feat: add sql solutions to lc problems: No.2159,2228,2308

solution/2100-2199/2159.Order Two Columns Independently/README.md

@@ -66,7 +66,24 @@ Data 表：
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT
+            first_col,
+            row_number() OVER (ORDER BY first_col) AS rk
+        FROM Data
+    ),
+    T AS (
+        SELECT
+            second_col,
+            row_number() OVER (ORDER BY second_col DESC) AS rk
+        FROM Data
+    )
+SELECT first_col, second_col
+FROM
+    S
+    JOIN T USING (rk);
 ```
 
 <!-- tabs:end -->

solution/2100-2199/2159.Order Two Columns Independently/README_EN.md

@@ -59,7 +59,24 @@ Data table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT
+            first_col,
+            row_number() OVER (ORDER BY first_col) AS rk
+        FROM Data
+    ),
+    T AS (
+        SELECT
+            second_col,
+            row_number() OVER (ORDER BY second_col DESC) AS rk
+        FROM Data
+    )
+SELECT first_col, second_col
+FROM
+    S
+    JOIN T USING (rk);
 ```
 
 <!-- tabs:end -->

solution/2100-2199/2159.Order Two Columns Independently/Solution.sql

@@ -0,0 +1,18 @@
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT
+            first_col,
+            row_number() OVER (ORDER BY first_col) AS rk
+        FROM Data
+    ),
+    T AS (
+        SELECT
+            second_col,
+            row_number() OVER (ORDER BY second_col DESC) AS rk
+        FROM Data
+    )
+SELECT first_col, second_col
+FROM
+    S
+    JOIN T USING (rk);

solution/2200-2299/2228.Users With Two Purchases Within Seven Days/README.md

@@ -68,7 +68,24 @@ Purchases 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            user_id,
+            datediff(
+                purchase_date,
+                lag(purchase_date, 1) OVER (
+                    PARTITION BY user_id
+                    ORDER BY purchase_date
+                )
+            ) AS d
+        FROM Purchases
+    )
+SELECT DISTINCT user_id
+FROM t
+WHERE d <= 7
+ORDER BY user_id;
 ```
 
 <!-- tabs:end -->

solution/2200-2299/2228.Users With Two Purchases Within Seven Days/README_EN.md

@@ -62,7 +62,24 @@ User 7 had two purchases on the same day so we add their ID.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            user_id,
+            datediff(
+                purchase_date,
+                lag(purchase_date, 1) OVER (
+                    PARTITION BY user_id
+                    ORDER BY purchase_date
+                )
+            ) AS d
+        FROM Purchases
+    )
+SELECT DISTINCT user_id
+FROM t
+WHERE d <= 7
+ORDER BY user_id;
 ```
 
 <!-- tabs:end -->

solution/2200-2299/2228.Users With Two Purchases Within Seven Days/Solution.sql

@@ -0,0 +1,18 @@
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            user_id,
+            datediff(
+                purchase_date,
+                lag(purchase_date, 1) OVER (
+                    PARTITION BY user_id
+                    ORDER BY purchase_date
+                )
+            ) AS d
+        FROM Purchases
+    )
+SELECT DISTINCT user_id
+FROM t
+WHERE d <= 7
+ORDER BY user_id;

solution/2200-2299/2298.Tasks Count in the Weekend/README.md

@@ -70,14 +70,22 @@ Task 6 是在周日提交的。
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：WEEKDAY() 函数**
+
+`WEEKDAY()` 函数返回日期的工作日编号，从 0 开始，0 表示星期一，1 表示星期二，以此类推，6 表示星期日。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    sum(weekday(submit_date) IN (5, 6)) AS weekend_cnt,
+    sum(weekday(submit_date) NOT IN (5, 6)) AS working_cnt
+FROM Tasks;
 ```
 
 <!-- tabs:end -->

solution/2200-2299/2298.Tasks Count in the Weekend/README_EN.md

@@ -71,7 +71,11 @@ Task 6 was submitted on Sunday.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    sum(weekday(submit_date) IN (5, 6)) AS weekend_cnt,
+    sum(weekday(submit_date) NOT IN (5, 6)) AS working_cnt
+FROM Tasks;
 ```
 
 <!-- tabs:end -->

solution/2200-2299/2298.Tasks Count in the Weekend/Solution.sql

@@ -0,0 +1,5 @@
+# Write your MySQL query statement below
+SELECT
+    sum(weekday(submit_date) IN (5, 6)) AS weekend_cnt,
+    sum(weekday(submit_date) NOT IN (5, 6)) AS working_cnt
+FROM Tasks;

solution/2300-2399/2308.Arrange Table by Gender/README.md

@@ -80,7 +80,25 @@ Genders 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            *,
+            rank() OVER (
+                PARTITION BY gender
+                ORDER BY user_id
+            ) AS rk1,
+            CASE
+                WHEN gender = 'female' THEN 0
+                WHEN gender = 'other' THEN 1
+                ELSE 2
+            END AS rk2
+        FROM Genders
+    )
+SELECT user_id, gender
+FROM t
+ORDER BY rk1, rk2;
 ```
 
 <!-- tabs:end -->

solution/2300-2399/2308.Arrange Table by Gender/README_EN.md

@@ -75,7 +75,25 @@ Note that the IDs of each gender are sorted in ascending order.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            *,
+            rank() OVER (
+                PARTITION BY gender
+                ORDER BY user_id
+            ) AS rk1,
+            CASE
+                WHEN gender = 'female' THEN 0
+                WHEN gender = 'other' THEN 1
+                ELSE 2
+            END AS rk2
+        FROM Genders
+    )
+SELECT user_id, gender
+FROM t
+ORDER BY rk1, rk2;
 ```
 
 <!-- tabs:end -->

solution/2300-2399/2308.Arrange Table by Gender/Solution.sql

@@ -0,0 +1,19 @@
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            *,
+            rank() OVER (
+                PARTITION BY gender
+                ORDER BY user_id
+            ) AS rk1,
+            CASE
+                WHEN gender = 'female' THEN 0
+                WHEN gender = 'other' THEN 1
+                ELSE 2
+            END AS rk2
+        FROM Genders
+    )
+SELECT user_id, gender
+FROM t
+ORDER BY rk1, rk2;