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LeetCode 1050~1077 mysql
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solution/1000-1099/1068.Product Sales Analysis I/README.md

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## 解法
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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### **SQL**
2016

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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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SELECT p.product_name,
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s.year,
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s.price
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FROM Sales s
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JOIN Product p
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ON s.product_id = p.product_id
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```
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solution/1000-1099/1068.Product Sales Analysis I/README_EN.md

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<!-- tabs:start -->
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### **Python3**
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### **SQL**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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SELECT p.product_name,
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s.year,
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s.price
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FROM Sales s
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JOIN Product p
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ON s.product_id = p.product_id
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```
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SELECT p.product_name,
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s.year,
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s.price
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FROM Sales s
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JOIN Product p
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ON s.product_id = p.product_id

solution/1000-1099/1069.Product Sales Analysis II/README.md

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### **Python3**
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### **SQL**
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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SELECT product_id,
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sum(quantity) AS total_quantity
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FROM Sales
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GROUP BY product_id
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```
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<!-- tabs:end -->

solution/1000-1099/1069.Product Sales Analysis II/README_EN.md

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## Solutions
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### **Python3**
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```python
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### **SQL**
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```
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### **Java**
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```java
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```
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### **...**
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```
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SELECT product_id,
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sum(quantity) AS total_quantity
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FROM Sales
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GROUP BY product_id
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```
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SELECT product_id,
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sum(quantity) AS total_quantity
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FROM Sales
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GROUP BY product_id

solution/1000-1099/1070.Product Sales Analysis III/README.md

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### **Python3**
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### **SQL**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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SELECT s.product_id,
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s.year AS first_year,
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s.quantity,
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s.price
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FROM Sales s
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WHERE (product_id, year) IN
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(SELECT product_id,
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min(year)
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FROM Sales
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GROUP BY product_id)
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```
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<!-- tabs:end -->

solution/1000-1099/1070.Product Sales Analysis III/README_EN.md

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### **Python3**
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### **SQL**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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SELECT s.product_id,
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s.year AS first_year,
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s.quantity,
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s.price
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FROM Sales s
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WHERE (product_id, year) IN
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(SELECT product_id,
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min(year)
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FROM Sales
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GROUP BY product_id)
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```
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SELECT s.product_id,
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s.year AS first_year,
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s.quantity,
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s.price
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FROM Sales s
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WHERE (product_id, year) IN
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(SELECT product_id,
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min(year)
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FROM Sales
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GROUP BY product_id)

solution/1000-1099/1075.Project Employees I/README.md

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### **Python3**
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### **SQL**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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SELECT p.project_id,
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round(avg(e.experience_years),
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2) AS average_years
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FROM Project p
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JOIN Employee e
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ON p.employee_id = e.employee_id
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GROUP BY p.project_id
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```
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<!-- tabs:end -->

solution/1000-1099/1075.Project Employees I/README_EN.md

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### **Python3**
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### **SQL**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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SELECT p.project_id,
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round(avg(e.experience_years),
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2) AS average_years
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FROM Project p
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JOIN Employee e
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ON p.employee_id = e.employee_id
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GROUP BY p.project_id
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```
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SELECT p.project_id,
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round(avg(e.experience_years),
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2) AS average_years
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FROM Project p
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JOIN Employee e
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ON p.employee_id = e.employee_id
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GROUP BY p.project_id

solution/1000-1099/1076.Project Employees II/README.md

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### **Python3**
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### **SQL**
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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SELECT project_id
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FROM Project
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GROUP BY project_id
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HAVING count(*) >= all
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(SELECT count(*)
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FROM Project
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GROUP BY project_id )
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```
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<!-- tabs:end -->

solution/1000-1099/1076.Project Employees II/README_EN.md

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### **Python3**
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### **SQL**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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SELECT project_id
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FROM Project
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GROUP BY project_id
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HAVING count(*) >= all
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(SELECT count(*)
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FROM Project
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GROUP BY project_id )
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```
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SELECT project_id
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FROM Project
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GROUP BY project_id
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HAVING count(*) >= all
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(SELECT count(*)
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FROM Project
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GROUP BY project_id )

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