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MySQL key_len 大小的计算

时间:2014-06-13 18:51:16      阅读:421      评论:0      收藏:0      [点我收藏+]

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背景:

      当用Explain查看SQL的执行计划时,里面有列显示了 key_len 的值,根据这个值可以判断索引的长度,在组合索引里面可以更清楚的了解到了哪部分字段使用到了索引。

环境:

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CREATE TABLE `tmp_0612` (
  `id` int(11) NOT NULL,
  `name` varchar(10) DEFAULT NULL,
  `age` int(11) DEFAULT NULL,
  `address` varchar(100) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8
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插入数据:

insert into tmp_0612 values(1,a,11,hz),(2,b,22,gz),(3,c,33,aa);

创建索引:

alter table tmp_0612 add index idx_name(name);
alter table tmp_0612 add index idx_age(age);

测试:

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explain select * from tmp_0612 where name =a;
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
| id | select_type | table    | type | possible_keys | key      | key_len | ref   | rows | Extra                 |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_name      | idx_name | 33      | const |    1 | Using index condition |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
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从上面的执行计划可知,索引的长度是33。比预想的30(10*3(utf8))要高出3字节,为什么呢?进一步测试:

修改name 成 not null

alter table tmp_0612 modify name varchar(10) not null;

再看执行计划:

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explain select * from tmp_0612 where name =a;
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
| id | select_type | table    | type | possible_keys | key      | key_len | ref   | rows | Extra                 |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_name      | idx_name | 32      | const |    1 | Using index condition |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
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发现上面的执行计划和第一次的有区别(key_len),经过多次测试,发现字段允许NULL的会多出一个字节。想到了之前的一篇文章,NULL是需要一个标志位的,占用1个字符。那还有2个字节怎么算?这里想到的是会不会和 多字节字符集 相关?那改字符集试试:

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alter table tmp_0612 convert to charset latin1;

explain select * from tmp_0612 where name =a;
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
| id | select_type | table    | type | possible_keys | key      | key_len | ref   | rows | Extra                 |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_name      | idx_name | 12      | const |    1 | Using index condition |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
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发现还是多了2个字节,看来和多字节字符集没有关系了。那会不会和 变长字段 有关系?再试试:

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alter table tmp_0612 convert to charset utf8;

alter table tmp_0612 modify name char(10) not null;

explain select * from tmp_0612 where name =a;
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
| id | select_type | table    | type | possible_keys | key      | key_len | ref   | rows | Extra                 |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_name      | idx_name | 30      | const |    1 | Using index condition |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
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和预料的一样了,是30=10*3。到这里相信大家都已经很清楚了,要是还比较模糊就看反推到33字节。

改成允许NULL,应该会变成31。

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alter table tmp_0612 modify name char(10);

explain select * from tmp_0612 where name =a;
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
| id | select_type | table    | type | possible_keys | key      | key_len | ref   | rows | Extra                 |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_name      | idx_name | 31      | const |    1 | Using index condition |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
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改成变长字段类型,应该会变成33。

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alter table tmp_0612 modify name varchar(10);

explain select * from tmp_0612 where name =a;
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
| id | select_type | table    | type | possible_keys | key      | key_len | ref   | rows | Extra                 |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_name      | idx_name | 33      | const |    1 | Using index condition |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
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改成单字节字符集,还是还是需要额外的3字节(1:null ;变长字段:2),和字符集无关。

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alter table tmp_0612 convert to charset latin1;

explain select * from tmp_0612 where name =a;
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
| id | select_type | table    | type | possible_keys | key      | key_len | ref   | rows | Extra                 |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_name      | idx_name | 13      | const |    1 | Using index condition |
+----+-------------+----------+------+---------------+----------+---------+-------+------+-----------------------+
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反推上去都和预测的一样。

其他测试:

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explain select * from tmp_0612 where age = 11;
+----+-------------+----------+------+---------------+---------+---------+-------+------+-------+
| id | select_type | table    | type | possible_keys | key     | key_len | ref   | rows | Extra |
+----+-------------+----------+------+---------------+---------+---------+-------+------+-------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_age       | idx_age | 5       | const |    1 | NULL  |
+----+-------------+----------+------+---------------+---------+---------+-------+------+-------+

alter table tmp_0612 modify age int not null;

explain select * from tmp_0612 where age = 11;
+----+-------------+----------+------+---------------+---------+---------+-------+------+-------+
| id | select_type | table    | type | possible_keys | key     | key_len | ref   | rows | Extra |
+----+-------------+----------+------+---------------+---------+---------+-------+------+-------+
|  1 | SIMPLE      | tmp_0612 | ref  | idx_age       | idx_age | 4       | const |    1 | NULL  |
+----+-------------+----------+------+---------------+---------+---------+-------+------+-------+
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int 是占4个字符的,上面key_len的也是符合预期。关于组合索引的,可以自己去测试玩。

总结:

      变长字段需要额外的2个字节,固定长度字段不需要额外的字节。而null都需要1个字节的额外空间,所以以前有个说法:索引字段最好不要为NULL,因为NULL让统计更加复杂,并且需要额外一个字节的存储空间。这个结论在此得到了证实。

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key_len的长度计算公式:

varchr(10)变长字段且允许NULL      : 10*(Character Set:utf8=3,gbk=2,latin1=1)+1(NULL)+2(变长字段)
varchr(10)变长字段且不允许NULL    : 10*(Character Set:utf8=3,gbk=2,latin1=1)+2(变长字段)

char(10)固定字段且允许NULL        : 10*(Character Set:utf8=3,gbk=2,latin1=1)+1(NULL)
char(10)固定字段且允许NULL        : 10*(Character Set:utf8=3,gbk=2,latin1=1)
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MySQL key_len 大小的计算,布布扣,bubuko.com

MySQL key_len 大小的计算

标签:style   class   blog   code   java   http   

原文地址:http://www.cnblogs.com/zhoujinyi/p/3784450.html

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