如何选择一天中每小时的最新记录 [英] How to select the latest record of each hour in a day

问题描述

我想基于datetime列"reading_on"选择给定日期中每个小时的最新记录.我已经执行了以下查询

I want to select the latest record of each hour in a given date, based on the datetime column 'reading_on'. I have executed the below query

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

In the above query I am not getting the required result. What is the proper way to select the latest record of each hour in a day. Below is the table structure

推荐答案

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;

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