refactor 1076

fishercoder1534 · fishercoder1534 · commit b0219a2b7556 · 2020-04-25T13:51:20.000-07:00
diff --git a/database/_1076.sql b/database/_1076.sql
@@ -1,61 +1,3 @@
---1076. Project Employees II
---
---Table: Project
---
---+-------------+---------+
---| Column Name | Type    |
---+-------------+---------+
---| project_id  | int     |
---| employee_id | int     |
---+-------------+---------+
---(project_id, employee_id) is the primary key of this table.
---employee_id is a foreign key to Employee table.
---Table: Employee
---
---+------------------+---------+
---| Column Name      | Type    |
---+------------------+---------+
---| employee_id      | int     |
---| name             | varchar |
---| experience_years | int     |
---+------------------+---------+
---employee_id is the primary key of this table.
---
---
---Write an SQL query that reports all the projects that have the most employees.
---
---The query result format is in the following example:
---
---Project table:
---+-------------+-------------+
---| project_id  | employee_id |
---+-------------+-------------+
---| 1           | 1           |
---| 1           | 2           |
---| 1           | 3           |
---| 2           | 1           |
---| 2           | 4           |
---+-------------+-------------+
---
---Employee table:
---+-------------+--------+------------------+
---| employee_id | name   | experience_years |
---+-------------+--------+------------------+
---| 1           | Khaled | 3                |
---| 2           | Ali    | 2                |
---| 3           | John   | 1                |
---| 4           | Doe    | 2                |
---+-------------+--------+------------------+
---
---Result table:
---+-------------+
---| project_id  |
---+-------------+
---| 1           |
---+-------------+
---The first project has 3 employees while the second one has 2.
-
---# Write your MySQL query statement below
 select project_id from Project
 group by project_id
 having count(employee_id) =
