以下查询使用临时文件和filesort。如果可能,我想避免这种情况。有没有更好的索引来加速这个查询?
SELECT lib_name, description, count(seq_id), floor(avg(size))
FROM libraries l JOIN sequence s ON (l.lib_id=s.lib_id)
WHERE s.is_contig=0 and foreign_seqs=0 GROUP BY lib_name;
的EXPLAIN
说:
id,select_type,table,type,possible_keys,key,key_len,ref,rows,Extra
1,SIMPLE,s,ref,libseq,contigs,contigs,4,const,28447,Using temporary; Using filesort
1,SIMPLE,l,eq_ref,PRIMARY,PRIMARY,4,s.lib_id,1,Using where
的表是这样的:
库
CREATE TABLE `libraries` (
`lib_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`lib_name` varchar(30) NOT NULL,
`method_id` int(10) unsigned DEFAULT NULL,
`lib_efficiency` decimal(4,2) unsigned DEFAULT NULL,
`insert_avg` decimal(5,2) DEFAULT NULL,
`insert_high` decimal(5,2) DEFAULT NULL,
`insert_low` decimal(5,2) DEFAULT NULL,
`amtvector` decimal(4,2) unsigned DEFAULT NULL,
`description` text,
`foreign_seqs` tinyint(1) NOT NULL DEFAULT '0' COMMENT '1 means the sequences in this library are not ours',
PRIMARY KEY (`lib_id`),
UNIQUE KEY `lib_name` (`lib_name`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=latin1;
序列
CREATE TABLE `sequence` (
`seq_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`seq_name` varchar(40) NOT NULL DEFAULT '',
`lib_id` int(10) unsigned DEFAULT NULL,
`size` int(10) unsigned DEFAULT NULL,
`add_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`sequencing_date` date DEFAULT '0000-00-00',
`comment` text DEFAULT NULL,
`is_contig` int(10) unsigned NOT NULL DEFAULT '0',
`fasta_seq` longtext,
`primer` varchar(15) DEFAULT NULL,
`gc_count` int(10) DEFAULT NULL,
PRIMARY KEY (`seq_id`),
UNIQUE KEY `seq_name` (`seq_name`),
UNIQUE KEY `libseq` (`lib_id`,`seq_id`),
KEY `primer` (`primer`),
KEY `sgitnoc` (`seq_name`,`is_contig`),
KEY `contigs` (`is_contig`,`seq_name`) USING BTREE,
CONSTRAINT `FK_sequence_1` FOREIGN KEY (`lib_id`) REFERENCES `libraries` (`lib_id`)
) ENGINE=InnoDB AUTO_INCREMENT=61508 DEFAULT CHARSET=latin1 ROW_FORMAT=DYNAMIC;
我能做些什么来改变查询速度吗?如果不是,那么当(对于Web应用程序)将上述查询的结果放入MEMORY表中时值得吗?
感谢您提供非常明确和完整的答案。在这种情况下,Reindexing不起作用。 sort_buffer_size = 2M,我认为它在正常的高端 - 或者至少这是我的阅读建议。所以我打算用一个汇总表。 – dnagirl 2010-09-08 13:25:44