SQL для тестировщика

Дана следующая таблица:

| OrderID | CustomerID |  OrderDate | TotalAmount |
|---------|------------|------------|-------------|
|    1    |    101     | 2022-05-15 |    500      |
|    2    |    102     | 2022-06-20 |    750      |
|    3    |    103     | 2022-06-25 |   1000      |
|    4    |    101     | 2022-07-10 |    300      |
|    5    |    104     | 2022-08-05 |    900      |

Какой из следующих SQL запросов вернет CustomerID и максимальный TotalAmount среди всех заказов?

A)

SELECT CustomerID, MAX(TotalAmount) FROM Orders

SELECT CustomerID, TotalAmount FROM Orders WHERE TotalAmount = MAX(TotalAmount)

SELECT CustomerID, TotalAmount FROM Orders WHERE TotalAmount = (SELECT MAX(TotalAmount) FROM Orders)

D) SELECT CustomerID, MAX(TotalAmount) FROM Orders GROUP BY CustomerID

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2.2K viewsJun 15, 2023 at 13:56

tg-me.com/sql_for_qa/108

Create: 2023-06-15
Last Update: 2025-07-01 22:12:40

Дана следующая таблица:

| OrderID | CustomerID | OrderDate | TotalAmount | |---------|------------|------------|-------------| | 1 | 101 | 2022-05-15 | 500 | | 2 | 102 | 2022-06-20 | 750 | | 3 | 103 | 2022-06-25 | 1000 | | 4 | 101 | 2022-07-10 | 300 | | 5 | 104 | 2022-08-05 | 900 |Какой из следующих SQL запросов вернет CustomerID и максимальный TotalAmount среди всех заказов?

A) SELECT CustomerID, MAX(TotalAmount) FROM OrdersB) SELECT CustomerID, TotalAmount FROM Orders WHERE TotalAmount = MAX(TotalAmount)C) SELECT CustomerID, TotalAmount FROM Orders WHERE TotalAmount = (SELECT MAX(TotalAmount) FROM Orders)D) SELECT CustomerID, MAX(TotalAmount) FROM Orders GROUP BY CustomerID

BY SQL для тестировщика

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