Examine this statement which executes successfully: The table is populated with a range of values including jobs for Robert, John, and Katie. Now, examine this statement and output: Why is an empty…

Question

Exhibit

Options

• AThe select requires json_unquoteo in the where clause.
• BThe virtual values in the name column must be accessed using functions.
• CThe JSON datatype cannot be used in virtual columns.
• DThe json_extract() function requires a length value that matches the field length in the schema.
• ETable statistics must be updated to generate values for the name column.

• A
  4% (1)
• B
  29% (8)
• C
  46% (13)
• D
  14% (4)
• E
  7% (2)

Explanation

Option C is correct because MySQL does not support the JSON datatype as the defined type for a virtual/generated column - when you attempt to use it, the column produces no usable values, resulting in an empty result set even though the underlying table contains data. Option A is wrong because json_unquote() removes surrounding quotes from extracted JSON strings but is not a requirement for a WHERE clause to execute. Option B is wrong because virtual column values behave like regular columns and are directly accessible without wrapping them in functions at query time. Option D is wrong because json_extract() takes a JSON document and a path expression - there is no length parameter involved. Option E is wrong because table statistics affect the query optimizer's execution plan, not whether column values are physically accessible or returned.

Memory tip: Think of it as JSON is for storing, not for typing virtual columns - you can extract JSON values into a virtual column, but the virtual column itself must be declared as a concrete scalar type like VARCHAR or TEXT, never JSON.

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