SQL Exercise: Customers who holds a grade less than 300
SQL JOINS: Exercise-9 with Solution
From the following tables write a SQL query to find those customers with a grade less than 300. Return cust_name, customer city, grade, Salesman, salesmancity. The result should be ordered by ascending customer_id.
Sample table: customer
Sample table: salesman
SELECT a.cust_name,a.city,a.grade, b.name AS "Salesman", b.city FROM customer a LEFT OUTER JOIN salesman b ON a.salesman_id=b.salesman_id WHERE a.grade<300 ORDER BY a.customer_id;
Output of the Query:
cust_name city grade Salesman city Nick Rimando New York 100 James Hoog New York Jozy Altidor Moscow 200 Paul Adam Rome Graham Zusi California 200 Nail Knite Paris Brad Davis New York 200 James Hoog New York Geoff Cameron Berlin 100 Lauson Hen San Jose
The said SQL query is selecting the customer name, city, and grade from the customer table aliased as a and the salesman name and city from the salesman table aliased as b.
It is joining these tables on the 'salesman_id' column and only selecting the customers whose grade is less than 300.
The results are ordered by the 'customer_id' 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.
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Previous SQL Exercise: Customer who works either through a salesman or by own.
Next SQL Exercise: Customer have placed no order or one or more orders.
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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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