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# 1795.(Easy)Rearrange Products Table

Table: Products

```
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| product_id  | int     |
| store1      | int     |
| store2      | int     |
| store3      | int     |
+-------------+---------+
product_id is the primary key for this table.
Each row in this table indicates the product's price in 3 different stores: store1, store2, and store3.
If the product is not available in a store, the price will be null in that store's column.
```

Write an SQL query to rearrange the Products table so that each row has (product\_id, store, price). If a product is not available in a store, do not include a row with that product\_id and store combination in the result table.

Return the result table in any order.

The query result format is in the following example:

```
Products table:
+------------+--------+--------+--------+
| product_id | store1 | store2 | store3 |
+------------+--------+--------+--------+
| 0          | 95     | 100    | 105    |
| 1          | 70     | null   | 80     |
+------------+--------+--------+--------+

Result table:
+------------+--------+-------+
| product_id | store  | price |
+------------+--------+-------+
| 0          | store1 | 95    |
| 0          | store2 | 100   |
| 0          | store3 | 105   |
| 1          | store1 | 70    |
| 1          | store3 | 80    |
+------------+--------+-------+

Product 0 is available in all three stores with prices 95, 100, and 105 respectively.
Product 1 is available in store1 with price 70 and store3 with price 80. The product is not available in store2.
```

来源：力扣（LeetCode）

链接：<https://leetcode-cn.com/problems/rearrange-products-table>

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

## Solution

[1777.(Easy)每家商店的产品价格](/leetcode/sql/1777.products-price-for-each-store.md) 的同系列题。long/wide table 的相互转换。

```sql
select * from 
(
    select product_id, 'store1' store, store1 price from Products
    union 
    select product_id, 'store2' store, store2 price from Products
    union
    select product_id, 'store3' store, store3 price from Products
)t 
where price is not null;
```