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What will be the result of the following query? SELECT customer_id, AVG(total_amount) AS avg_amount FROM Orders GROUP BY customer_id HAVING AVG(total_amount) > 200;OptionsReturns the customer IDs and average order amounts for all customers.Returns the customer IDs and average order amounts for customers with an average order amount greater than 200.Returns the customer IDs and total order amounts for customers with an average order amount greater than 200.Returns an error because the HAVING clause is not allowed in this context.

Question

What will be the result of the following query? SELECT customer_id, AVG(total_amount) AS avg_amount FROM Orders GROUP BY customer_id HAVING AVG(total_amount) > 200;OptionsReturns the customer IDs and average order amounts for all customers.Returns the customer IDs and average order amounts for customers with an average order amount greater than 200.Returns the customer IDs and total order amounts for customers with an average order amount greater than 200.Returns an error because the HAVING clause is not allowed in this context.

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Solution

The result of the query will be: "Returns the customer IDs and average order amounts for customers with an average order amount greater than 200."

Here's a step-by-step explanation:

  1. The SELECT statement is used to select data from a database. In this case, it's selecting the customer_id and the average (AVG) of the total_amount from the Orders table.

  2. The GROUP BY statement is used in conjunction with the aggregate function (like AVG), to group the result-set by one or more columns. Here, it's grouping the results by customer_id.

  3. The HAVING clause was added to SQL because the WHERE keyword could not be used with aggregate functions. Here, it's being used to filter the results to only include those where the average total_amount is greater than 200.

So, the query is returning the customer IDs and their corresponding average order amounts, but only for those customers whose average order amount is greater than 200.

This problem has been solved

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