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feat: add sql solutions to lc problems (doocs#1199)
* No.1097.Game Play Analysis V * No.1098.Unpopular Books * No.1107.New Users Daily Count * No.1112.Highest Grade For Each Student
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‎solution/1000-1099/1097.Game Play Analysis V/README.md

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@@ -68,7 +68,7 @@ Result 表：
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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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T AS (
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SELECT
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player_id,
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event_date,
@@ -79,13 +79,11 @@ SELECT
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install_dt,
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count(DISTINCT player_id) AS installs,
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round(
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sum(if(datediff(event_date, install_dt) = 1, 1, 0)) / count(
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DISTINCT player_id
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),
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sum(datediff(event_date, install_dt) = 1) / count(DISTINCT player_id),
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2
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) AS day1_retention
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FROM t
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GROUP BY install_dt;
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FROM T
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GROUP BY 1;
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```
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<!-- tabs:end -->

‎solution/1000-1099/1097.Game Play Analysis V/README_EN.md

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@@ -68,7 +68,7 @@ Player 2 installed the game on 2017-06-25 but didn&#39;t log back in on 2017-06-
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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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T AS (
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SELECT
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player_id,
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event_date,
@@ -79,13 +79,11 @@ SELECT
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install_dt,
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count(DISTINCT player_id) AS installs,
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round(
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sum(if(datediff(event_date, install_dt) = 1, 1, 0)) / count(
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DISTINCT player_id
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),
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sum(datediff(event_date, install_dt) = 1) / count(DISTINCT player_id),
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2
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) AS day1_retention
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FROM t
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GROUP BY install_dt;
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FROM T
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GROUP BY 1;
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```
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<!-- tabs:end -->

‎solution/1000-1099/1097.Game Play Analysis V/Solution.sql

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@@ -1,6 +1,6 @@
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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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T AS (
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SELECT
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player_id,
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event_date,
@@ -11,10 +11,8 @@ SELECT
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install_dt,
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count(DISTINCT player_id) AS installs,
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round(
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sum(if(datediff(event_date, install_dt) = 1, 1, 0)) / count(
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DISTINCT player_id
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),
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sum(datediff(event_date, install_dt) = 1) / count(DISTINCT player_id),
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2
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) AS day1_retention
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FROM t
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GROUP BY install_dt;
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FROM T
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GROUP BY 1;

‎solution/1000-1099/1098.Unpopular Books/README.md

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@@ -86,20 +86,13 @@ Result 表：
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```sql
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# Write your MySQL query statement below
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SELECT
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b.book_id,
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b.NAME
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SELECT book_id, name
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FROM
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books b
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LEFT JOIN orders o ON b.book_id = o.book_id
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WHERE
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b.available_from < '2019-05-23'
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GROUP BY
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b.book_id
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HAVING
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ifnull( sum( IF ( o.dispatch_date < '2018-06-23', 0, quantity )), 0 )< 10
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ORDER BY
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b.book_id
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Books
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LEFT JOIN Orders USING (book_id)
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WHERE available_from < '2019-05-23'
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GROUP BY 1
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HAVING sum(if(dispatch_date >= '2018-06-23', quantity, 0)) < 10;
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```
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<!-- tabs:end -->

‎solution/1000-1099/1098.Unpopular Books/README_EN.md

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```sql
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# Write your MySQL query statement below
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SELECT
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b.book_id,
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b.NAME
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SELECT book_id, name
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FROM
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books b
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LEFT JOIN orders o ON b.book_id = o.book_id
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WHERE
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b.available_from < '2019-05-23'
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GROUP BY
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b.book_id
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HAVING
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ifnull( sum( IF ( o.dispatch_date < '2018-06-23', 0, quantity )), 0 )< 10
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ORDER BY
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b.book_id
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Books
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LEFT JOIN Orders USING (book_id)
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WHERE available_from < '2019-05-23'
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GROUP BY 1
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HAVING sum(if(dispatch_date >= '2018-06-23', quantity, 0)) < 10;
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```
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<!-- tabs:end -->

