feat: add sql solution to lc problems: No.0182,1050,1084,1587

- No.0182.Duplicate Emails
- No.1050.Actors and Directors Who Cooperated At Least Three Times
- No.1084.Sales Analysis III
- No.1587.Bank Account Summary II

Author: YangFong

solution/0100-0199/0182.Duplicate Emails/README.md

@@ -39,17 +39,18 @@
 ### **SQL**
 
 ```sql
-select Email from Person group by Email having count(Email) > 1
+SELECT Email
+FROM Person
+GROUP BY Email
+HAVING count(Email) > 1;
 ```
 
 ```sql
-# Write your MySQL query statement below
-SELECT DISTINCT
-	p1.email AS "Email"
-FROM
-	person AS p1
-	JOIN person AS p2 ON p1.id != p2.id
-	AND p1.email = p2.email
+SELECT DISTINCT p1.email
+FROM person AS p1,
+    person AS p2
+WHERE p1.id != p2.id
+    AND p1.email = p2.email;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0182.Duplicate Emails/README_EN.md

@@ -54,17 +54,18 @@ Person table:
 ### **SQL**
 
 ```sql
-select Email from Person group by Email having count(Email) > 1
+SELECT Email
+FROM Person
+GROUP BY Email
+HAVING count(Email) > 1;
 ```
 
 ```sql
-# Write your MySQL query statement below
-SELECT DISTINCT
-	p1.email AS "Email"
-FROM
-	person AS p1
-	JOIN person AS p2 ON p1.id != p2.id
-	AND p1.email = p2.email
+SELECT DISTINCT p1.email
+FROM person AS p1,
+    person AS p2
+WHERE p1.id != p2.id
+    AND p1.email = p2.email;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0182.Duplicate Emails/Solution.sql

@@ -1,2 +1,4 @@
-select Name as Employee from Employee Curr where
-    Salary > (select Salary from Employee where Id = Curr.ManagerId)
+SELECT Email
+FROM Person
+GROUP BY Email
+HAVING count(Email) > 1;

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/README.md

@@ -58,7 +58,6 @@ Result 表：
 ### **SQL**
 
 ```sql
-# Write your MySQL query statement below
 SELECT
     actor_id, director_id
 FROM

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/README_EN.md

@@ -60,7 +60,6 @@ Use `GROUP BY` & `HAVING`.
 ### **SQL**
 
 ```sql
-# Write your MySQL query statement below
 SELECT
     actor_id, director_id
 FROM

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/Solution.sql

@@ -1,4 +1,3 @@
-# Write your MySQL query statement below
 SELECT
     actor_id, director_id
 FROM

solution/1000-1099/1084.Sales Analysis III/README.md

@@ -90,7 +90,13 @@ id 3的产品在2019年春季之后销售。
 ### **SQL**
 
 ```sql
-
+SELECT p.product_id,
+    P.product_name
+FROM product AS p
+    JOIN sales AS s ON p.product_id = s.product_id
+GROUP BY p.product_id
+HAVING SUM(sale_date < '2019-01-01') = 0
+    AND SUM(sale_date > '2019-03-31') = 0;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1084.Sales Analysis III/README_EN.md

@@ -86,7 +86,13 @@ We return only product 1 as it is the product that was only sold in the spring o
 ### **SQL**
 
 ```sql
-
+SELECT p.product_id,
+    P.product_name
+FROM product AS p
+    JOIN sales AS s ON p.product_id = s.product_id
+GROUP BY p.product_id
+HAVING SUM(sale_date < '2019-01-01') = 0
+    AND SUM(sale_date > '2019-03-31') = 0;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1084.Sales Analysis III/Solution.sql

@@ -0,0 +1,7 @@
+SELECT p.product_id,
+    P.product_name
+FROM product AS p
+    JOIN sales AS s ON p.product_id = s.product_id
+GROUP BY p.product_id
+HAVING SUM(sale_date < '2019-01-01') = 0
+    AND SUM(sale_date > '2019-03-31') = 0;

solution/1500-1599/1587.Bank Account Summary II/README.md

@@ -85,26 +85,21 @@ Charlie 的余额为(6000 + 6000 - 4000) = 8000.
 
 <!-- tabs:start -->
 
-### **Python3**
+### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
-```python
-
-```
-
-### **Java**
-
-<!-- 这里可写当前语言的特殊实现逻辑 -->
-
-```java
-
-```
-
-### **...**
-
-```
-
+```sql
+SELECT
+    u.name,
+    SUM(t.amount) AS balance
+FROM
+    users AS u
+    JOIN transactions AS t ON u.account = t.account
+GROUP BY
+    name
+HAVING
+    SUM(t.amount) > 10000;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1587.Bank Account Summary II/README_EN.md

@@ -85,22 +85,19 @@ Charlie's balance is (6000 + 6000 - 4000) = 8000.
 
 <!-- tabs:start -->
 
-### **Python3**
-
-```python
-
-```
-
-### **Java**
-
-```java
-
-```
-
-### **...**
-
-```
-
+### **SQL**
+
+```sql
+SELECT
+    u.name,
+    SUM(t.amount) AS balance
+FROM
+    users AS u
+    JOIN transactions AS t ON u.account = t.account
+GROUP BY
+    name
+HAVING
+    SUM(t.amount) > 10000;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1587.Bank Account Summary II/Solution.sql

@@ -0,0 +1,10 @@
+SELECT
+    u.name,
+    SUM(t.amount) AS balance
+FROM
+    users AS u
+    JOIN transactions AS t ON u.account = t.account
+GROUP BY
+    name
+HAVING
+    SUM(t.amount) > 10000;