1

如何在 MySQL 中简化此代码？

SELECT name,
  MAX(IF(to_days(thedate) - to_days('2009-06-13') = 0, price, '')) AS date1,
  MAX(IF(to_days(thedate) - to_days('2009-06-13') = 1, price, '')) AS date2,
  MAX(IF(to_days(thedate) - to_days('2009-06-13') = 2, price, '')) AS date3,
  MAX(IF(to_days(thedate) - to_days('2009-06-13') = 3, price, '')) AS date4,
  MAX(IF(to_days(thedate) - to_days('2009-06-13') = 4, price, '')) AS date5,
  MAX(IF(to_days(thedate) - to_days('2009-06-13') = 5, price, '')) AS date6,
  MAX(IF(to_days(thedate) - to_days('2009-06-13') = 6, price, '')) AS date7,
AVG(price),SUM(price)
FROM `personals`
WHERE personal_id = '1234'
GROUP BY name

这样计算的日期数是动态的吗？

4

3 回答 3

1

最好的方法可能是将其拆分为两个查询，一个用于加载平均价格和每人价格的总和：

SELECT
  AVG(price), SUM(price)
FROM `personals`
WHERE personal_id = '1234'
GROUP BY name;

第二个是你想知道的最大值：

SELECT
  MAX(price)
FROM `personals`
WHERE personal_id = '1234'
GROUP BY name, to_days(thedate) - to_days('2009-06-13');

如果您真的希望所有列都在同一个查询中，请为第一个查询使用子查询（在大型数据库上可能效率不高）

SELECT
  MAX(price),
  AVG(price),
  SUM(price)
FROM `personals`
LEFT JOIN (
  SELECT
    AVG(price), SUM(price), name
  FROM `personals`
  WHERE personal_id = '1234' -- # this line is optional
  GROUP BY name
) totals
ON totals.name = personals.name
WHERE personal_id = '1234'
GROUP BY name, to_days(thedate) - to_days('2009-06-13');
于 2009-06-11T10:25:35.933 回答
0

您不能动态更改列数，但您可以很容易地为每个日期获取单独的行：

SELECT to_days(thedate) - to_days('2009-06-13') as interval,
   max(price) FROM `personals`
WHERE personal_id = '1234'
GROUP BY name, thedate

你需要做一个单独的查询来获取平均数据：

SELECT name,
   AVG(price),SUM(price)
FROM `personals`
WHERE personal_id = '1234'
GROUP BY name
于 2009-06-11T09:36:08.357 回答
0

如果我对您的理解正确，那么我认为您不能做您想做的事：SQL 不能那样工作。如果您的查询正在从另一个程序中使用，并且您知道您感兴趣的天数集，那么您可以在提交之前将其构建为字符串。如果您想要所有可能的天数并且必须有一个 SQL 查询，那么我想 UNION 会这样做（尽管我更喜欢两个查询）：

SELECT
  name
, to_days(thedate) - to_days('2009-06-13') AS num_days
, MAX(price) As max_price
, NULL AS avg_price
, NULL AS sum_price
FROM `personals`
WHERE personal_id = '1234'
GROUP BY
  name
, to_days(thedate) - to_days('2009-06-13') 
UNION ALL
SELECT
  name
, NULL
, NULL
, AVG(price)
, SUM(price)
FROM `personals`
WHERE personal_id = '1234'
GROUP BY name
于 2009-06-11T09:39:18.180 回答