Which Snowflake function will interpret an input string as a JSON document, and produce a VARIANT value?

The correct answer is A. parse_json(). The parse_json() function in Snowflake interprets an input string as a JSON document and produces a VARIANT value containing the JSON document. This function is specifically designed for parsing strings that contain valid JSON information.

Data Transformations

Question

Options

Aparse_json()
Bjson_extract_path_text()
Cobject_construct()
Dflatten

How the community answered

(30 responses)

A
87% (26)
B
3% (1)
C
7% (2)
D
3% (1)

Explanation

The parse_json() function in Snowflake interprets an input string as a JSON document and produces a VARIANT value containing the JSON document. This function is specifically designed for parsing strings that contain valid JSON information.

Topics

#JSON#VARIANT data type#Semi-structured data#Data transformation

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