feat: add sql solutions to lc problems: No.2993~2995 (doocs#2182)

* No.2993.Friday Purchases I
* No.2994.Friday Purchases II
* No.2995.Viewers Turned Streamers

Author: yanglbme

.prettierrc

@@ -49,7 +49,8 @@
                 "solution/2600-2699/2686.Immediate Food Delivery III/Solution.sql",
                 "solution/2400-2499/2494.Merge Overlapping Events in the Same Hall/Solution.sql",
                 "solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/Solution.sql",
-                "solution/2700-2799/2793.Status of Flight Tickets/Solution.sql"
+                "solution/2700-2799/2793.Status of Flight Tickets/Solution.sql",
+                "solution/2900-2999/2994.Friday Purchases II/Solution.sql"
             ],
             "options": {
                 "parser": "bigquery"

package-lock.json

@@ -16,7 +16,7 @@
     "lint-staged": "^13.2.2",
     "prettier": "^2.8.8",
     "prettier-plugin-rust": "^0.1.9",
-    "prettier-plugin-sql-cst": "^0.8.2"
+    "prettier-plugin-sql-cst": "^0.10.0"
   },
   "lint-staged": {
     "*.{js,ts,php,sql,rs,md}": "prettier --write",

package.json

@@ -63,14 +63,32 @@ Output table is ordered by week_of_month in ascending order.</pre>
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：日期函数**
+
+我们用到的日期函数有：
+
+-   `DATE_FORMAT(date, format)`：将日期格式化为字符串
+-   `DAYOFWEEK(date)`：返回日期对应的星期几，1 代表星期日，2 代表星期一，以此类推
+-   `DAYOFMONTH(date)`：返回日期对应的月份中的第几天
+
+我们先用 `DATE_FORMAT` 函数将日期格式化为 `YYYYMM` 的形式，然后筛选出 2023 年 11 月且是星期五的记录，然后将记录按照 `purchase_date` 分组，计算出每个星期五的总消费金额。
+
 <!-- 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
+    CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
+    purchase_date,
+    SUM(amount_spend) AS total_amount
+FROM Purchases
+WHERE DATE_FORMAT(purchase_date, '%Y%m') = '202311' AND DAYOFWEEK(purchase_date) = 6
+GROUP BY 2
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2993.Friday Purchases I/README.md

@@ -59,12 +59,30 @@ Output table is ordered by week_of_month in ascending order.</pre>
 
 ## Solutions
 
+**Solution 1: Date Functions**
+
+The date functions we use include:
+
+-   `DATE_FORMAT(date, format)`: Formats a date as a string
+-   `DAYOFWEEK(date)`: Returns the day of the week for a date, where 1 represents Sunday, 2 represents Monday, and so on
+-   `DAYOFMONTH(date)`: Returns the day of the month for a date
+
+First, we use the `DATE_FORMAT` function to format the date in the form of `YYYYMM`, then filter out the records of November 2023 that fall on a Friday. Next, we group the records by `purchase_date` and calculate the total consumption amount for each Friday.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
+    purchase_date,
+    SUM(amount_spend) AS total_amount
+FROM Purchases
+WHERE DATE_FORMAT(purchase_date, '%Y%m') = '202311' AND DAYOFWEEK(purchase_date) = 6
+GROUP BY 2
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2993.Friday Purchases I/README_EN.md

@@ -0,0 +1,9 @@
+# Write your MySQL query statement below
+SELECT
+    CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
+    purchase_date,
+    SUM(amount_spend) AS total_amount
+FROM Purchases
+WHERE DATE_FORMAT(purchase_date, '%Y%m') = '202311' AND DAYOFWEEK(purchase_date) = 6
+GROUP BY 2
+ORDER BY 1;

solution/2900-2999/2993.Friday Purchases I/Solution.sql

@@ -65,14 +65,35 @@ Output table is ordered by week_of_month in ascending order.</pre>
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：递归 + 左连接 + 日期函数**
+
+我们可以使用递归生成一个包含 2023 年 11 月所有日期的表 `T`，然后使用左连接将 `T` 与 `Purchases` 表按照日期进行连接，最后按照题目要求进行分组求和即可。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+WITH RECURSIVE
+    T AS (
+        SELECT '2023-11-01' AS purchase_date
+        UNION
+        SELECT purchase_date + INTERVAL 1 DAY
+        FROM T
+        WHERE purchase_date < '2023-11-30'
+    )
+SELECT
+    CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
+    purchase_date,
+    IFNULL(SUM(amount_spend), 0) AS total_amount
+FROM
+    T
+    LEFT JOIN Purchases USING (purchase_date)
+WHERE DAYOFWEEK(purchase_date) = 6
+GROUP BY 2
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2994.Friday Purchases II/README.md

