mysql - 查找每年每天的平均交易数量

Question

我有一个 epos 系统的订单表。

我有几年的数据，我想了解每年一周中每天的平均交易数。即 2014 年与 2013 年等相比，哪一天是一周中最繁忙的一天。

我可以这样获取信息:

SELECT ROUND(AVG(sales),1) AS dayaverage, oday, oyear FROM (
SELECT COUNT(oid) AS sales, DAYNAME(odate) AS oday , YEAR(odate) AS oyear
FROM orders 
WHERE ostatus = 'completed'
GROUP BY DATE(odate)
ORDER BY YEAR(odate), DAYOFWEEK(odate)) t1
GROUP BY oday, oyear
ORDER BY oyear DESC, AVG(sales) DESC

这会产生所需的数据，但格式不正确。

我明白了:

"dayaverage"    "oday"      "oyear"
"28.9"          "Saturday"  "2015"
"17.1"          "Sunday"    "2015"
"15.0"          "Tuesday"   "2015"
"14.3"          "Monday"    "2015"
"13.1"          "Wednesday" "2015"
"13.0"          "Friday"    "2015"

我创建了这个声明:

SELECT tsun.pyear, sunday,monday,tuesday, wednesday, thursday, friday, saturday FROM (
SELECT ROUND(AVG(payments),1) AS Sunday, pyear FROM (
SELECT COUNT(paymentid) AS payments, YEAR(DATETIME) AS pyear
FROM payments 
WHERE DAYNAME(DATETIME) = "Sunday"
GROUP BY DATE(DATETIME)
ORDER BY YEAR(DATETIME), DAYOFWEEK(DATETIME)) t1
GROUP BY pyear
ORDER BY pyear DESC, AVG(payments) DESC) AS tsun
LEFT JOIN (SELECT * FROM (SELECT ROUND(AVG(payments),1) AS Monday, pyear FROM (
SELECT COUNT(paymentid) AS payments, YEAR(DATETIME) AS pyear
FROM payments 
WHERE DAYNAME(DATETIME) = "Monday"
GROUP BY DATE(DATETIME)
ORDER BY YEAR(DATETIME), DAYOFWEEK(DATETIME)) t2
GROUP BY pyear
ORDER BY pyear DESC, AVG(payments) DESC) AS tmon) tmon2 ON tsun.pyear = tmon2.pyear
LEFT JOIN (SELECT * FROM (SELECT ROUND(AVG(payments),1) AS Tuesday, pyear FROM (
SELECT COUNT(paymentid) AS payments, YEAR(DATETIME) AS pyear
FROM payments 
WHERE DAYNAME(DATETIME) = "Tuesday"
GROUP BY DATE(DATETIME)
ORDER BY YEAR(DATETIME), DAYOFWEEK(DATETIME)) t2
GROUP BY pyear
ORDER BY pyear DESC, AVG(payments) DESC) AS ttue) ttue2 ON tsun.pyear = ttue2.pyear
LEFT JOIN (SELECT * FROM (SELECT ROUND(AVG(payments),1) AS Wednesday, pyear FROM (
SELECT COUNT(paymentid) AS payments, YEAR(DATETIME) AS pyear
FROM payments 
WHERE DAYNAME(DATETIME) = "Wednesday"
GROUP BY DATE(DATETIME)
ORDER BY YEAR(DATETIME), DAYOFWEEK(DATETIME)) t2
GROUP BY pyear
ORDER BY pyear DESC, AVG(payments) DESC) AS twed) twed2 ON tsun.pyear = twed2.pyear
LEFT JOIN (SELECT * FROM (SELECT ROUND(AVG(payments),1) AS Thursday, pyear FROM (
SELECT COUNT(paymentid) AS payments, YEAR(DATETIME) AS pyear
FROM payments 
WHERE DAYNAME(DATETIME) = "Thursday"
GROUP BY DATE(DATETIME)
ORDER BY YEAR(DATETIME), DAYOFWEEK(DATETIME)) t2
GROUP BY pyear
ORDER BY pyear DESC, AVG(payments) DESC) AS tthu) tthu2 ON tsun.pyear = tthu2.pyear
LEFT JOIN (SELECT * FROM (SELECT ROUND(AVG(payments),1) AS Friday, pyear FROM (
SELECT COUNT(paymentid) AS payments, YEAR(DATETIME) AS pyear
FROM payments 
WHERE DAYNAME(DATETIME) = "Friday"
GROUP BY DATE(DATETIME)
ORDER BY YEAR(DATETIME), DAYOFWEEK(DATETIME)) t2
GROUP BY pyear
ORDER BY pyear DESC, AVG(payments) DESC) AS tfri) tfri2 ON tsun.pyear = tfri2.pyear
LEFT JOIN (SELECT * FROM (SELECT ROUND(AVG(payments),1) AS Saturday, pyear FROM (
SELECT COUNT(paymentid) AS payments, YEAR(DATETIME) AS pyear
FROM payments 
WHERE DAYNAME(DATETIME) = "Saturday"
GROUP BY DATE(DATETIME)
ORDER BY YEAR(DATETIME), DAYOFWEEK(DATETIME)) t2
GROUP BY pyear
ORDER BY pyear DESC, AVG(payments) DESC) AS tsat) tsat2 ON tsun.pyear = tsat2.pyear

