Which query converts an array of project names into individual rows without using lateral?

• A. select emp_id, emp_name, p.value::string as project_names from employee_project_details,table(flatten(employee_project_details.project_names)) p;
• B. select emp_id, emp_name, p.value::string as project_names from employee_project_details,lateral(flatten(employee_project_details.project_names)) p;
• C. select emp_id, emp_name, p.value::string as project_names from employee_project_details, flatten(employee_project_details.project_names) p;
• D. select emp_id, emp_name, p.value::string as project_names from employee_project_details, flatten(employee_project_details.project_names) p;

Answer: (A)

Flattening an array into rows without using LATERAL is done by treating the flatten operation as a table-valued function and wrapping it with TABLE(). This makes the flattened result a real table that you can join with the original rows, producing one output row for each element of the array while keeping the parent row’s columns intact. In practice, you pull the elements via the alias and its value field (p.value) and cast as needed, so each project name becomes its own row alongside the corresponding emp_id and emp_name.

This approach fits the requirement of not using LATERAL, since TABLE(FLATTEN(INPUT => ...)) is a standalone table expression that can be joined in the FROM clause like a regular table. The other patterns rely on LATERAL or omit the proper TABLE wrapper around FLATTEN, which either violates the constraint or isn’t valid syntax in Snowflake for turning an array into rows.