Question

是否有可能以某种方式存储值，然后在需要时使用它。也许您对以下问题有任何其他解决方案。我有每小时的水读数，我想把它作为一天消费，但有时可能会丢失一些读数。

SELECT Readings.dte AS day,
       Ifnull(NextReadings.reading, (SELECT Max(consumption)
                                     FROM   meter_readings)) - Readings.reading AS Consumption
FROM   (SELECT Date(readdate)   AS dte,
               Min(consumption) AS Reading
        FROM   meter_readings
        GROUP  BY Date(readdate)) AS Readings
     LEFT JOIN 
       (SELECT Date(readdate)   AS dte,
               Min(consumption) AS Reading
        FROM   meter_readings
        GROUP  BY Date(readdate)) AS NextReadings
     ON NextReadings.dte = date_add(Readings.dte, INTERVAL +1 DAY)

如果有

...
insert into meter_readings values ('2013-07-30 00:00:31',    143.860);
insert into meter_readings values ('2013-07-30 01:00:32',    143.870);
insert into meter_readings values ('2013-07-30 02:00:31',    143.870);
insert into meter_readings values ('2013-07-30 03:00:32',    143.870);
insert into meter_readings values ('2013-07-30 04:00:31',    143.870);

insert into meter_readings values ('2013-07-31 02:00:31',    143.890);

insert into meter_readings values ('2013-08-03 00:00:31',    143.900);
insert into meter_readings values ('2013-08-03 01:00:32',    143.920);
insert into meter_readings values ('2013-08-03 02:00:31',    143.920);

http://sqlfiddle.com/#!2/dd66cd/2＆lt; - 例如

我可以看到：

July, 30 2013   0.03
July, 31 2013   0.03 
August, 03 2013 0.02

2013-07-31的使用情况为0.03，由于SELECT Max(consumption) FROM meter_readings)而不是这样。以某种方式存储最后一个正确的值（143.870）然后使用而不是这个Max（消耗）将是很好的。 2013-07-31的用法应为：143.890 - Variable，其中变量为143.870

Answer 1

我想你想拥有每一天的最后阅读。使用此语法来获取它：

SELECT M1.*

FROM meter_readings M1
LEFT JOIN meter_readings M2
ON Date(M1.ReadDate) = Date(M2.ReadDate)
AND M1.ReadDate < M2.ReadDate
WHERE M2.ReadDate IS NULL

或者每天读一读：

SELECT M1.*

FROM meter_readings M1
LEFT JOIN meter_readings M2
ON Date(M1.ReadDate) = Date(M2.ReadDate)
AND M1.ReadDate > M2.ReadDate
WHERE M2.ReadDate IS NULL

并获得当天的使用。我不太使用MySql。所以也许排名函数可以更好地设计。

SET @curRank1 = 0;
SET @curRank2 = 0;

SELECT
D1.ReadDate
,D1.Consumption AS D1Consumption
,D2.Consumption AS D2Consumption
,(D2.Consumption - D1.Consumption) AS UsageOfDay
FROM
(
  SELECT M1.*, @curRank1 := @curRank1 + 1 AS rank

FROM meter_readings M1
LEFT JOIN meter_readings M2
ON Date(M1.ReadDate) = Date(M2.ReadDate)
AND M1.ReadDate < M2.ReadDate
WHERE M2.ReadDate IS NULL
  ) D2
LEFT JOIN
(
  SELECT M1.*, @curRank2 := @curRank2 + 1 AS rank

FROM meter_readings M1
LEFT JOIN meter_readings M2
ON Date(M1.ReadDate) = Date(M2.ReadDate)
AND M1.ReadDate < M2.ReadDate
WHERE M2.ReadDate IS NULL
  ) D1
ON D2.rank = (D1.rank + 1)

Answer 2

根据Wietze314的代码，我找到了解决方案。它有效，但它很重

SET @curRank1 = 0;
SET @curRank2 = 0;

SELECT
D1.ReadDate
,D1.Consumption AS D1Consumption
,D2.Consumption AS D2Consumption
,(D2.Consumption - D1.Consumption) AS UsageOfDay
FROM
(
  SELECT M1.*, @curRank1 := @curRank1 + 1 AS rank

FROM meter_readings M1
LEFT JOIN meter_readings M2
ON Date(M1.ReadDate) = Date(M2.ReadDate)
AND M1.ReadDate > M2.ReadDate
WHERE M2.ReadDate IS NULL
  ) D2
LEFT JOIN
(
  SELECT M1.*, @curRank2 := @curRank2 + 1 AS rank

FROM meter_readings M1
LEFT JOIN meter_readings M2
ON Date(M1.ReadDate) = Date(M2.ReadDate)
AND M1.ReadDate > M2.ReadDate
WHERE M2.ReadDate IS NULL
  ) D1
ON D2.rank = (D1.rank + 1)
WHERE D1.ReadDate IS NOT NULL
UNION ALL
(
SELECT DISTINCT
 Date(My.ReadDate),
  MIN(My.Consumption) AS D1Consumption,
  MAX(My.Consumption)AS D2Consumption,
  (MAX(My.Consumption) - MIN(My.Consumption)) AS UsageOfDay
  FROM
  meter_readings MY
GROUP  BY Date(My.ReadDate) DESC
LIMIT 1
  )

SQL存储值，每日读数

2 个答案: