Question

查询

$(document).ready(function(){
    $.ajax({
        url : "http://192.100.1.100/content.php",
        type : "GET",
        success : function(data){
            console.log(data);

            var userid = [];
            var facebook_follower = [];
            var twitter_follower = [];
            var googleplus_follower = [];

            for(var i in data) {
                userid.push("UserID " + data[i].userid);
                facebook_follower.push(data[i].facebook);
                twitter_follower.push(data[i].twitter);
                googleplus_follower.push(data[i].googleplus);
            }

            var chartdata = {
                labels: userid,
                datasets: [
                    {
                        label: "facebook",
                        fill: false,
                        lineTension: 0.1,
                        backgroundColor: "rgba(59, 89, 152, 0.75)",
                        borderColor: "rgba(59, 89, 152, 1)",
                        pointHoverBackgroundColor: "rgba(59, 89, 152, 1)",
                        pointHoverBorderColor: "rgba(59, 89, 152, 1)",
                        data: facebook_follower
                    },
                    {
                        label: "twitter",
                        fill: false,
                        lineTension: 0.1,
                        backgroundColor: "rgba(29, 202, 255, 0.75)",
                        borderColor: "rgba(29, 202, 255, 1)",
                        pointHoverBackgroundColor: "rgba(29, 202, 255, 1)",
                        pointHoverBorderColor: "rgba(29, 202, 255, 1)",
                        data: twitter_follower
                    },
                    {
                        label: "googleplus",
                        fill: false,
                        lineTension: 0.1,
                        backgroundColor: "rgba(211, 72, 54, 0.75)",
                        borderColor: "rgba(211, 72, 54, 1)",
                        pointHoverBackgroundColor: "rgba(211, 72, 54, 1)",
                        pointHoverBorderColor: "rgba(211, 72, 54, 1)",
                        data: googleplus_follower
                    }
                ]
            };

            var ctx = $("#mycanvas");

            var LineGraph = new Chart(ctx, {
                type: 'line',
                data: chartdata
                });
            },
        error : function(data) {
            }
        });
    });

给出每个工作日每小时的订单总数

结果

SELECT 
       DATENAME(dw, O1.CreationDateTime) AS 'Day of the week', 
       DATENAME(hh, O1.CreationDateTime) AS hour, 
       COUNT(*) AS 'Nr. Orders'
FROM 
       Orders AS O1
WHERE 
       O1.CreationDateTime >= '2015-10-01' 
GROUP BY 
       DATENAME(dw, O1.CreationDateTime), DATENAME(hh, O1.CreationDateTime)
ORDER BY 
       [Day of the week], CAST(DATENAME(hh, O1.CreationDateTime) as int)

目标是获得当天每小时的平均订单数量。但有些日子没有订单那些日子不应该算在内。为了解决这个问题，我在下面编写了select distinct查询。

查询

Weekday Hour    Nr.Orders
Friday  0   10
Friday  1   11
Friday  2   2

例如，它具有如下所示的订单周五的数量。

结果

SELECT 
       DATENAME(dw,Formatteddate.Date) AS "Day of the week",
       COUNT(*) AS "Weekdays in period"
FROM
       (SELECT DISTINCT
       ((CONVERT (date,O2.CreationDateTime,112))) AS Date
FROM Orders AS O2
WHERE 
       O2.CreationDateTime >= '2015-10-01' 
       ) AS Formatteddate
GROUP BY DATENAME(dw,Formatteddate.Dates)

基本上我的问题是两个如何合并我使用外部的那两个查询适用于变通方法。然而，由于事实它很慢检查每一行。我想我需要使用over子句但我没有想法如何使用datename函数。

最终结果看起来应该像平均一样容易计算。

Day of the week | Week days in period (with order)
    Friday            32
    Monday            30
    Saturday          21

Answer 1

您可以在一周的某一天将两张桌子加在一起：

SELECT t1.[Day of the week] AS Weekday,
       t1.hour,
       t1.[Nr. Orders],
       COALESCE(t2.[Weekdays in period], 0)
FROM
(
    SELECT DATENAME(dw, O1.CreationDateTime) AS 'Day of the week', 
           DATENAME(hh, O1.CreationDateTime) AS hour, 
           COUNT(*) AS 'Nr. Orders'
    FROM Orders AS O1
    WHERE O1.CreationDateTime >= '2015-10-01' 
    GROUP BY DATENAME(dw, O1.CreationDateTime),
             DATENAME(hh, O1.CreationDateTime)
) t1
LEFT JOIN
(
    SELECT DATENAME(dw,Formatteddate.Date) AS "Day of the week",
           COUNT(*) AS "Weekdays in period"
    FROM
    (
        SELECT DISTINCT CONVERT (date,O2.CreationDateTime, 112) AS Date
        FROM Orders AS O2
        WHERE O2.CreationDateTime >= '2015-10-01' 
    ) AS Formatteddate
    GROUP BY DATENAME(dw,Formatteddate.Dates)
) t2
    ON t1.[Day of the week] = t2.[Day of the week]

Answer 2

尝试加入当天：

WITH DayO AS (
SELECT 
       DATENAME(dw,Formatteddate.Date) AS DayOfWeek,
       COUNT(*) AS PeriodDays
FROM
       (SELECT DISTINCT
       ((CONVERT (date,O2.CreationDateTime,112))) AS Date
FROM Orders AS O2
WHERE 
       O2.CreationDateTime >= '2015-10-01' 
       ) AS Formatteddate
GROUP BY DATENAME(dw,Formatteddate.Dates) 
)

, HourO AS (
SELECT 
       DATENAME(dw, O1.CreationDateTime) AS DayOfWeek, 
       DATENAME(hh, O1.CreationDateTime) AS OrdHour, 
       COUNT(*) AS NumOrd
FROM 
       Orders AS O1
WHERE 
       O1.CreationDateTime >= '2015-10-01' 
GROUP BY 
       DATENAME(dw, O1.CreationDateTime), DATENAME(hh, O1.CreationDateTime)
ORDER BY 
       [DayOfWeek], CAST(DATENAME(hh, O1.CreationDateTime) as int)
)

SELECT ho.DayOfWeek as Weekday, ho.OrdHour as Hour, ho.NumOrd as [Nr. Orders], DayO.PeriodDays as [Weekdays in period]
from HourO ho
inner join DayO
on ho.DayOfWeek = DayO.DayOfWeek

合并两个计数日期名称查询

2 个答案: