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leetcode-problemset/leetcode-cn/problem (English)/项目员工 I(English) [project-employees-i].html
2023-12-09 18:53:53 +08:00

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<p>Table: <code>Project</code></p>
<pre>
+-------------+---------+
| 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 <code>Employee</code> table.
Each row of this table indicates that the employee with employee_id is working on the project with project_id.
</pre>
<p>&nbsp;</p>
<p>Table: <code>Employee</code></p>
<pre>
+------------------+---------+
| Column Name | Type |
+------------------+---------+
| employee_id | int |
| name | varchar |
| experience_years | int |
+------------------+---------+
employee_id is the primary key of this table. It&#39;s guaranteed that experience_years is not NULL.
Each row of this table contains information about one employee.
</pre>
<p>&nbsp;</p>
<p>Write an SQL query that reports the <strong>average</strong> experience years of all the employees for each project, <strong>rounded to 2 digits</strong>.</p>
<p>Return the result table in <strong>any order</strong>.</p>
<p>The query result format is in the following example.</p>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong>
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 |
+-------------+--------+------------------+
<strong>Output:</strong>
+-------------+---------------+
| project_id | average_years |
+-------------+---------------+
| 1 | 2.00 |
| 2 | 2.50 |
+-------------+---------------+
<strong>Explanation:</strong> The average experience years for the first project is (3 + 2 + 1) / 3 = 2.00 and for the second project is (3 + 2) / 2 = 2.50
</pre>