SQL Exercise: Customer who is neither in the list nor have a grade
SQL JOINS: Exercise-16 with Solution
Write a SQL statement to generate a report with the customer name, city, order no. order date, purchase amount for only those customers on the list who must have a grade and placed one or more orders or which order(s) have been placed by the customer who neither is on the list nor has a grade.
Sample table: customer
Sample table: orders
SELECT a.cust_name,a.city, b.ord_no, b.ord_date,b.purch_amt AS "Order Amount" FROM customer a FULL OUTER JOIN orders b ON a.customer_id=b.customer_id WHERE a.grade IS NOT NULL;
Output of the Query:
cust_name city ord_no ord_date Order Amount Brad Guzan London 70009 2012-09-10 270.65 Nick Rimando New York 70002 2012-10-05 65.26 Geoff Cameron Berlin 70004 2012-08-17 110.50 Brad Davis New York 70005 2012-07-27 2400.60 Nick Rimando New York 70008 2012-09-10 5760.00 Fabian Johnson Paris 70010 2012-10-10 1983.43 Geoff Cameron Berlin 70003 2012-10-10 2480.40 Jozy Altidor Moscow 70011 2012-08-17 75.29 Nick Rimando New York 70013 2012-04-25 3045.60 Graham Zusi California 70001 2012-10-05 150.50 Graham Zusi California 70007 2012-09-10 948.50 Julian Green London 70012 2012-06-27 250.45
The said SQL query that is joining two tables: customer alias a, and orders alias b. The query is using a FULL OUTER JOIN on customer and orders tables on the 'customer_id' column.
The query is then selecting several columns from each table: 'cust_name' and 'city' from customer table, 'ord_no', 'ord_date', and 'purch_amt' aliased as 'Order Amount' from orders table.
The FULL OUTER JOIN ensures that all rows from both the customer table and the orders table are included in the result set, with NULL values in the columns of the table that doesn't have any matching rows for a customer.
The query also filters the results based on the condition "a.grade IS NOT NULL", which means it will only return the rows where the grade of the customer is not null. This means that rows where the customer doesn't have any grade will not be included in the result set.
N.B.: In certain instances not null is removed in table structure, so results may vary.
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Previous SQL Exercise: One or more customers ordered from the existing list
Next SQL Exercise: Salesman will appear for all customer and vice versa.
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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;
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