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If creating a secondary index fails (typically, ADD UNIQUE INDEX fails due to duplicate key), it is possible that concurrently running UPDATE or DELETE will access the index stub and hit the debug assertion. It does not make any sense to keep updating an uncommitted index whose creation has failed. dict_index_t::is_corrupted(): Replaces dict_index_is_corrupted(). Also take online_status into account. Replace some calls to dict_index_is_clust() with calls to dict_index_t::is_primary().
27 lines
720 B
Text
27 lines
720 B
Text
CREATE TABLE t1 (a INT, b VARCHAR(10)) ENGINE=InnoDB
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PARTITION BY RANGE(a)
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(PARTITION pa VALUES LESS THAN (3),
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PARTITION pb VALUES LESS THAN (5));
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INSERT INTO t1 VALUES(2,'two'),(2,'two'),(4,'four');
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connect ddl,localhost,root,,test;
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SET DEBUG_SYNC = 'inplace_after_index_build SIGNAL go WAIT_FOR done';
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ALTER TABLE t1 ADD UNIQUE KEY (a,b(3));
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connection default;
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SET DEBUG_SYNC = 'now WAIT_FOR go';
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BEGIN;
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SELECT * FROM t1 FOR UPDATE;
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a b
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2 two
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2 two
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4 four
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SET DEBUG_SYNC = 'now SIGNAL done';
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connection ddl;
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ERROR 23000: Duplicate entry '2-two' for key 'a'
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connection default;
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DELETE FROM t1;
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disconnect ddl;
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SET DEBUG_SYNC = 'RESET';
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CHECK TABLE t1;
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Table Op Msg_type Msg_text
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test.t1 check status OK
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DROP TABLE t1;
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