如何选择一天中每小时的最新记录

Question

我想根据日期时间列'reading_on'选择给定日期中每个小时的最新记录。我执行了下面的查询

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, 
date_format(a.reading_on,'%j-%Y-%k') 
FROM 
    InverterReadings a 
LEFT JOIN 
    InverterReadings b 
ON 
     date_format(a.reading_on,'%j-%Y-%k') = date_format(b.reading_on,'%j-%Y-%k') 
AND 
    a.reading_on < b.reading_on 
WHERE 
    b.reading_on is null 
group by a.reading_on; 

Answer 2

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