solved

akscoder082003 · akscoder082003 · commit ac715d15fd20 · 2024-01-06T13:52:54.000+05:30
diff --git a/MySQL/577. Employee Bonus.sql b/MySQL/577. Employee Bonus.sql
@@ -0,0 +1,33 @@
+-- Table: Employee
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | empId       | int     |
+-- | name        | varchar |
+-- | supervisor  | int     |
+-- | salary      | int     |
+-- +-------------+---------+
+-- empId is the column with unique values for this table.
+-- Each row of this table indicates the name and the ID of an employee in addition to their salary and the id of their manager.
+
+-- Table: Bonus
+
+-- +-------------+------+
+-- | Column Name | Type |
+-- +-------------+------+
+-- | empId       | int  |
+-- | bonus       | int  |
+-- +-------------+------+
+-- empId is the column of unique values for this table.
+-- empId is a foreign key (reference column) to empId from the Employee table.
+-- Each row of this table contains the id of an employee and their respective bonus.
+
+-- Write a solution to report the name and bonus amount of each employee with a bonus less than 1000.
+
+select Employee.name, 
+Bonus.bonus 
+from Employee left join Bonus 
+on Employee.empId = Bonus.empId 
+where Bonus.bonus < 1000 
+or Bonus.bonus is null;
