SQL Exercise: Find customers who have placed no order or one or more
SQL JOINS: Exercise-11 with Solution
SQL statement to generate a report with customer name, city, order number, order date, order amount, salesperson name, and commission to determine if any of the existing customers have not placed orders or if they have placed orders through their salesman or by themselves.
Sample table: customer
Sample table: orders
Sample table: salesman
Sample Solution:
SELECT a.cust_name,a.city, b.ord_no,
b.ord_date,b.purch_amt AS "Order Amount",
c.name,c.commission
FROM customer a
LEFT OUTER JOIN orders b
ON a.customer_id=b.customer_id
LEFT OUTER JOIN salesman c
ON c.salesman_id=b.salesman_id;
Output of the Query:
cust_name city ord_no ord_date Order Amount name commission Brad Guzan London 70009 2012-09-10 270.65 Pit Alex 0.11 Nick Rimando New York 70002 2012-10-05 65.26 James Hoog 0.15 Geoff Cameron Berlin 70004 2012-08-17 110.50 Lauson Hen 0.12 Brad Davis New York 70005 2012-07-27 2400.60 James Hoog 0.15 Nick Rimando New York 70008 2012-09-10 5760.00 James Hoog 0.15 Fabian Johnson Paris 70010 2012-10-10 1983.43 Mc Lyon 0.14 Geoff Cameron Berlin 70003 2012-10-10 2480.40 Lauson Hen 0.12 Jozy Altidor Moscow 70011 2012-08-17 75.29 Paul Adam 0.13 Nick Rimando New York 70013 2012-04-25 3045.60 James Hoog 0.15 Graham Zusi California 70001 2012-10-05 150.50 Nail Knite 0.13 Graham Zusi California 70007 2012-09-10 948.50 Nail Knite 0.13 Julian Green London 70012 2012-06-27 250.45 Nail Knite 0.13
Explanation:
The said SQL query is selecting the customer name, city, order number, order date, purchase amount, salesman name and commission from 3 tables customer aliased as a, orders aliased as b, and salesman aliased as c. It is joining these tables on the 'customer_id' and 'salesman_id' column respectively.
Additionally, it is using two LEFT OUTER JOINs, which will retrieve all records from the left table and the matching records from the right table. If no match is found on the right table, it will return NULL for the right table's fields.
This query will retrieve all customer details, order details and salesman details along with their commission even if some customer or salesman doesn't have any orders.
Visual Explanation:
Practice Online
Query Visualization:
Duration:
Rows:
Cost:
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Previous SQL Exercise: Customer have placed no order or one or more orders.
Next SQL Exercise: Salesmen works either for one or more customer or not.
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SQL: Tips of the Day
Grouped LIMIT in PostgreSQL: Show the first N rows for each group?
db=# SELECT * FROM xxx; id | section_id | name ----+------------+------ 1 | 1 | A 2 | 1 | B 3 | 1 | C 4 | 1 | D 5 | 2 | E 6 | 2 | F 7 | 3 | G 8 | 2 | H (8 rows)
I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:
id | section_id | name ----+------------+------ 1 | 1 | A 2 | 1 | B 5 | 2 | E 6 | 2 | F 7 | 3 | G (5 rows)
PostgreSQL v9.3 you can do a lateral join
select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
select * from t t_inner
where t_inner.section_id = t_outer.section_id
order by t_inner.name
limit 2
) t_top on true
order by t_outer.section_id;
Database: PostgreSQL
Ref: https://bit.ly/3AfYwZI
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