feat: add sql formatter (doocs#1039)

Author: yanglbme

.prettierignore

@@ -12,4 +12,12 @@ node_modules/
 /solution/problem_readme_template.md
 /solution/problem_readme_template_en.md
 /solution/bash_problem_readme_template.md
-/solution/bash_problem_readme_template_en.md
+/solution/bash_problem_readme_template_en.md
+/solution/0100-0199/0177.Nth Highest Salary/Solution.sql
+/solution/1000-1099/1076.Project Employees II/Solution.sql
+/solution/1000-1099/1082.Sales Analysis I/Solution.sql
+/solution/1500-1599/1511.Customer Order Frequency
+/solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
+/solution/1600-1699/1613.Find the Missing IDs/Solution.sql
+/solution/1400-1499/1484.Group Sold Products By The Date/Solution.sql
+/solution/1100-1199/1173.Immediate Food Delivery I/Solution.sql

.prettierrc

@@ -8,5 +8,29 @@
     "arrowParens": "avoid",
     "phpVersion": "8.1",
     "braceStyle": "1tbs",
-    "endOfLine": "lf"
+    "endOfLine": "lf",
+    "sqlKeywordCase": "upper",
+    "overrides": [
+        {
+            "files": [
+                "solution/0500-0599/0578.Get Highest Answer Rate Question/Solution.sql",
+                "solution/0600-0699/0618.Students Report By Geography/Solution.sql",
+                "solution/0600-0699/0626.Exchange Seats/Solution.sql",
+                "solution/1000-1099/1098.Unpopular Books/Solution.sql",
+                "solution/1100-1199/1113.Reported Posts/Solution.sql",
+                "solution/1100-1199/1174.Immediate Food Delivery II/Solution.sql",
+                "solution/1100-1199/1193.Monthly Transactions I/Solution.sql",
+                "solution/1300-1399/1384.Total Sales Amount by Year/Solution.sql",
+                "solution/1300-1399/1322.Ads Performance/Solution.sql",
+                "solution/1300-1399/1398.Customers Who Bought Products A and B but Not C/Solution.sql",
+                "solution/1500-1599/1501.Countries You Can Safely Invest In/Solution.sql",
+                "solution/1600-1699/1667.Fix Names in a Table/Solution.sql",
+                "solution/1900-1999/1972.First and Last Call On the Same Day/Solution.sql",
+                "solution/2600-2699/2686.Immediate Food Delivery III/Solution.sql"
+            ],
+            "options": {
+                "parser": "bigquery"
+            }
+        }
+    ]
 }

package-lock.json

@@ -14,10 +14,11 @@
     "clang-format": "1.8.0",
     "husky": "^8.0.3",
     "lint-staged": "^13.2.2",
-    "prettier": "^2.8.8"
+    "prettier": "^2.8.8",
+    "prettier-plugin-sql-cst": "^0.8.2"
   },
   "lint-staged": {
-    "*.{js,ts,php,md}": "prettier --write",
+    "*.{js,ts,php,md,sql}": "prettier --write",
     "*.py": "black -S"
   }
 }

package.json

@@ -86,14 +86,14 @@ addressId = 1 包含了 personId = 2 的地址信息。</pre>
 
 ```sql
 # Write your MySQL query statement below
-select
+SELECT
     firstName,
     lastName,
     city,
     state
-from
-    Person p
-    left join Address a on p.personId = a.personId
+FROM
+    Person AS p
+    LEFT JOIN Address AS a ON p.personId = a.personId;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0175.Combine Two Tables/README.md

@@ -82,14 +82,14 @@ addressId = 1 contains information about the address of personId = 2.
 
 ```sql
 # Write your MySQL query statement below
-select
+SELECT
     firstName,
     lastName,
     city,
     state
-from
-    Person p
-    left join Address a on p.personId = a.personId
+FROM
+    Person AS p
+    LEFT JOIN Address AS a ON p.personId = a.personId;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0175.Combine Two Tables/README_EN.md

@@ -1,9 +1,9 @@
 # Write your MySQL query statement below
-select
+SELECT
     firstName,
     lastName,
     city,
     state
-from
-    Person p
-    left join Address a on p.personId = a.personId
+FROM
+    Person AS p
+    LEFT JOIN Address AS a ON p.personId = a.personId;

solution/0100-0199/0175.Combine Two Tables/Solution.sql

@@ -82,12 +82,12 @@ Employee 表：
 ```sql
 # Write your MySQL query statement below
 SELECT
-(
-    SELECT DISTINCT Salary
-    FROM Employee
-    ORDER BY Salary DESC
-    LIMIT 1 OFFSET 1
-) AS SecondHighestSalary;
+    (
+        SELECT DISTINCT Salary
+        FROM Employee
+        ORDER BY Salary DESC
+        LIMIT 1 OFFSET 1
+    ) AS SecondHighestSalary;
 ```
 
 解法 2：使用 `MAX()` 函数，从小于 `MAX()` 的 Salary 中挑选最大值 `MAX()` 即可。
@@ -96,10 +96,11 @@ SELECT
 # Write your MySQL query statement below
 SELECT MAX(Salary) AS SecondHighestSalary
 FROM Employee
-WHERE Salary < (
-    SELECT MAX(Salary)
-    FROM Employee
-);
+WHERE
+    Salary < (
+        SELECT MAX(Salary)
+        FROM Employee
+    );
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0176.Second Highest Salary/README.md

