Question

我目前有一个给定表的前X个结果列表，给定的一天，其中'top'被定义为具有最高聚合计数的结果。我希望能够看到的不仅仅是那个特定日子的数量，还有它与昨天的数量以及上周数量的平均值的比较。

这是我当前的查询：

SELECT foo, bar, SUM(count) as today
FROM tablename
WHERE col1 = 'asdf' and date = '2012-08-23'
GROUP BY foo, bar
ORDER BY count desc
LIMIT 5

结果如下：

|foo        | bar        | today |
|something1 | something3 | 2345  |
|something2 | something4 | 1234  |
 ....

但我想拥有的是这样的：

|foo        | bar        | today | yesterday | week_avg |
|something1 | something3 | 2345  | 2273      | 2193     |
|something2 | something4 | 1234  | 935       | 1001     |
 ....

作为奖励积分，我不介意在某种GROUP_CONCAT结果中拥有过去20天的所有日期（因此我可以很容易地用它来制作迷你图）

Answer 1

您可以获取上周的所有行，然后使用条件聚合（通过CASE expressions）仅将当前日期的count值和昨天的count值相加：

SELECT   foo, 
         bar,
         SUM(CASE WHEN date = CURDATE() THEN count ELSE 0 END) AS today,
         SUM(CASE WHEN date = CURDATE() - INTERVAL 1 DAY THEN count ELSE 0 END) AS yesterday,
         AVG(count) AS week_avg
FROM     tablename
WHERE    col1  = 'asdf' AND
         date >= CURDATE() - INTERVAL 1 WEEK
GROUP BY foo,
         bar
ORDER BY today DESC
LIMIT    5

最后，我们按今天count值的总和排序，并根据该订单选择前五名。

Answer 2

一旦您准备了包含日期和日期金额的汇总表，并假设您正在寻找与昨天的差异，您可以使用COALESCE。下面的tbl表引用了您的摘要表，如下所示：

SELECT date, foo, bar, SUM(count) as today
FROM tablename
WHERE col1 = 'asdf'
GROUP BY date, foo, bar
ORDER BY count desc
LIMIT 5

查询

SELECT a.ddate, a.amount AS today, 
COALESCE(a.amount - 
  (SELECT b.amount 
   FROM tbl b 
   WHERE b.ddate = a.ddate - INTERVAL 1 DAY), a.amount) AS yesterday
FROM tbl a

请参阅fiddle。

Answer 3

你的第一个问题。试试这个：

 SELECT day0.foo, day0.bar, 
        day0.addup as today, 
        day1.addup as yesterday, 
        week0.addup/7 as week_avg
   FROM 
  ( 
    SELECT foo, bar, SUM(count) as addup
      FROM tablename
     WHERE col1 = 'asdf'
       AND date >= CURDATE()
       AND date < CURDATE() + INTERVAL 1 DAY
  GROUP BY foo, bar
  ) day0
  JOIN ( 
    SELECT foo, bar, SUM(count) as addup
      FROM tablename
     WHERE col1 = 'asdf'
       AND date >= CURDATE() - INTERVAL 1 DAY
       AND date < CURDATE()
  GROUP BY foo, bar
  ) day1 ON (day0.foo = day1.foo AND day0.bar=day1.bar)
  JOIN ( 
    SELECT foo, bar, SUM(count) as addup
      FROM tablename
     WHERE col1 = 'asdf'
       AND date >= CURDATE() - INTERVAL 7 DAY
       AND date < CURDATE()
  GROUP BY foo, bar
  ) week0 ON (day0.foo = week0.foo AND day0.bar = week0.bar)
 ORDER BY day0.addup desc
    LIMIT 5

它看起来很毛茸茸，但它是三个不同的摘要查询的连接，每天一个。

你的第二个问题：

SELECT foo, bar, GROUP_CONCAT(addup)
  FROM (
    SELECT foo, bar, 
           DATEDIFF(CURDATE(),date) AS daysago, 
           SUM(count) as addup
      FROM tablename
     WHERE col1 = 'asdf'
       AND date >= CURDATE() - INTERVAL 20 DAY
       AND date < CURDATE() )A
  GROUP BY foo, bar, date
  ORDER BY foo, bar, date

获取上周每个结果的计数摘要

3 个答案: