SQL Exercises: Ratings of all customers with a comment string
SQL UNION : Exercise-8 with Solution
8. From the following table, write a SQL query to create a union of two queries that shows the customer id, cities, and ratings of all customers. Those with a rating of 300 or greater will have the words 'High Rating', while the others will have the words 'Low Rating'.
Sample table: Customer
Sample Solution:
SELECT customer_id, city, grade, 'High Rating'
FROM customer
WHERE grade >= 300
UNION
(SELECT customer_id, city, grade, 'Low Rating'
FROM customer
WHERE grade < 300)
Sample Output:
customer_id city grade ?column? 3002 New York 100 Low Rating 3003 Moscow 200 Low Rating 3004 Paris 300 High Rating 3008 London 300 High Rating 3005 California 200 Low Rating 3007 New York 200 Low Rating 3009 Berlin 100 Low Rating
Code Explanation:
The said query in SQL which selects the customer ID, city, and grade columns from the customer table and adds a new column called 'High Rating' or 'Low Rating' based on the value of the grade column.
The query checks if the grade is greater than or equal to 300, and if so, the value of the column 'Rating' is set to the value 'High Rating' as a result. When not done, it set the value of the 'Rating' column to 'Low Rating'.
Then the results of both queries are then combined using the UNION operator.
The results are then sorted by the customer ID.
Relational Algebra Expression:

Relational Algebra Tree:

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Previous SQL Exercise: Any salesman was matched to the city of any customer.
Next SQL Exercise: Customer with more than one current order.
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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;
Database: PostgreSQL
Ref: https://bit.ly/3AfYwZI
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