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Commit e55e434

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feat: update sql solutions to lc problems: No.0511,0512,0534 (doocs#1182)
* No.0511.Game Play Analysis I * No.0512.Game Play Analysis II * No.0534.Game Play Analysis III
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‎.github/workflows/black-lint.yml‎

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name: black-linter
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on:
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branches: [main]
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types:
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- opened
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branches:
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‎.github/workflows/clang-format-lint.yml‎

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name: clang-format-linter
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branches: [main]
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pull_request_target:
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types:
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branches:
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‎.github/workflows/pr-checker.yml‎

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In addition, the coding style (such as the naming of variables and functions) should be as consistent as possible with the existing code in the project.
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<!-- Sticky Pull Request Comment -->
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emojis: '+1, laugh'
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body-include: '<!-- Sticky Pull Request Comment -->'

‎solution/0500-0599/0511.Game Play Analysis I/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**
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我们可以用 `GROUP BY` 对 `player_id` 进行分组,然后取每一组中最小的 `event_date` 作为玩家第一次登录平台的日期。
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<!-- tabs:start -->
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### **SQL**
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```sql
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SELECT
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player_id,
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MIN(event_date) AS first_login
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# Write your MySQL query statement below
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SELECT player_id, min(event_date) AS first_login
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FROM Activity
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GROUP BY player_id;
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```

‎solution/0500-0599/0511.Game Play Analysis I/README_EN.md‎

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### **SQL**
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```sql
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SELECT
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player_id,
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MIN(event_date) AS first_login
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# Write your MySQL query statement below
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SELECT player_id, min(event_date) AS first_login
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FROM Activity
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GROUP BY player_id;
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```
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SELECT
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player_id,
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MIN(event_date) AS first_login
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# Write your MySQL query statement below
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SELECT player_id, min(event_date) AS first_login
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FROM Activity
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GROUP BY player_id;

‎solution/0500-0599/0512.Game Play Analysis II/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:子查询**
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我们可以使用 `GROUP BY` 和 `MIN` 函数来找到每个玩家的第一次登录日期,然后使用联合键子查询来找到每个玩家的第一次登录设备。
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**方法二:窗口函数**
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我们可以使用窗口函数 `rank()`,它可以为每个玩家的每个登录日期分配一个排名,然后我们可以选择排名为 1ドル$ 的行。
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### **SQL**
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);
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```
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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*,
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rank() OVER (
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PARTITION BY player_id
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ORDER BY event_date
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) AS rk
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FROM Activity
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)
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SELECT player_id, device_id
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FROM T
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WHERE rk = 1;
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```
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‎solution/0500-0599/0512.Game Play Analysis II/README_EN.md‎

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);
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```
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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*,
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rank() OVER (
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PARTITION BY player_id
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ORDER BY event_date
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) AS rk
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FROM Activity
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)
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SELECT player_id, device_id
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FROM T
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WHERE rk = 1;
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```
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‎solution/0500-0599/0534.Game Play Analysis III/README.md‎

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**方法一:SUM() OVER() 窗口函数**
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我们可以使用 `SUM() OVER()` 窗口函数来计算每个玩家到目前为止玩了多少游戏。在 `OVER()` 子句中,我们使用 `PARTITION BY` 子句将玩家分组,然后使用 `ORDER BY` 子句按日期排序。
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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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player_id,
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event_date,
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SUM(games_played) OVER (
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sum(games_played) OVER (
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PARTITION BY player_id
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ORDER BY event_date
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) AS games_played_so_far
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FROM Activity
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ORDER BY 1, 2;
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FROM Activity;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0534.Game Play Analysis III/README_EN.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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player_id,
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event_date,
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SUM(games_played) OVER (
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sum(games_played) OVER (
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PARTITION BY player_id
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ORDER BY event_date
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) AS games_played_so_far
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FROM Activity
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ORDER BY 1, 2;
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FROM Activity;
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```
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<!-- tabs:end -->

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