这也会以我想要的格式生成我想要的数据；

"pyear" "sunday"    "monday"    "tuesday"   "wednesday" "thursday"  "friday"    "saturday"
"2015"  "18.4"  "13.0"  "16.5"  "14.3"  "13.3"  "17.3"  "31.3"
"2014"  "17.8"  "19.0"  "17.6"  "15.3"  "15.5"  "20.2"  "32.0"
"2013"  "3.4"   "3.9"   "3.4"   "3.7"   "3.4"   "4.5"   "6.2"
"2012"  "2.8"   "4.8"   "4.7"   "4.8"   "3.7"   "5.7"   "7.1"
"2011"  "4.0"   "7.1"   "5.6"   "6.2"   "6.6"   "5.4"   "6.2"
"2010"  "3.0"   "5.5"   "5.7"   "5.2"   "5.3"   "4.6"   "6.6"
"2009"  "2.5"   "4.3"   "3.5"   "4.9"   "4.8"   "2.9"   "3.9"

但这不是很优雅，不是吗!

我想知道是否有更好的方法？

是否有类似于第一个语句的内容会产生类似于第二个语句的内容？

我确信有一种聪明的方法可以做到这一点，例如使用 case 语句，但目前我无法找到解决方案。有人有什么想法吗？

提前致谢，

马克

<小时/>

所以我就做到了，谢谢戈登。

SELECT oyear,
       ROUND(AVG(CASE WHEN oday = 'Monday' THEN sales END), 1) AS Mon,
       ROUND(AVG(CASE WHEN oday = 'Tuesday' THEN sales END), 1) AS Tue,
       ROUND(AVG(CASE WHEN oday = 'Wednesday' THEN sales END), 1) AS Wed,
       ROUND(AVG(CASE WHEN oday = 'Thursday' THEN sales END), 1) AS Thu,
       ROUND(AVG(CASE WHEN oday = 'Friday' THEN sales END), 1) AS Fri,
       ROUND(AVG(CASE WHEN oday = 'Saturday' THEN sales END), 1) AS Sat,
       ROUND(AVG(CASE WHEN oday = 'Sunday' THEN sales END), 1) AS Sun
FROM (SELECT COUNT(oid) AS sales, DAYNAME(odate) AS oday , YEAR(odate) AS oyear
      FROM orders 
      WHERE ostatus = 'completed'
      GROUP BY DATE(odate)
     ) d
GROUP BY oyear
ORDER BY oyear ASC

Answer 1

使用条件聚合:

SELECT ROUND(AVG(sales),1) AS dayaverage, oday, oyear,
       ROUND(AVG(CASE WHEN oday = 'Monday' THEN sales END), 1) as Mon,
       ROUND(AVG(CASE WHEN oday = 'Tuesday' THEN sales END), 1) as Tue,
       ROUND(AVG(CASE WHEN oday = 'Wednesday' THEN sales END), 1) as Wed,
       ROUND(AVG(CASE WHEN oday = 'Thursday' THEN sales END), 1) as Thu,
       ROUND(AVG(CASE WHEN oday = 'Friday' THEN sales END), 1) as Fri,
       ROUND(AVG(CASE WHEN oday = 'Saturday' THEN sales END), 1) as Sat,
       ROUND(AVG(CASE WHEN oday = 'Sun' THEN sales END), 1) as Sun
FROM (SELECT COUNT(oid) AS sales, DAYNAME(odate) AS oday , YEAR(odate) AS oyear
      FROM orders 
      WHERE ostatus = 'completed'
      GROUP BY DATE(odate)
     ) d
GROUP BY oyear
ORDER BY DESC;