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feat: add sql solutions to lc problems (doocs#1197)
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‎.prettierignore

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/solution/bash_problem_readme_template_en.md
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/solution/0100-0199/0177.Nth Highest Salary/Solution.sql
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/solution/0600-0699/0627.Swap Salary/Solution.sql
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/solution/1000-1099/1076.Project Employees II/Solution.sql
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/solution/1000-1099/1082.Sales Analysis I/Solution.sql
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/solution/1100-1199/1173.Immediate Food Delivery I/Solution.sql
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/solution/1400-1499/1454.Active Users/Solution.sql
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/solution/1400-1499/1484.Group Sold Products By The Date/Solution.sql

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

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```sql
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# Write your MySQL query statement below
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SELECT project_id
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FROM Project p
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GROUP BY project_id
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HAVING COUNT(employee_id) >= all(
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SELECT COUNT(employee_id)
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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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GROUP BY 1
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HAVING
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count(1) >= all(
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SELECT count(1)
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FROM Project
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GROUP BY project_id
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);
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```
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```sql
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# Write your MySQL query statement below
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SELECT project_id
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FROM
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(
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WITH
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T AS (
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SELECT
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project_id,
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dense_rank() OVER (ORDER BY COUNT(employee_id) DESC) AS rk
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rank() OVER (ORDER BY count(employee_id) DESC) AS rk
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FROM Project
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GROUP BY project_id
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) AS t
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GROUP BY 1
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)
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SELECT project_id
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FROM T
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WHERE rk = 1;
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```
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‎solution/1000-1099/1076.Project Employees II/README_EN.md

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```sql
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# Write your MySQL query statement below
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SELECT project_id
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FROM Project p
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GROUP BY project_id
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HAVING COUNT(employee_id) >= all(
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SELECT COUNT(employee_id)
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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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GROUP BY 1
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HAVING
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count(1) >= all(
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SELECT count(1)
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FROM Project
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GROUP BY project_id
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);
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```
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```sql
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# Write your MySQL query statement below
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SELECT project_id
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FROM
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(
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WITH
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T AS (
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SELECT
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project_id,
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dense_rank() OVER (ORDER BY COUNT(employee_id) DESC) AS rk
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rank() OVER (ORDER BY count(employee_id) DESC) AS rk
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FROM Project
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GROUP BY project_id
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) AS t
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GROUP BY 1
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)
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SELECT project_id
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FROM T
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WHERE rk = 1;
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```
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‎solution/1000-1099/1076.Project Employees II/Solution.sql

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# Write your MySQL query statement below
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SELECT project_id
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FROM Project p
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GROUP BY project_id
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HAVING COUNT(employee_id) >= all(
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SELECT COUNT(employee_id)
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FROM Project
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GROUP BY project_id )
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WITH
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T AS (
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SELECT
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project_id,
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rank() OVER (ORDER BY count(employee_id) DESC) AS rk
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FROM Project
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GROUP BY 1
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)
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SELECT project_id
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FROM T
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WHERE rk = 1;

‎solution/1000-1099/1077.Project Employees III/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：内连接 + 窗口函数**
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我们先将 `Project` 表和 `Employee` 表进行内连接，然后使用窗口函数 `rank()` 对 `Project` 表进行分组，按照 `experience_years` 降序排列，最后取出每个项目中经验最丰富的雇员。
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<!-- tabs:start -->
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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project_id,
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employee_id
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FROM
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(
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WITH
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T AS (
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SELECT
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p.project_id,
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p.employee_id,
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project_id,
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employee_id,
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rank() OVER (
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PARTITION BY p.project_id
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ORDER BY e.experience_years DESC
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PARTITION BY project_id
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ORDER BY experience_years DESC
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) AS rk
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FROM
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Project AS p
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LEFT JOIN Employee AS e ON p.employee_id = e.employee_id
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) AS t
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Project
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JOIN Employee USING (employee_id)
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)
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SELECT project_id, employee_id
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FROM T
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WHERE rk = 1;
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```
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‎solution/1000-1099/1077.Project Employees III/README_EN.md

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```sql
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# Write your MySQL query statement below
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SELECT
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project_id,
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employee_id
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FROM
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(
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WITH
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T AS (
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SELECT
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p.project_id,
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p.employee_id,
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project_id,
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employee_id,
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rank() OVER (
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PARTITION BY p.project_id
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ORDER BY e.experience_years DESC
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PARTITION BY project_id
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ORDER BY experience_years DESC
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) AS rk
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FROM
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Project AS p
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LEFT JOIN Employee AS e ON p.employee_id = e.employee_id
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) AS t
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Project
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JOIN Employee USING (employee_id)
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)
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SELECT project_id, employee_id
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FROM T
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WHERE rk = 1;
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```
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‎solution/1000-1099/1077.Project Employees III/Solution.sql

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# Write your MySQL query statement below
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SELECT
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project_id,
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employee_id
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FROM
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(
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WITH
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T AS (
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SELECT
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p.project_id,
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p.employee_id,
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project_id,
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employee_id,
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rank() OVER (
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PARTITION BY p.project_id
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ORDER BY e.experience_years DESC
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PARTITION BY project_id
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ORDER BY experience_years DESC
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) AS rk
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FROM
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Project AS p
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LEFT JOIN Employee AS e ON p.employee_id = e.employee_id
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) AS t
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Project
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JOIN Employee USING (employee_id)
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)
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SELECT project_id, employee_id
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FROM T
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WHERE rk = 1;

‎solution/1000-1099/1082.Sales Analysis I/README.md

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```sql
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# Write your MySQL query statement below
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SELECT seller_id
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SELECT seller_id
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FROM Sales
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GROUP BY seller_id
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HAVING SUM(price) >= ALL (
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SELECT SUM(price)
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FROM Sales
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GROUP BY seller_id )
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GROUP BY seller_id
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HAVING
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SUM(price) >= ALL(
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SELECT SUM(price)
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FROM Sales
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GROUP BY seller_id
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);
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```
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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seller_id,
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sum(price) AS tot,
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rank() OVER (ORDER BY sum(price) DESC) AS rk
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FROM Sales
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GROUP BY seller_id
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)
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SELECT seller_id
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FROM T
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WHERE rk = 1;
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```
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<!-- tabs:end -->

‎solution/1000-1099/1082.Sales Analysis I/README_EN.md

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```sql
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# Write your MySQL query statement below
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SELECT seller_id
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SELECT seller_id
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FROM Sales
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GROUP BY seller_id
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HAVING SUM(price) >= ALL (
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SELECT SUM(price)
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FROM Sales
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GROUP BY seller_id )
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GROUP BY seller_id
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HAVING
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SUM(price) >= ALL(
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SELECT SUM(price)
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FROM Sales
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GROUP BY seller_id
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);
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```
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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seller_id,
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sum(price) AS tot,
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rank() OVER (ORDER BY sum(price) DESC) AS rk
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FROM Sales
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GROUP BY seller_id
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)
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SELECT seller_id
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FROM T
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WHERE rk = 1;
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```
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<!-- tabs:end -->

‎solution/1000-1099/1082.Sales Analysis I/Solution.sql

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# Write your MySQL query statement below
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SELECT seller_id
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FROM Sales
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GROUP BY seller_id
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HAVING SUM(price) >= ALL (
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SELECT SUM(price)
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FROM Sales
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GROUP BY seller_id )
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WITH
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T AS (
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SELECT
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seller_id,
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sum(price) AS tot,
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rank() OVER (ORDER BY sum(price) DESC) AS rk
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FROM Sales
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GROUP BY seller_id
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)
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SELECT seller_id
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FROM T
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WHERE rk = 1;

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