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Question 1098

A column named “Data” contains VARIANT data and stores values as follows:

How will Snowflake extract the employee’s name from the column data?

Data:employee.name

DATA:employee.name

data:Employee.name

data:employee.name

Correct Answer: C

To correctly extract data from a VARIANT column in Snowflake, it's important to know that the path to the nested field is case-sensitive. In the given JSON structure,

Discussion

NachoPrendesOption: C

C, because "Employee" has the capital "E"

Jacobr5000Option: D

"Regardless of which notation you use, the column name is case-insensitive but element names are case-sensitive. For example, in the following list, the first two paths are equivalent, but the third is not: src:salesperson.name SRC:salesperson.name SRC:Salesperson.Name" https://docs.snowflake.com/en/user-guide/querying-semistructured#traversing-semi-structured-data

Lematthew31Option: C

It's C