leetcode 180~183

Author: LjyYano

solution/0100-0199/0180.Consecutive Numbers/README.md

@@ -39,7 +39,9 @@
 ### **SQL**
 
 ```
-
+select distinct(Num) as ConsecutiveNums from Logs Curr where 
+    Num = (select Num from Logs where id = Curr.id - 1) and
+    Num = (select Num from Logs where id = Curr.id - 2)
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

solution/0100-0199/0180.Consecutive Numbers/README_EN.md

@@ -64,7 +64,9 @@
 ### **SQL**
 
 ```
-
+select distinct(Num) as ConsecutiveNums from Logs Curr where
+    Num = (select Num from Logs where id = Curr.id - 1) and
+    Num = (select Num from Logs where id = Curr.id - 2)
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

solution/0100-0199/0180.Consecutive Numbers/Solution.sql

@@ -0,0 +1,3 @@
+select distinct(Num) as ConsecutiveNums from Logs Curr where
+    Num = (select Num from Logs where id = Curr.id - 1) and
+    Num = (select Num from Logs where id = Curr.id - 2)

solution/0100-0199/0181.Employees Earning More Than Their Managers/README.md

@@ -36,7 +36,8 @@
 ### **SQL**
 
 ```
-
+select Name as Employee from Employee Curr where
+    Salary > (select Salary from Employee where Id = Curr.ManagerId)
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

solution/0100-0199/0181.Employees Earning More Than Their Managers/README_EN.md

@@ -58,7 +58,8 @@
 ### **SQL**
 
 ```
-
+select Name as Employee from Employee Curr where
+    Salary > (select Salary from Employee where Id = Curr.ManagerId)
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.sql

@@ -0,0 +1,2 @@
+select Name as Employee from Employee Curr where
+    Salary > (select Salary from Employee where Id = Curr.ManagerId)

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

@@ -39,7 +39,7 @@
 ### **SQL**
 
 ```
-
+select Email from Person group by Email having count(Email) > 1
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

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

@@ -60,7 +60,7 @@
 ### **SQL**
 
 ```
-
+select Email from Person group by Email having count(Email) > 1
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

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

@@ -0,0 +1,2 @@
+select Name as Employee from Employee Curr where
+    Salary > (select Salary from Employee where Id = Curr.ManagerId)

solution/0100-0199/0183.Customers Who Never Order/README.md

@@ -49,7 +49,8 @@
 ### **SQL**
 
 ```
-
+select Name as Customers from Customers 
+    where id not in (select CustomerId from Orders)
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->