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feat: add sql solutions to lc problems (doocs#1195)
* No.0619.Biggest Single Number * No.0620.Not Boring Movies * No.1045.Customers Who Bought All Products * No.1068.Product Sales Analysis I * No.1069.Product Sales Analysis II * No.1070.Product Sales Analysis III * No.1075.Project Employees I
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‎solution/0600-0699/0619.Biggest Single Number/README.md

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### **SQL**
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```sql
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SELECT MAX(a.num) AS num
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# Write your MySQL query statement below
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SELECT max(num) AS num
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FROM
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(
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SELECT num
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FROM MyNumbers
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GROUP BY num
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HAVING count(*) = 1
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) AS a;
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HAVING count(1) = 1
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) AS t;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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CASE
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WHEN count(1) = 1 THEN num
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ELSE NULL
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END AS num
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FROM MyNumbers
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GROUP BY num
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ORDER BY 1 DESC
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LIMIT 1;
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```
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<!-- tabs:end -->

‎solution/0600-0699/0619.Biggest Single Number/README_EN.md

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### **SQL**
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```sql
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SELECT MAX(a.num) AS num
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# Write your MySQL query statement below
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SELECT max(num) AS num
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FROM
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(
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SELECT num
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FROM MyNumbers
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GROUP BY num
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HAVING count(*) = 1
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) AS a;
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HAVING count(1) = 1
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) AS t;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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CASE
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WHEN count(1) = 1 THEN num
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ELSE NULL
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END AS num
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FROM MyNumbers
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GROUP BY num
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ORDER BY 1 DESC
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LIMIT 1;
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```
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<!-- tabs:end -->

‎solution/0600-0699/0619.Biggest Single Number/Solution.sql

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SELECT MAX(a.num) AS num
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FROM
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(
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SELECT num
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FROM MyNumbers
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GROUP BY num
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HAVING count(*) = 1
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) AS a;
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# Write your MySQL query statement below
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SELECT
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CASE
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WHEN count(1) = 1 THEN num
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ELSE NULL
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END AS num
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FROM MyNumbers
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GROUP BY num
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ORDER BY 1 DESC
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LIMIT 1;

‎solution/0600-0699/0620.Not Boring Movies/README.md

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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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FROM cinema
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WHERE description NOT LIKE '%boring%' AND mod(id, 2) = 1
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FROM Cinema
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WHERE description != 'boring' AND id % 2 = 1
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ORDER BY rating DESC;
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```
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‎solution/0600-0699/0620.Not Boring Movies/README_EN.md

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@@ -61,9 +61,10 @@ We have three movies with odd-numbered IDs: 1, 3, and 5. The movie with ID = 3 i
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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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FROM cinema
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WHERE description NOT LIKE '%boring%' AND mod(id, 2) = 1
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FROM Cinema
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WHERE description != 'boring' AND id % 2 = 1
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ORDER BY rating DESC;
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```
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‎solution/0600-0699/0620.Not Boring Movies/Solution.sql

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# Write your MySQL query statement below
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SELECT *
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FROM cinema
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WHERE description NOT LIKE '%boring%' AND mod(id, 2) = 1
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FROM Cinema
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WHERE description != 'boring' AND id % 2 = 1
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ORDER BY rating DESC;

‎solution/0600-0699/0626.Exchange Seats/README.md

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seat
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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CASE
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WHEN id & 1 = 0 THEN id - 1
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WHEN row_number() OVER (ORDER BY id) != count(id) OVER () THEN id + 1
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ELSE id
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END AS id,
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student
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FROM Seat
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/0600-0699/0626.Exchange Seats/README_EN.md

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seat
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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CASE
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WHEN id & 1 = 0 THEN id - 1
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WHEN row_number() OVER (ORDER BY id) != count(id) OVER () THEN id + 1
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ELSE id
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END AS id,
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student
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FROM Seat
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/1000-1099/1045.Customers Who Bought All Products/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：GROUP BY + HAVING**
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我们将 `Customer` 表按照 `customer_id` 进行分组，然后使用 `HAVING` 子句筛选出购买了所有产品的客户。
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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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customer_id
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SELECT customer_id
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FROM Customer
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GROUP BY customer_id
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HAVING
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COUNT(DISTINCT product_key) = (
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SELECT
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COUNT(1)
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FROM Product
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);
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GROUP BY 1
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HAVING count(DISTINCT product_key) = (SELECT count(1) FROM Product);
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```
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<!-- tabs:end -->

‎solution/1000-1099/1045.Customers Who Bought All Products/README_EN.md

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@@ -79,16 +79,10 @@ The customers who bought all the products (5 and 6) are customers with IDs 1 and
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```sql
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# Write your MySQL query statement below
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SELECT
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customer_id
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SELECT customer_id
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FROM Customer
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GROUP BY customer_id
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HAVING
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COUNT(DISTINCT product_key) = (
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SELECT
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COUNT(1)
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FROM Product
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);
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GROUP BY 1
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HAVING count(DISTINCT product_key) = (SELECT count(1) FROM Product);
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```
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<!-- tabs:end -->

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