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feat: update sql solutions to lc problems (doocs#1187)
* No.0577.Employee Bonus * No.0578.Get Highest Answer Rate Question * No.0579.Find Cumulative Salary of an Employee * No.0584.Find Customer Referee
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‎solution/0500-0599/0577.Employee Bonus/README.md

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@@ -52,13 +52,12 @@ empId 是这张表单的主关键字
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### **SQL**
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```sql
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SELECT
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e.name,
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b.bonus
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# Write your MySQL query statement below
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SELECT name, bonus
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FROM
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Employee AS e
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LEFT JOIN Bonus AS b ON e.empid = b.empid
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WHERE b.bonus < 1000 OR b.bonus IS NULL;
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Employee
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LEFT JOIN Bonus USING (empId)
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WHERE ifnull(bonus, 0) < 1000;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0577.Employee Bonus/README_EN.md

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@@ -81,13 +81,12 @@ Bonus table:
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### **SQL**
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```sql
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SELECT
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e.name,
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b.bonus
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# Write your MySQL query statement below
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SELECT name, bonus
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FROM
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Employee AS e
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LEFT JOIN Bonus AS b ON e.empid = b.empid
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WHERE b.bonus < 1000 OR b.bonus IS NULL;
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Employee
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LEFT JOIN Bonus USING (empId)
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WHERE ifnull(bonus, 0) < 1000;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0577.Employee Bonus/Solution.sql

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@@ -1,7 +1,6 @@
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SELECT
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e.name,
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b.bonus
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# Write your MySQL query statement below
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SELECT name, bonus
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FROM
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Employee AS e
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LEFT JOIN Bonus AS b ON e.empid = b.empid
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WHERE b.bonus < 1000 OR b.bonus IS NULL;
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Employee
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LEFT JOIN Bonus USING (empId)
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WHERE ifnull(bonus, 0) < 1000;

‎solution/0500-0599/0578.Get Highest Answer Rate Question/README.md

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Original file line numberDiff line numberDiff line change
@@ -73,11 +73,28 @@ SurveyLog table:
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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 question_id AS survey_log
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FROM SurveyLog
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GROUP BY 1
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ORDER BY SUM(action = 'answer') / SUM(action = 'show') DESC
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LIMIT 1;
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FROM SurveyLog
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GROUP BY 1
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ORDER BY sum(action = 'answer') / sum(action = 'show') DESC, 1
81+
LIMIT 1;
82+
```
83+
84+
```sql
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WITH
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T AS (
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SELECT
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question_id AS survey_log,
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(sum(action = 'answer') OVER (PARTITION BY question_id)) / (
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sum(action = 'show') OVER (PARTITION BY question_id)
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) AS ratio
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FROM SurveyLog
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)
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SELECT survey_log
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FROM T
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ORDER BY ratio DESC, 1
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LIMIT 1;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0578.Get Highest Answer Rate Question/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -64,11 +64,28 @@ Question 285 has the highest answer rate.</pre>
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### **SQL**
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6666
```sql
67+
# Write your MySQL query statement below
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SELECT question_id AS survey_log
68-
FROM SurveyLog
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GROUP BY 1
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ORDER BY SUM(action = 'answer') / SUM(action = 'show') DESC
71-
LIMIT 1;
69+
FROM SurveyLog
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GROUP BY 1
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ORDER BY sum(action = 'answer') / sum(action = 'show') DESC, 1
72+
LIMIT 1;
73+
```
74+
75+
```sql
76+
WITH
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T AS (
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SELECT
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question_id AS survey_log,
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(sum(action = 'answer') OVER (PARTITION BY question_id)) / (
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sum(action = 'show') OVER (PARTITION BY question_id)
82+
) AS ratio
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FROM SurveyLog
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)
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SELECT survey_log
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FROM T
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ORDER BY ratio DESC, 1
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LIMIT 1;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0578.Get Highest Answer Rate Question/Solution.sql

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@@ -1,5 +1,13 @@
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SELECT question_id AS survey_log
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FROM SurveyLog
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GROUP BY 1
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ORDER BY SUM(action = 'answer') / SUM(action = 'show') DESC
1+
WITH
2+
T AS (
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SELECT
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question_id AS survey_log,
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(sum(action = 'answer') OVER (PARTITION BY question_id)) / (
6+
sum(action = 'show') OVER (PARTITION BY question_id)
7+
) AS ratio
8+
FROM SurveyLog
9+
)
10+
SELECT survey_log
11+
FROM T
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ORDER BY ratio DESC, 1
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LIMIT 1;

‎solution/0500-0599/0579.Find Cumulative Salary of an Employee/README.md

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@@ -109,4 +109,28 @@ WHERE
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ORDER BY id, month DESC;
110110
```
111111

112+
```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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id,
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month,
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sum(salary) OVER (
120+
PARTITION BY id
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ORDER BY month
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RANGE 2 PRECEDING
123+
) AS salary,
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rank() OVER (
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PARTITION BY id
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ORDER BY month DESC
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) AS rk
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FROM Employee
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)
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SELECT id, month, salary
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FROM T
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WHERE rk > 1
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ORDER BY 1, 2 DESC;
134+
```
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112136
<!-- tabs:end -->

