SQL Exercise: Customer have placed no order or one or more orders
SQL JOINS: Exercise-10 with Solution
Write a SQL statement to make a report with customer name, city, order number, order date, and order amount in ascending order according to the order date to determine whether any of the existing customers have placed an order or not.
Sample table: orders
Sample table: customer
Sample Solution:
SELECT a.cust_name,a.city, b.ord_no,
b.ord_date,b.purch_amt AS "Order Amount"
FROM customer a
LEFT OUTER JOIN orders b
ON a.customer_id=b.customer_id
order by b.ord_date;
Output of the Query:
cust_name city ord_no ord_date Order Amount Nick Rimando New York 70013 2012-04-25 3045.60 Julian Green London 70012 2012-06-27 250.45 Brad Davis New York 70005 2012-07-27 2400.60 Jozy Altidor Moscow 70011 2012-08-17 75.29 Geoff Cameron Berlin 70004 2012-08-17 110.50 Brad Guzan London 70009 2012-09-10 270.65 Nick Rimando New York 70008 2012-09-10 5760.00 Graham Zusi California 70007 2012-09-10 948.50 Graham Zusi California 70001 2012-10-05 150.50 Nick Rimando New York 70002 2012-10-05 65.26 Fabian Johnson Paris 70010 2012-10-10 1983.43 Geoff Cameron Berlin 70003 2012-10-10 2480.40
Explanation:
The said SQL query is selecting the customer name, city from the customer table aliased as a and the order number, order date and purchase amount from the orders table aliased as b. It is joining these tables on the 'customer_id' column, and the results are ordered by the 'ord_date' column.
Additionally, it is using a LEFT OUTER JOIN 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.
Visual Explanation:
Practice Online
Query Visualization:
Duration:
Rows:
Cost:
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Previous SQL Exercise: Customers who holds a grade less than 300.
Next SQL Exercise: Find customers who have placed no order or one or more.
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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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