@@ -61,12 +61,33 @@ Output table is ordered by week_of_month in ascending order.</pre>
 
 ## Solutions
 
+**Solution 1: Recursion + Left Join + Date Functions**
+
+We can generate a table `T` that contains all dates in November 2023 using recursion, then use a left join to connect `T` and the `Purchases` table by date. Finally, group and sum according to the requirements of the problem.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+WITH RECURSIVE
+    T AS (
+        SELECT '2023-11-01' AS purchase_date
+        UNION
+        SELECT purchase_date + INTERVAL 1 DAY
+        FROM T
+        WHERE purchase_date < '2023-11-30'
+    )
+SELECT
+    CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
+    purchase_date,
+    IFNULL(SUM(amount_spend), 0) AS total_amount
+FROM
+    T
+    LEFT JOIN Purchases USING (purchase_date)
+WHERE DAYOFWEEK(purchase_date) = 6
+GROUP BY 2
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2994.Friday Purchases II/README_EN.md

@@ -0,0 +1,18 @@
+WITH RECURSIVE
+    T AS (
+        SELECT '2023-11-01' AS purchase_date
+        UNION
+        SELECT purchase_date + INTERVAL 1 DAY
+        FROM T
+        WHERE purchase_date < '2023-11-30'
+    )
+SELECT
+    CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month,
+    purchase_date,
+    IFNULL(SUM(amount_spend), 0) AS total_amount
+FROM
+    T
+    LEFT JOIN Purchases USING (purchase_date)
+WHERE DAYOFWEEK(purchase_date) = 6
+GROUP BY 2
+ORDER BY 1;

solution/2900-2999/2994.Friday Purchases II/Solution.sql

@@ -65,14 +65,36 @@ Sessions table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：窗口函数 + 等值连接**
+
+我们可以用窗口函数 `RANK()` 按照 `user_id` 维度，对每个会话进行排名，记录在表 `T` 中，然后再将 `T` 与 `Sessions` 表按照 `user_id` 进行等值连接，并且筛选出 `T` 中排名为 1 的记录，并且 `session_type` 为 `Viewer`，`Sessions` 表中 `session_type` 为 `Streamer` 的记录，最后按照 `user_id` 进行分组求和即可。
+
 <!-- 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
+WITH
+    T AS (
+        SELECT
+            user_id,
+            session_type,
+            RANK() OVER (
+                PARTITION BY user_id
+                ORDER BY session_start
+            ) AS rk
+        FROM Sessions
+    )
+SELECT user_id, COUNT(1) AS sessions_count
+FROM
+    T AS t
+    JOIN Sessions AS s USING (user_id)
+WHERE rk = 1 AND t.session_type = 'Viewer' AND s.session_type = 'Streamer'
+GROUP BY 1
+ORDER BY 2 DESC, 1 DESC;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2995.Viewers Turned Streamers/README.md

@@ -60,12 +60,34 @@ Output table is ordered by sessions count and user_id in descending order.
 
 ## Solutions
 
+**Solution 1: Window Function + Equi-Join**
+
+We can use the window function `RANK()` to rank each session by `user_id` dimension, and record it in table `T`. Then, we equi-join `T` and the `Sessions` table by `user_id`, and filter out the records in `T` where the rank is 1, and `session_type` is `Viewer`, and `session_type` in the `Sessions` table is `Streamer`. Finally, we group by `user_id` and sum up.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            user_id,
+            session_type,
+            RANK() OVER (
+                PARTITION BY user_id
+                ORDER BY session_start
+            ) AS rk
+        FROM Sessions
+    )
+SELECT user_id, COUNT(1) AS sessions_count
+FROM
+    T AS t
+    JOIN Sessions AS s USING (user_id)
+WHERE rk = 1 AND t.session_type = 'Viewer' AND s.session_type = 'Streamer'
+GROUP BY 1
+ORDER BY 2 DESC, 1 DESC;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2995.Viewers Turned Streamers/README_EN.md

@@ -0,0 +1,19 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            user_id,
+            session_type,
+            RANK() OVER (
+                PARTITION BY user_id
+                ORDER BY session_start
+            ) AS rk
+        FROM Sessions
+    )
+SELECT user_id, COUNT(1) AS sessions_count
+FROM
+    T AS t
+    JOIN Sessions AS s USING (user_id)
+WHERE rk = 1 AND t.session_type = 'Viewer' AND s.session_type = 'Streamer'
+GROUP BY 1
+ORDER BY 2 DESC, 1 DESC;