Oracle Database MySQL 5.6 Developer — Question 30

Consider the statement:
CREATE TABLE t1 (a INT) PARTITION BY KEY
/*150611 ALGORITHM = 1*/
What does this statement do?

Answer options

• A. Create the t1 table partitioned by KEY with the default algorithm in all versions.
• B. Create the t1 table partitioned by KEY using algorithm 1 only in MYSQL version 5.6.11 and the default algorithm in other versions.
• C. Create the t1 table partitioned by KEY using algorithm 1 only in MYSQL versions 5.6.11 or newer and the default algorithm in older versions.
• D. Create the t1 table partitioned by KEY using algorithm 1 only if the preceding statement returned error condition 50611.

Correct answer: C

Explanation

The correct answer is C because it accurately reflects that algorithm 1 is used for MYSQL versions 5.6.11 and later, while the default algorithm applies to earlier versions. Option A is incorrect as it suggests the default algorithm is used in all versions, which is not true. Option B limits algorithm 1's usage to only version 5.6.11, ignoring newer versions, and D introduces an irrelevant error condition that does not apply to the statement.