Question

我有一个epos系统订单表。

我有几年的数据，我想知道每年每周的平均交易次数。即2014年与2013年等最繁忙的一天。

我可以通过这种方式获取信息：

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

这需要数据，但不是正确的格式。

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

这也以我想要的格式产生我想要的数据;

"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"

但它不是很优雅！

我想知道的是，有更好的方法吗？

是否有类似第一个声明会产生类似第二个声明的东西？

我确信有一种聪明的方法可以做到这一点，就像使用案例陈述一样，但解决方案暂时让我失望了。有人有任何想法吗？

提前致谢，

马克

所以我得到了这个，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;

查找一天中平均的交易数量

1 个答案: