SQL Exercises: Customer with more than one current order
SQL UNION: Exercise-9 with Solution
9. From the following table, write a SQL query to find those salespersons and customers who have placed more than one order. Return ID, name.
Sample table: customer
Sample table: salesman
Sample table: orders
SELECT customer_id as “ID”, cust_name as “NAME” FROM customer a WHERE 1< (SELECT COUNT (*) FROM orders b WHERE a.customer_id = b.customer_id) UNION (SELECT salesman_idas “ID”, nameas “NAME” FROM salesman a WHERE 1 < (SELECT COUNT (*) FROM orders b WHERE a.salesman_id = b.salesman_id)) ORDER BY 2
ID NAME 3009 Geoff Cameron 3005 Graham Zusi 5001 James Hoog 5003 Lauson Hen 5002 Nail Knite 3002 Nick Rimando
The said query in SQL which selects customer_id and cust_name columns from the customer table and renames them as "ID" and "NAME" respectively.
The subquery checks whether at least one order placed by each customer or not. If the count of orders for a customer is greater than 1, then that customer is included in the results.
The UNION operator is used to combine the results of this query with another query that selects salesman_id and name columns from the salesman table, renames them as "ID" and "NAME" respectively, and checks if there is at least one order placed by each salesman. The final result is then ordered by the "NAME" column.
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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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