‎solution/1000-1099/1098.Unpopular Books/Solution.sql

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# Write your MySQL query statement below
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SELECT
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b.book_id,
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b.NAME
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SELECT book_id, name
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FROM
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books AS b
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LEFT JOIN orders AS o ON b.book_id = o.book_id
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WHERE b.available_from < '2019-05-23'
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GROUP BY b.book_id
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HAVING ifnull(sum(IF(o.dispatch_date < '2018-06-23', 0, quantity)), 0) < 10
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ORDER BY b.book_id;
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Books
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LEFT JOIN Orders USING (book_id)
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WHERE available_from < '2019-05-23'
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GROUP BY 1
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HAVING sum(if(dispatch_date >= '2018-06-23', quantity, 0)) < 10;

‎solution/1100-1199/1107.New Users Daily Count/README.md

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@@ -67,18 +67,18 @@ ID 为 5 的用户第一次登陆于 2019-03-01，因此他不算在 2019-06-21
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```sql
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# Write your MySQL query statement below
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SELECT
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login_date,
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count(user_id) AS user_count
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FROM
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(
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SELECT min(activity_date) AS login_date, user_id
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WITH
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T AS (
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SELECT
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user_id,
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min(activity_date) OVER (PARTITION BY user_id) AS login_date
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FROM Traffic
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WHERE activity = 'login'
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GROUP BY user_id
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) AS t
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WHERE DATEDIFF('2019-6-30', login_date) <= 90
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GROUP BY login_date;
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)
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SELECT login_date, count(DISTINCT user_id) AS user_count
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FROM T
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WHERE datediff('2019-06-30', login_date) <= 90
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GROUP BY 1;
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```
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<!-- tabs:end -->

‎solution/1100-1199/1107.New Users Daily Count/README_EN.md

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@@ -71,18 +71,18 @@ The user with id 5 first logged in on 2019-03-01 so he&#39;s not counted on 2019
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```sql
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# Write your MySQL query statement below
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SELECT
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login_date,
76-
count(user_id) AS user_count
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FROM
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(
79-
SELECT min(activity_date) AS login_date, user_id
74+
WITH
75+
T AS (
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SELECT
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user_id,
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min(activity_date) OVER (PARTITION BY user_id) AS login_date
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FROM Traffic
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WHERE activity = 'login'
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GROUP BY user_id
83-
) AS t
84-
WHERE DATEDIFF('2019-6-30', login_date) <= 90
85-
GROUP BY login_date;
81+
)
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SELECT login_date, count(DISTINCT user_id) AS user_count
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FROM T
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WHERE datediff('2019-06-30', login_date) <= 90
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GROUP BY 1;
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```
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<!-- tabs:end -->

‎solution/1100-1199/1107.New Users Daily Count/Solution.sql

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Original file line numberDiff line numberDiff line change
@@ -1,13 +1,13 @@
11
# Write your MySQL query statement below
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SELECT
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login_date,
4-
count(user_id) AS user_count
5-
FROM
6-
(
7-
SELECT min(activity_date) AS login_date, user_id
2+
WITH
3+
T AS (
4+
SELECT
5+
user_id,
6+
min(activity_date) OVER (PARTITION BY user_id) AS login_date
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FROM Traffic
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WHERE activity = 'login'
10-
GROUP BY user_id
11-
) AS t
12-
WHERE DATEDIFF('2019-6-30', login_date) <= 90
13-
GROUP BY login_date;
9+
)
10+
SELECT login_date, count(DISTINCT user_id) AS user_count
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FROM T
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WHERE datediff('2019-06-30', login_date) <= 90
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GROUP BY 1;

‎solution/1100-1199/1112.Highest Grade For Each Student/README.md

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@@ -63,21 +63,21 @@ Enrollments 表：
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### **SQL**
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```sql
66-
SELECT
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student_id,
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course_id,
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grade
70-
FROM
71-
(
66+
# Write your MySQL query statement below
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WITH
68+
T AS (
7269
SELECT
7370
*,
74-
RANK() OVER (
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rank() OVER (
7572
PARTITION BY student_id
7673
ORDER BY grade DESC, course_id
7774
) AS rk
7875
FROM Enrollments
79-
) AS a
80-
WHERE a.rk = 1;
76+
)
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SELECT student_id, course_id, grade
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FROM T
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WHERE rk = 1
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ORDER BY student_id;
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```
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<!-- tabs:end -->

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