SQL Exercises: Find all customers with orders on October 5, 2012
SQL Query on multiple tables : Exercise-8 with Solution
From the following table, write a SQL query to find those customers who placed orders on October 5, 2012. Return customer_id, cust_name, city, grade, salesman_id, ord_no, purch_amt, ord_date, customer_id and salesman_id.
Sample table: Customer
Sample table: Orders
SELECT * FROM customer a,orders b WHERE a.customer_id=b.customer_id AND b.ord_date='2012-10-05';
Output of the Query:
customer_id cust_name city grade salesman_id ord_no purch_amt ord_date customer_id salesman_id 3002 Nick Rimando New York 100 5001 70002 65.26 2012-10-05 3002 5001 3005 Graham Zusi California 200 5002 70001 150.50 2012-10-05 3005 5002
The said query in SQL that performs a join between two tables: 'customer' and 'orders'. It returns all columns (*) from both tables where the customer_id column in table 'customer' matches the customer_id column in table 'orders' and the ord_date in table 'orders' is equal to '2012-10-05'.
Relational Algebra Expression:
Relational Algebra Tree:
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Previous SQL Exercise: Find salesman commission details of given customer.
Next SQL Exercise: SQL Exercises, Practice, Solution - JOINS.
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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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