【发布时间】：2014-05-17 14:02:33
【问题描述】：
我有一张简单的桌子
ID usr_id card_amount state created
----------------------------------------------------------------
1 a1 10.000 2 2014-03-13 14:33:39
2 a2 30.000 2 2014-03-11 14:33:39
3 a3 50.000 1 2014-03-10 14:33:39
4 a4 20.000 2 2014-04-13 14:33:39
5 a5 40.000 2 2014-03-19 14:33:39
----------------------------------------------------------------
我有这个查询，但它需要很长时间！帮我解决一下，让这更快
SELECT
DATE_FORMAT(`created`, '%Y-%m-%d') AS `key_date`,
(
SELECT
SUM(`card_amount`)
FROM `payment`
WHERE `card_amount`> 0 AND `state` = 2 AND `created` >= `key_date` AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `totalamount`,
(
SELECT
COUNT(`id`)
FROM `payment`
WHERE `card_amount`> 0 AND `state` =2 AND `created` >= `key_date` AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `turncharge`,
(
SELECT
COUNT(DISTINCT `usr_id`)
FROM `payment`
WHERE `card_amount`> 0 AND `state` =2 AND `created` >= `key_date` AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `pu`,
(
SELECT
SUM(`card_amount`)
FROM `payment`
WHERE `card_amount`> 0 AND `state` = 2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 6 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `totalamount7`,
(
SELECT
COUNT(`id`)
FROM `payment` WHERE `card_amount`> 0 AND `state` =2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 6 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `turncharge7`,
(
SELECT
COUNT(DISTINCT `usr_id`)
FROM `payment`
WHERE `card_amount`> 0 AND `state` =2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 6 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `pu7`,
(
SELECT
SUM(`card_amount`)
FROM `payment` WHERE `card_amount`> 0 AND `state` = 2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 29 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `totalamount30`,
(
SELECT
COUNT(`id`)
FROM `payment` WHERE `card_amount`> 0 AND `state` =2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 29 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `turncharge30`,
(
SELECT
COUNT(DISTINCT `usr_id`)
FROM `payment`
WHERE `card_amount`> 0 AND `state` =2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 29 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY)
) AS `pu30`
FROM `payment`
WHERE MONTH(`created`)=$month AND YEAR(`created`)=$year
GROUP BY `key_date`
ORDER BY `key_date` DESC
我使用 LEFT JOIN 但它没有给出我需要的东西：c1.totalamount = c2 totalmount7, c1.turncharge =c2.turncharge7 ........（不对）
SELECT DATE_FORMAT(`created`, '%Y-%m-%d') AS `key_date`,c1.totalamount,c1.turncharge,c1.pu,c2.totalamount7,c2.turncharge7,c2.pu7
FROM `payment`
LEFT JOIN (SELECT DATE_FORMAT(`created`, '%Y-%m-%d') AS `date1`,SUM(`card_amount`) AS `totalamount`, COUNT(`id`) AS `turncharge`,COUNT(DISTINCT `usr_id`) AS `pu`
FROM `payment`
WHERE `card_amount`> 0 AND `state` = 2 AND `created` >= DATE_FORMAT(`created`, '%Y-%m-%d')
AND `created` < DATE_SUB(DATE_FORMAT(`created`, '%Y-%m-%d'),INTERVAL -1 DAY) GROUP BY date1) AS c1 ON date1 = DATE_FORMAT(`created`, '%Y-%m-%d')
LEFT JOIN (SELECT DATE_FORMAT(`created`, '%Y-%m-%d') AS `date2`,SUM(`card_amount`) AS `totalamount7`, COUNT(`id`) AS `turnchargep7`,COUNT(DISTINCT `usr_id`) AS `pu7`
FROM `payment`
WHERE `card_amount`> 0 AND `state` = 2 AND created >= DATE_SUB(DATE_FORMAT(`created`, '%Y-%m-%d'),INTERVAL 6 DAY)
AND `created` < DATE_SUB(DATE_FORMAT(`created`, '%Y-%m-%d'),INTERVAL -1 DAY) GROUP BY date2) AS c2 ON date2 = DATE_FORMAT(`created`, '%Y-%m-%d')
WHERE MONTH(`created`)=4 AND YEAR(`created`)=2014
GROUP BY `key_date`
ORDER BY `key_date` DESC
感谢 Mladen Prajdic 我使用 CASE 表达式，它工作得更快
SELECT date.key_date,
SUM(case when `card_amount`> 0 AND `state` = 2 AND `created` >= `key_date` AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY) then `card_amount` else 0 end) AS totalamount,
COUNT(DISTINCT(case when `card_amount`> 0 AND `state` =2 AND `created` >= date.`key_date` AND `created` < DATE_SUB(date.`key_date`,INTERVAL -1 DAY) then payment.`id` else 0 end)) AS `turncharge`,
COUNT(DISTINCT(case when `card_amount`> 0 AND `state` =2 AND `created` >= `key_date` AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY) then `usr_id` else 0 end) )AS `pu` ,
SUM(case when `card_amount`> 0 AND `state` =2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 6 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY) then `card_amount` else 0 end) AS `totalamount7`,
COUNT(DISTINCT(case when `card_amount`> 0 AND `state` =2 AND `created` >= DATE_SUB(date.`key_date`,INTERVAL 6 DAY) AND `created` < DATE_SUB(date.`key_date`,INTERVAL -1 DAY) then payment.`id` else 0 end)) AS `turncharge7`,
COUNT(DISTINCT(case when `card_amount`> 0 AND `state` = 2 AND `created` >= DATE_SUB(date.`key_date`,INTERVAL 6 DAY) AND `created` < DATE_SUB(date.`key_date`,INTERVAL -1 DAY) then `usr_id` else 0 end) )AS `pu7` ,
SUM(case when `card_amount`> 0 AND `state` = 2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 29 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY) then `card_amount` else 0 end) AS `totalamount30`,
COUNT(DISTINCT(case when `card_amount`> 0 AND `state` = 2 AND `created` >= DATE_SUB(date.`key_date`,INTERVAL 29 DAY) AND `created` < DATE_SUB(date.`key_date`,INTERVAL -1 DAY) then payment.`id` else 0 end)) AS `turncharge30`,
COUNT(DISTINCT(case when `card_amount`> 0 AND `state` = 2 AND `created` >= DATE_SUB(`key_date`,INTERVAL 29 DAY) AND `created` < DATE_SUB(`key_date`,INTERVAL -1 DAY) then `usr_id` else 0 end) )AS `pu30`
FROM `payment`,(SELECT DATE_FORMAT(`created`, '%Y-%m-%d') AS `key_date` FROM `payment`
WHERE MONTH(`created`)=3 AND YEAR(`created`)=2014
GROUP BY `key_date`) AS date
GROUP BY date.key_date
ORDER BY date.key_date DESC
【问题讨论】：
-
当您进行这样的子选择时，它将永远花费很长时间，因为对于每个子选择，它必须与表中的每一行进行比较。因此，例如，如果您只有 1000 行付款，那么只需进行选择，就必须将每一行与 1000 行进行比较……进行 9000 次比较……甚至不考虑您在其中所做的所有其他事情。我建议为此查看多个查询，或进行联接。
标签： mysql sql database select where-clause