SQL Exercise: One or more customers ordered from the existing list
SQL JOINS: Exercise-15 with Solution
For those customers from the existing list who put one or more orders, or which orders have been placed by the customer who is not on the list, create a report containing the customer name, city, order number, order date, and purchase amount
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 LEFT OUTER JOIN orders b ON a.customer_id=b.customer_id;
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 LEFT 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 LEFT OUTER JOIN ensures that all rows from the customer table are included in the result set, with NULL values in the columns of the orders table for those customers that do not have any matching rows in the orders table.
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Next SQL Exercise: Customer who is neither in the list nor have a grade.
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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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