MySQL查询以获取每日差异值 [英] MySQL query to get daily differential values

问题描述

我想制作一个 MySQL 从一个看起来像这样的表中获取每日差异值:

I want to make a MySQL to get daily differential values from a table who looks like this:

Date               | VALUE
--------------------------------
"2011-01-14 19:30" |   5
"2011-01-15 13:30" |   6
"2011-01-15 23:50" |   9
"2011-01-16 9:30"  |   10
"2011-01-16 18:30" |   15

我做了两个子查询.第一个是获取最后一个每日值，因为我想从这些数据中计算差值:

I have made two subqueries. The first one is to get the last daily value, because I want to compute the difference values from this data:

SELECT r.Date, r.VALUE
    FROM table AS r
    JOIN (
    SELECT DISTINCT max(t.Date) AS Date
    FROM table AS t
    WHERE t.Date < CURDATE() 
    GROUP BY DATE(t.Date) 
    ) AS x USING (Date)

第二个是为了从第一个结果中获取差分值(我用表"名称显示):

The second one is made to get the differential values from the result of the first one (I show it with "table" name):

SELECT Date, VALUE - IFNULL( 
    (SELECT MAX( VALUE ) 
    FROM table
    WHERE Date < t1.table) , 0) AS diff
    FROM table AS t1
    ORDER BY Date

起初，我试图将第一次查询的结果保存在一个临时表中，但是 第二个查询不能使用临时表.如果我在 () 之间的第二个查询的 FROM 中使用第一个查询和别名，则服务器对表别名的投诉不存在.如何得到这样的东西:

At first, I tried to save the result of first query in a temporary table but it's not possible to use temporary tables with the second query. If I use the first query inside the FROM of second one between () with an alias, the server complaints about table alias doesn't exist. How can get a something like this:

 Date               | VALUE
 ---------------------------
 "2011-01-15 00:00" |   4
 "2011-01-16 00:00" |   6

推荐答案

试试这个查询 -

SELECT
  t1.dt AS date,
  t1.value - t2.value AS value
FROM
  (SELECT DATE(date) dt, MAX(value) value FROM table GROUP BY dt) t1
JOIN
  (SELECT DATE(date) dt, MAX(value) value FROM table GROUP BY dt) t2
    ON t1.dt = t2.dt + INTERVAL 1 DAY

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