https://docs.snowflake.com/en/sql-reference/functions/to_variant
"The TO_VARIANT function cannot be used directly in an INSERT statement. Instead, use INSERT INTO ... SELECT....
Hence, ANSWER B is not correct.
As per the right answer, or A, or D. I am saying D, only because I have to select an answer. Please, see below syntax, for JSON:
INSERT INTO to_variant_example (v_varchar, v_number, v_timestamp, v_array, v_object)
SELECT
TO_VARIANT('Skiing is fun!'),
TO_VARIANT(3.14),
TO_VARIANT('2024-01-25 01:02:03'),
TO_VARIANT(ARRAY_CONSTRUCT('San Mateo', 'Seattle', 'Berlin')),
PARSE_JSON(' { "key1": "value1", "key2": "value2" } ');
SELECT * FROM to_variant_example; Not very confident that this will help here.
A voting comment increases the vote count for the chosen answer by one.
Upvoting a comment with a selected answer will also increase the vote count towards that answer by one.
So if you see a comment that you already agree with, you can upvote it instead of posting a new comment.
37ceea2
2 months, 2 weeks ago37ceea2
2 months, 2 weeks ago