mysql - 如何选取一天中每个小时的最新记录

Question

我想根据日期时间列“reading_on”选择给定日期每小时的最新记录。我已经执行了以下查询

hourly_max =   InverterReading
               .where("DATE(reading_on) = ? AND imei = ?", Date.today, "770000000000126")
               .group("HOUR(reading_on)")
               .having("max(HOUR(reading_on))")

hourly_max.group_by(&:id).each { |k,v| puts v.last.reading_on }

在上面的查询中，我没有得到所需的结果。选择一天中每个小时的最新记录的正确方法是什么。下面是表结构

Answer 1

SELECT 
   HOUR(a.reading_on) As hr, max(a.id),a.reading_on
FROM 
   InverterReadings a
LEFT JOIN
   InverterReadings b
ON
        YEAR(a.reading_on)=YEAR(b.reading_on)
        AND MONTH(a.reading_on)=MONTH(b.reading_on)
        AND day(a.reading_on)=day(b.reading_on)
        AND hour(a.reading_on)=hour(b.reading_on)
AND 
    a.reading_on < b.reading_on
WHERE 
    b.reading_on is null
group by a.reading_on;

演示:http://sqlfiddle.com/#!2/49a69/14