feat: add sql solution to lc problems: No.0175,1581,1795

- No.0175.Combine Two Tables
- No.1581.Customer Who Visited but Did Not Make Any Transactions
- No.1795.Rearrange Products Table

Author: YangFong

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

@@ -51,11 +51,12 @@ AddressId 是上表主键
 ### **SQL**
 
 ```sql
-# Write your MySQL query statement below
-SELECT p.FirstName, p.LastName, a.City, a.State
+SELECT p.FirstName,
+    p.LastName,
+    a.City,
+    a.State
 FROM Person p
-LEFT JOIN Address a
-ON p.PersonId = a.PersonId;
+    LEFT JOIN Address a ON p.PersonId = a.PersonId;
 ```
 
 <!-- tabs:end -->

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

@@ -81,11 +81,12 @@ addressId = 1 contains information about the address of personId = 2.
 ### **SQL**
 
 ```sql
-# Write your MySQL query statement below
-SELECT p.FirstName, p.LastName, a.City, a.State
+SELECT p.FirstName,
+    p.LastName,
+    a.City,
+    a.State
 FROM Person p
-LEFT JOIN Address a
-ON p.PersonId = a.PersonId;
+    LEFT JOIN Address a ON p.PersonId = a.PersonId;
 ```
 
 <!-- tabs:end -->

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

@@ -1,5 +1,6 @@
-# Write your MySQL query statement below
-SELECT p.FirstName, p.LastName, a.City, a.State
+SELECT p.FirstName,
+    p.LastName,
+    a.City,
+    a.State
 FROM Person p
-LEFT JOIN Address a
-ON p.PersonId = a.PersonId;
+    LEFT JOIN Address a ON p.PersonId = a.PersonId;

solution/1500-1599/1581.Customer Who Visited but Did Not Make Any Transactions/README.md

@@ -88,26 +88,19 @@ ID = 96 的顾客曾经去过购物中心，并且没有进行任何交易。
 
 <!-- tabs:start -->
 
-### **Python3**
+### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
-```python
-
-```
-
-### **Java**
-
-<!-- 这里可写当前语言的特殊实现逻辑 -->
-
-```java
-
-```
-
-### **...**
-
-```
-
+```sql
+SELECT customer_id,
+    COUNT(*) AS count_no_trans
+FROM Visits
+WHERE visit_id NOT IN (
+        SELECT visit_id
+        FROM Transactions
+    )
+GROUP BY customer_id;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1581.Customer Who Visited but Did Not Make Any Transactions/README_EN.md

@@ -89,22 +89,19 @@ As we can see, users with IDs 30 and 96 visited the mall one time without making
 
 <!-- tabs:start -->
 
-### **Python3**
-
-```python
-
-```
-
-### **Java**
-
-```java
-
-```
-
-### **...**
-
-```
-
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+SELECT customer_id,
+    COUNT(*) AS count_no_trans
+FROM Visits
+WHERE visit_id NOT IN (
+        SELECT visit_id
+        FROM Transactions
+    )
+GROUP BY customer_id;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1581.Customer Who Visited but Did Not Make Any Transactions/Solution.sql

@@ -0,0 +1,8 @@
+SELECT customer_id,
+    COUNT(*) AS count_no_trans
+FROM Visits
+WHERE visit_id NOT IN (
+        SELECT visit_id
+        FROM Transactions
+    )
+GROUP BY customer_id;

solution/1700-1799/1795.Rearrange Products Table/README.md

@@ -68,7 +68,23 @@ Products table:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+SELECT product_id,
+    'store1' AS store,
+    store1 AS price
+FROM products
+WHERE store1 IS NOT NULL
+UNION
+SELECT product_id,
+    'store2' AS store,
+    store2 AS price
+FROM products
+WHERE store2 IS NOT NULL
+UNION
+SELECT product_id,
+    'store3' AS store,
+    store3 AS price
+FROM products
+WHERE store3 IS NOT NULL;
 ```
 
 <!-- tabs:end -->

solution/1700-1799/1795.Rearrange Products Table/README_EN.md

@@ -62,7 +62,23 @@ Product 1 is available in store1 with price 70 and store3 with price 80. The pro
 ### **SQL**
 
 ```sql
-
+SELECT product_id,
+    'store1' AS store,
+    store1 AS price
+FROM products
+WHERE store1 IS NOT NULL
+UNION
+SELECT product_id,
+    'store2' AS store,
+    store2 AS price
+FROM products
+WHERE store2 IS NOT NULL
+UNION
+SELECT product_id,
+    'store3' AS store,
+    store3 AS price
+FROM products
+WHERE store3 IS NOT NULL;
 ```
 
 <!-- tabs:end -->

solution/1700-1799/1795.Rearrange Products Table/Solution.sql

@@ -0,0 +1,17 @@
+SELECT product_id,
+    'store1' AS store,
+    store1 AS price
+FROM products
+WHERE store1 IS NOT NULL
+UNION
+SELECT product_id,
+    'store2' AS store,
+    store2 AS price
+FROM products
+WHERE store2 IS NOT NULL
+UNION
+SELECT product_id,
+    'store3' AS store,
+    store3 AS price
+FROM products
+WHERE store3 IS NOT NULL;