DEA-C02 Data Transformation Practice Question
A data engineer is working with a table that contains a column of type VARCHAR storing JSON strings. The engineer needs to transform this column into a VARIANT type to leverage Snowflake's semi-structured data functions. Which function should be used to convert the VARCHAR column to VARIANT?
⚠ Common exam trap
The trap here is assuming that TO_VARIANT or CAST can parse JSON; they only convert the string as a scalar, not as structured JSON.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
PARSE_JSON
PARSE_JSON is the dedicated function for converting a JSON-formatted string into a VARIANT. It parses the string and creates a structured object that can be queried using Snowflake's semi-structured data functions. This transformation is essential for enabling access to nested JSON elements and is the recommended approach for converting VARCHAR JSON to VARIANT.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
CAST
Why it's wrong here
CAST can convert between data types, but casting a VARCHAR to VARIANT does not parse the JSON; it simply wraps the string as a VARIANT scalar. The result would not be a structured JSON object, and semi-structured functions would not work as intended. CAST is not suitable for JSON parsing.
- ✗
TO_VARIANT
Why it's wrong here
TO_VARIANT converts a value to VARIANT, but it does not parse JSON strings. If applied to a VARCHAR containing JSON, it would treat the entire string as a scalar string within a VARIANT, not as a structured JSON object. Thus, it does not achieve the desired parsing.
- ✗
TRY_PARSE_JSON
Why it's wrong here
TRY_PARSE_JSON is similar to PARSE_JSON but returns NULL if the input is not valid JSON, instead of raising an error. While it is useful for error handling, the scenario does not indicate invalid JSON, so PARSE_JSON is the standard choice. TRY_PARSE_JSON is a variant but not the primary function for this task.
- ✓
PARSE_JSON
Why this is correct
PARSE_JSON interprets a string as a JSON document and returns a VARIANT. It is specifically designed to convert JSON-formatted strings into Snowflake's semi-structured VARIANT type, allowing access to nested fields using colon notation. This is the correct function for this transformation.
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Written and reviewed by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
Last reviewed September 2026 · checked against the official Snowflake exam blueprint
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