A Data Analyst is working with three tables:Which query would return a list of all...

To achieve the desired result, an analyst must understand the fundamental behavior of different JOIN types within Snowflake and how they affect the retention of records from the "left" or primary table. The goal here is to list all brokers, even those who have zero customers (like "Drew") or customers with zero orders (like "Debby").

In SQL, an INNER JOIN only returns rows when there is a match in both tables. If we were to use an INNER JOIN between BROKER and CUSTOMER, Drew would be excluded from the results because he has no associated records in the CUSTOMER table. Similarly, an INNER JOIN with the ORDERS table would exclude any broker whose customers haven't placed an order.

Evaluating the Join Logic:

Option C is the correct solution because it utilizes a chain of LEFT JOINs. A LEFT JOIN (or LEFT OUTER JOIN) ensures that every record from the left table (BROKER) is preserved in the result set. If no matching record exists in the joined table (CUSTOMER or ORDERS), Snowflake populates the columns with NULL. This is why "Drew" appears with a CUST_COUNT of 0 and "Debby" appears with a NULL for the total order amount.

Option A fails because it uses an INNER JOIN for the CUSTOMER table, which would immediately filter out "Drew."

Option B and Option D fail because they use INNER JOINs at different stages of the query, which would strip away brokers or customers that do not have matching order activity.

Additionally, the query correctly uses COUNT(DISTINCT c.customer_id) to ensure that customers are not double-counted if they have multiple orders, and GROUP BY 1 (referencing b.broker_name) to aggregate the data at the broker level. This pattern is essential for accurate Data Analysis in Snowflake when dealing with "optional" relationships in a star or snowflake schema.