‎solution/0500-0599/0579.Find Cumulative Salary of an Employee/README_EN.md

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@@ -134,4 +134,28 @@ WHERE
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ORDER BY id, month DESC;
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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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id,
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month,
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sum(salary) OVER (
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PARTITION BY id
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ORDER BY month
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RANGE 2 PRECEDING
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) AS salary,
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rank() OVER (
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PARTITION BY id
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ORDER BY month DESC
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) AS rk
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FROM Employee
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)
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SELECT id, month, salary
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FROM T
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WHERE rk > 1
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ORDER BY 1, 2 DESC;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0579.Find Cumulative Salary of an Employee/Solution.sql

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Original file line numberDiff line numberDiff line change
@@ -1,19 +1,21 @@
11
# Write your MySQL query statement below
2-
SELECT
3-
id,
4-
month,
5-
sum(salary) OVER (
6-
PARTITION BY id
7-
ORDER BY month
8-
RANGE 2 PRECEDING
9-
) AS Salary
10-
FROM employee
11-
WHERE
12-
(id, month) NOT IN (
2+
WITH
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T AS (
134
SELECT
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id,
15-
max(month)
6+
month,
7+
sum(salary) OVER (
8+
PARTITION BY id
9+
ORDER BY month
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RANGE 2 PRECEDING
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) AS salary,
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rank() OVER (
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PARTITION BY id
14+
ORDER BY month DESC
15+
) AS rk
1616
FROM Employee
17-
GROUP BY id
1817
)
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ORDER BY id, month DESC;
18+
SELECT id, month, salary
19+
FROM T
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WHERE rk > 1
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ORDER BY 1, 2 DESC;

‎solution/0500-0599/0580.Count Student Number in Departments/README.md

+11-7
Original file line numberDiff line numberDiff line change
@@ -84,14 +84,18 @@ Department 表:
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### **SQL**
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```sql
87-
SELECT
88-
department.dept_name,
89-
COUNT(student.dept_id) AS student_number
87+
# Write your MySQL query statement below
88+
WITH
89+
S AS (
90+
SELECT dept_id, count(1) AS cnt
91+
FROM Student
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GROUP BY dept_id
93+
)
94+
SELECT dept_name, ifnull(cnt, 0) AS student_number
9095
FROM
91-
Student
92-
RIGHT JOIN Department ON student.dept_id = department.dept_id
93-
GROUP BY dept_name
94-
ORDER BY student_number DESC, dept_name;
96+
S
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RIGHT JOIN Department USING (dept_id)
98+
ORDER BY 2 DESC, 1;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0580.Count Student Number in Departments/README_EN.md

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@@ -81,14 +81,18 @@ Department table:
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### **SQL**
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```sql
84-
SELECT
85-
department.dept_name,
86-
COUNT(student.dept_id) AS student_number
84+
# Write your MySQL query statement below
85+
WITH
86+
S AS (
87+
SELECT dept_id, count(1) AS cnt
88+
FROM Student
89+
GROUP BY dept_id
90+
)
91+
SELECT dept_name, ifnull(cnt, 0) AS student_number
8792
FROM
88-
Student
89-
RIGHT JOIN Department ON student.dept_id = department.dept_id
90-
GROUP BY dept_name
91-
ORDER BY student_number DESC, dept_name;
93+
S
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RIGHT JOIN Department USING (dept_id)
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ORDER BY 2 DESC, 1;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0580.Count Student Number in Departments/Solution.sql

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@@ -1,8 +1,12 @@
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SELECT
2-
department.dept_name,
3-
COUNT(student.dept_id) AS student_number
1+
# Write your MySQL query statement below
2+
WITH
3+
S AS (
4+
SELECT dept_id, count(1) AS cnt
5+
FROM Student
6+
GROUP BY dept_id
7+
)
8+
SELECT dept_name, ifnull(cnt, 0) AS student_number
49
FROM
5-
Student
6-
RIGHT JOIN Department ON student.dept_id = department.dept_id
7-
GROUP BY dept_name
8-
ORDER BY student_number DESC, dept_name;
10+
S
11+
RIGHT JOIN Department USING (dept_id)
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ORDER BY 2 DESC, 1;

‎solution/0500-0599/0584.Find Customer Referee/README.md

+3-12
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@@ -45,19 +45,10 @@
4545
### **SQL**
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4747
```sql
48-
SELECT
49-
name
48+
# Write your MySQL query statement below
49+
SELECT name
5050
FROM Customer
51-
WHERE referee_id != 2 OR referee_id IS NULL;
52-
```
53-
54-
MySQL 可使用 `IFNULL()`：
55-
56-
```sql
57-
SELECT
58-
name
59-
FROM customer
60-
WHERE IFNULL(referee_id, 0) != 2;
51+
WHERE ifnull(referee_id, 0) != 2;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0584.Find Customer Referee/README_EN.md

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@@ -60,19 +60,10 @@ Customer table:
6060
### **SQL**
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6262
```sql
63-
SELECT
64-
name
63+
# Write your MySQL query statement below
64+
SELECT name
6565
FROM Customer
66-
WHERE referee_id != 2 OR referee_id IS NULL;
67-
```
68-
69-
MySQL can use `IFNULL()`:
70-
71-
```sql
72-
SELECT
73-
name
74-
FROM customer
75-
WHERE IFNULL(referee_id, 0) != 2;
66+
WHERE ifnull(referee_id, 0) != 2;
7667
```
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7869
<!-- tabs:end -->

‎solution/0500-0599/0584.Find Customer Referee/Solution.sql

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@@ -1,4 +1,4 @@
1-
SELECT
2-
name
1+
# Write your MySQL query statement below
2+
SELECT name
33
FROM Customer
4-
WHERE referee_id != 2 OR referee_id IS NULL;
4+
WHERE ifnull(referee_id, 0) != 2;

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