按年度最快的选择方式？

Question

根据this comment，至少有8种不同的方式按年和月选择记录：

1. 其中日期类似于＆＃39; 2005-01 - ％＆＃39;
2. 其中DATE_FORMAT（日期，＆＃39;％Y-％m＆＃39;）=＆＃39; 2005-01＆＃39;
3. 其中EXTRACT（YEAR_MONTH FROM date）=＆＃39; 200501＆＃39;
4. 其中YEAR（日期）=＆＃39; 2005＆＃39;和MONTH（日期）=＆＃39; 1＆＃39;
5. 其中substring（date，1,7）=＆＃39; 2005-01＆＃39;
6. 其中日期介于2005-01-01＆＃39;和＆＃39; 2005-01-31＆＃39;
7. 其中日期＆gt; =＆＃39; 2005-01-01＆＃39;和日期＆lt; =＆＃39; 2005-01-31＆＃39;
8. 其中日期为IN（＆＃39; 2005-01-01＆＃39;，＆＃39; 2005-01-02＆＃39;，＆＃39; 2005-01-03＆＃39;，＆＃39; 2005-01-04＆＃39;，＆＃39; 2005-01-05＆＃39;，＆＃39; 2005-01-06＆＃39;，＆＃39; 2005-01-07＆＃39;，＆＃ 39; 2005-01-08＆＃39;，＆＃39; 2005-01-09＆＃39;，＆＃39; 2005-01-10＆＃39;，＆＃39; 2005-01-11＆＃39;， ＆＃39; 2005-01-12＆＃39;，＆＃39; 2005-01-13＆＃39;，＆＃39; 2005-01-14＆＃39;，＆＃39; 2005-01-15＆＃39 ;，＆＃39; 2005-01-16＆＃39;，＆＃39; 2005-01-17＆＃39;，＆＃39; 2005-01-18＆＃39;，＆＃39; 2005-01-19＆ ＃39;，＆＃39; 2005-01-20＆＃39;，＆＃39; 2005-01-21＆＃39;，＆＃39; 2005-01-22＆＃39;，＆＃39; 2005-01 -23＆＃39;，＆＃39; 2005-01-24＆＃39;，＆＃39; 2005-01-25＆＃39;，＆＃39; 2005-01-26＆＃39;，＆＃39; 2005 -01-27＆＃39;，＆＃39; 2005-01-28＆＃39;，＆＃39; 2005-01-29＆＃39;，＆＃39; 2005-01-30＆＃39;，＆＃39 ; 2005-01-31＆＃39）

哪个是 WHERE 子句中最快的？哪个 GROUP BY 最快？他们中的任何一个都可以利用索引吗？

我的特定查询如下：

select count(*) 
from quotes quote
where EXTRACT(YEAR_MONTH FROM qtime) = :date

qtime是TIMESTAMP。

但我还需要获得最多报价的月份：

select date_format(qtime,'%Y-%m') d,count(*) c
from quotes
group by EXTRACT(YEAR_MONTH FROM qtime)
order by c desc, qid asc limit 1;

在这种情况下，我无法使用比较运算符。

Answer 1

我分析了所有这些查询并得到了结果。

between => Showing rows 0 - 29 (158 total, Query took 0.0030 sec)

like => Showing rows 0 - 29 (167 total, Query took 0.0031 sec)

DATE_FORMAT => Showing rows 0 - 29 (167 total, Query took 0.0237 sec)

EXTRACT => Showing rows 0 - 29 (167 total, Query took 0.0031 sec)

YEAR, MONTH => Showing rows 0 - 29 (167 total, Query took 0.0187 sec)

substr => Showing rows 0 - 29 (167 total, Query took 0.0224 sec)

> < => Showing rows 0 - 29 (158 total, Query took 0.0030 sec)

所以我可以说中间的使用是最好的，而且速度更快。虽然使用大于和小于是相同的结果，但我认为介于两者之间非常聪明。