter paneer 500
10 Sugar mutter paneer 200
V_VALUES varchar2(4000);
sql_query varchar2(4000);
SELECT
LISTAGG(''''||ITEMNAME||'''',',')WITHIN GROUP (ORDER BY ITEMNAME)as ITEMNAME
INTO V_LIST
FROM(SELECT DISTINCT ITEMNAME FROM tblItem);
sql_query : = 'SELECT * FROM (
SELECT ItemName,RecipeName,Sum(Quantity) as Quantity from tblItem group by ItemName,RecipeName)
PIVOT (
sum(Quantity) for ItemName in (' ||V_LIST|| ')
)';
OPEN p_cursor
FOR sql_query;

How to group by category, year so as to include zero sums for each year per category?

My imaginary results would look like:
Category | Year | sum |
--------- ------ --------
A 2008 200
A 2009 0
B 2008 100
B 2009 5
... ... ...
i.e. the sum of the transactions per year and per category.
There are cases where a category does not have any transaction for one year. in those cases the 2nd line of the results will not appear. How do I have to re-write the above query in order to include 2008, 2009 for every category?
select category, to_char(trans_date, 'YYYY') year, sum(trans_value)
from transaction
group by category, to_char(trans_date, 'YYYY')
order by 1, 2;
With a partitioned outer join, you don't need a categories table.
I used the same transactions table as "dcp" used:
SQL> create table transactions
2 ( category varchar(1)
3 , trans_date date
4 , trans_value number(25,8)
5 );
Table created.
SQL> insert into transactions values ('A',to_date('2008-01-01','yyyy-mm-dd'),100.0);
1 row created.
SQL> insert into transactions values ('A',to_date('2008-02-01','yyyy-mm-dd'),100.0);
1 row created.
SQL> insert into transactions values ('B',to_date('2008-01-01','yyyy-mm-dd'),50.0);
1 row created.
SQL> insert into transactions values ('B',to_date('2008-02-01','yyyy-mm-dd'),50.0);
1 row created.
SQL> insert into transactions values ('B',to_date('2009-08-01','yyyy-mm-dd'),5.0);
1 row created.
For the partitioned outer join you only need a set of years to partition outer join against. In the query below I used 2 years (2008 and 2009), but you can easily adjust that set.
SQL> with the_years as
2 ( select 2007 + level year
3 , trunc(to_date(2007 + level,'yyyy'),'yy') start_of_year
4 , trunc(to_date(2007 + level + 1,'yyyy'),'yy') - interval '1' second end_of_year
5 from dual
6 connect by level <= 2
7 )
8 select t.category "Category"
9 , y.year "Year"
10 , nvl(sum(t.trans_value),0) "sum"
11 from the_years y
12 left outer join transactions t
13 partition by (t.category)
14 on (t.trans_date between y.start_of_year and y.end_of_year)
15 group by t.category
16 , y.year
17 order by t.category
18 , y.year
19 /
Category Year sum
-------- ---------- ----------
A 2008 200
A 2009 0
B 2008 100
B 2009 5
4 rows selected.
Also note that I used start_of_year and end_of_year, so if you want to filter on trans_date and you have an index on that column, it could be used. Another option is to simply use trunc(t.trans_date) = y.year as on-condition.
Hope this helps.
Regards,
Rob.
You ideally need a table of categories and a table of years:
select c.category, y.year, nvl(sum(t.trans_value),0)
from categories c
cross join years y
left outer join transaction t
on to_char(t.trans_date, 'YYYY') = y.year
and t.category = c.category
group by c.category, y.year
order by 1, 2;
Hopefully you do have a table of categories, but you may well not have a table of years, in which case you can "fake" one like this:
with years as
( select 2007+rownum year
from dual
connect by rownum < 10) -- returns 2008, 2009, ..., 2017
select c.category, y.year, nvl(sum(t.trans_value),0)
from categories c
cross join years y
left outer join transaction t
on to_char(t.trans_date, 'YYYY') = y.year
and t.category = c.category
group by c.category, y.year
order by 1, 2;
Here's a complete, working example:
CREATE TABLE transactions (CATEGORY VARCHAR(1), trans_date DATE, trans_value NUMBER(25,8));
CREATE TABLE YEAR (YEAR NUMBER(4));
CREATE TABLE categories (CATEGORY VARCHAR(1));
INSERT INTO categories VALUES ('A');
INSERT INTO categories VALUES ('B');
INSERT INTO transactions VALUES ('A',to_date('2008-01-01','YYYY-MM-DD'),100.0);
INSERT INTO transactions VALUES ('A',to_date('2008-02-01','YYYY-MM-DD'),100.0);
INSERT INTO transactions VALUES ('B',to_date('2008-01-01','YYYY-MM-DD'),50.0);
INSERT INTO transactions VALUES ('B',to_date('2008-02-01','YYYY-MM-DD'),50.0);
INSERT INTO transactions VALUES ('B',to_date('2009-08-01','YYYY-MM-DD'),5.0);
INSERT INTO YEAR VALUES (2008);
INSERT INTO YEAR VALUES (2009);
SELECT b.category
, b.year
, SUM(nvl(a.trans_value,0))
FROM (SELECT to_char(a.trans_date,'YYYY') YEAR
, CATEGORY
, SUM(NVL(trans_value,0)) trans_value
FROM transactions a
GROUP BY to_char(a.trans_date,'YYYY')
, a.category ) a
, (SELECT
DISTINCT a.category
, b.year
FROM categories a
, YEAR b ) b
WHERE b.year = to_char(a.year(+))
AND b.category = a.category(+)
GROUP BY
b.category
, b.year
ORDER BY 1
,2;
Output:
CATEGORY YEAR SUM(NVL(A.TRANS_VALUE,0))
1 A 2008 200
2 A 2009 0
3 B 2008 100
4 B 2009 5

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