@@ -73,12 +73,12 @@ Solution 1: Use Sub Query and LIMIT.
 ```sql
 # Write your MySQL query statement below
 SELECT
-(
-    SELECT DISTINCT Salary
-    FROM Employee
-    ORDER BY Salary DESC
-    LIMIT 1 OFFSET 1
-) AS SecondHighestSalary;
+    (
+        SELECT DISTINCT Salary
+        FROM Employee
+        ORDER BY Salary DESC
+        LIMIT 1 OFFSET 1
+    ) AS SecondHighestSalary;
 ```
 
 Solution 2: Use `MAX()` function.
@@ -87,10 +87,11 @@ Solution 2: Use `MAX()` function.
 # Write your MySQL query statement below
 SELECT MAX(Salary) AS SecondHighestSalary
 FROM Employee
-WHERE Salary < (
-    SELECT MAX(Salary)
-    FROM Employee
-);
+WHERE
+    Salary < (
+        SELECT MAX(Salary)
+        FROM Employee
+    );
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0176.Second Highest Salary/README_EN.md

@@ -1,8 +1,8 @@
 # Write your MySQL query statement below
 SELECT
-(
-    SELECT DISTINCT Salary
-    FROM Employee
-    ORDER BY Salary DESC
-    LIMIT 1 OFFSET 1
-) AS SecondHighestSalary;
+    (
+        SELECT DISTINCT Salary
+        FROM Employee
+        ORDER BY Salary DESC
+        LIMIT 1 OFFSET 1
+    ) AS SecondHighestSalary;

solution/0100-0199/0176.Second Highest Salary/Solution.sql

@@ -92,12 +92,8 @@ DENSE_RANK() OVER (
 # Write your MySQL query statement below
 SELECT
     Score,
-    DENSE_RANK() OVER (
-        ORDER BY
-            Score DESC
-    ) 'Rank'
-FROM
-    Scores;
+    DENSE_RANK() OVER (ORDER BY Score DESC) AS 'Rank'
+FROM Scores;
 ```
 
 ### **MySQL5**

solution/0100-0199/0178.Rank Scores/README.md

@@ -81,12 +81,8 @@ Solution:
 # Write your MySQL query statement below
 SELECT
     Score,
-    DENSE_RANK() OVER (
-        ORDER BY
-            Score DESC
-    ) 'Rank'
-FROM
-    Scores;
+    DENSE_RANK() OVER (ORDER BY Score DESC) AS 'Rank'
+FROM Scores;
 ```
 
 ### **MySQL5**

solution/0100-0199/0178.Rank Scores/README_EN.md

@@ -1,9 +1,5 @@
 # Write your MySQL query statement below
 SELECT
     Score,
-    DENSE_RANK() OVER (
-        ORDER BY
-            Score DESC
-    ) 'Rank'
-FROM
-    Scores;
+    DENSE_RANK() OVER (ORDER BY Score DESC) AS 'Rank'
+FROM Scores;

solution/0100-0199/0178.Rank Scores/Solution.sql

@@ -61,9 +61,11 @@ Result 表：
 ### **SQL**
 
 ```sql
-select distinct(Num) as ConsecutiveNums from Logs Curr where
-    Num = (select Num from Logs where id = Curr.id - 1) and
-    Num = (select Num from Logs where id = Curr.id - 2)
+SELECT DISTINCT (Num) AS ConsecutiveNums
+FROM Logs AS Curr
+WHERE
+    Num = (SELECT Num FROM Logs WHERE id = Curr.id - 1)
+    AND Num = (SELECT Num FROM Logs WHERE id = Curr.id - 2);
 ```
 
 ```sql
@@ -75,12 +77,9 @@ FROM
     logs AS l3
 WHERE
     l1.id = l2.id - 1
-    AND
-    l2.id = l3.id - 1
-    AND
-    l1.num = l2.num
-    AND
-    l2.num = l3.num
+    AND l2.id = l3.id - 1
+    AND l1.num = l2.num
+    AND l2.num = l3.num;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0180.Consecutive Numbers/README.md

@@ -58,9 +58,11 @@ Logs table:
 ### **SQL**
 
 ```sql
-select distinct(Num) as ConsecutiveNums from Logs Curr where
-    Num = (select Num from Logs where id = Curr.id - 1) and
-    Num = (select Num from Logs where id = Curr.id - 2)
+SELECT DISTINCT (Num) AS ConsecutiveNums
+FROM Logs AS Curr
+WHERE
+    Num = (SELECT Num FROM Logs WHERE id = Curr.id - 1)
+    AND Num = (SELECT Num FROM Logs WHERE id = Curr.id - 2);
 ```
 
 ```sql
@@ -72,12 +74,9 @@ FROM
     logs AS l3
 WHERE
     l1.id = l2.id - 1
-    AND
-    l2.id = l3.id - 1
-    AND
-    l1.num = l2.num
-    AND
-    l2.num = l3.num
+    AND l2.id = l3.id - 1
+    AND l1.num = l2.num
+    AND l2.num = l3.num;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0180.Consecutive Numbers/README_EN.md

@@ -1,3 +1,5 @@
-select distinct(Num) as ConsecutiveNums from Logs Curr where
-    Num = (select Num from Logs where id = Curr.id - 1) and
-    Num = (select Num from Logs where id = Curr.id - 2)
+SELECT DISTINCT (Num) AS ConsecutiveNums
+FROM Logs AS Curr
+WHERE
+    Num = (SELECT Num FROM Logs WHERE id = Curr.id - 1)
+    AND Num = (SELECT Num FROM Logs WHERE id = Curr.id - 2);