Question

对于一个优秀的MySQL编码器来说，这是一个挑战。我们非常感谢有关创建在html / PHP页面中显示的MySQL报告的查询的一些帮助。

我们知道如何收集数据并存储它，并拥有PHP报告布局，但是如何构建查询则会丢失。

我们有两个MySQL表 - “访客”和“买家”。

访客表 - 访问我们网站的访问者标识sales_rep的特定URL和许多其中一个他们自己创造的营销代码。
买家表 - 适合那些买东西的人。

访问者可以通过以下链接访问我们的网站，例如：

www.sales_rep.website.com/index.php?marketing=code1/code2/code3

在“访客”表中，我们包含三个字段：

SALES_REP
日期
marketing_CODE - （在这种情况下，它将是字符串中的“code1 / code2 / code3”，但每个sales_rep可以基于其广告活动具有多个marketing_CODES;例如“code1a / codeTX / codeM2”和“market- Z /广告-123" ）

当访问者购买某物时，我们会将其添加到“买家”表中，其中包括：

SALES_REP
日期
marketing_TYPE - （可以是“网址”，“引荐”或其他类型）
marketing_VALUE - （如果marketing_TYPE是网址，则为marketing_CODE）

我们需要查看针对任何指定月份的每个特定销售代表的报告，按营销代码排序，仅针对marketing_TYPE ==“URL”。

此报告需要显示每个marketing_CODE的访客和买家活动的逐周细分，如下所示：

Table Example

为了确保清晰，在上表中，在第1周，8名访客通过“code1 / code2 /”链接进入网站，其中5人购买了东西。

我们尝试使用以下内容开始查询，但不知道该去哪里：

$chosen_month = $_POST['chosen_month'];
SELECT * FROM visitors INNER JOIN buyers ON sales_rep WHERE marketing_type = "URL" AND sales_rep = '$sales_rep' AND date('Y-m') = $'chosen_month' ;
使用while语句，我们将为找到的每条记录生成表格单元格。使用它，我们知道如何列出每条记录，但我们无法弄清楚如何显示非独特记录一次，显示它们的计数，然后显示不同的记录。

感谢您的时间和帮助。

Answer 1

这是我对您的查询的建议。

首先，我需要根据日期字段找出周数，所以我在this question中使用了答案：

SELECT  WEEK(dateField, 5) -
        WEEK(DATE_SUB(dateField, INTERVAL DAYOFMONTH(dateField) - 1 DAY), 5) + 1

我将从现在开始将上述结果称为@week，以使事情更具可读性：）

然后，我从一张桌子开始，试图看看销售代表每个月有多少访客：

select rep, 
  , SUM(CASE @week WHEN 1 THEN 1 ELSE 0 END) as Week1
  , SUM(CASE @week WHEN 2 THEN 1 ELSE 0 END) as Week2
  , SUM(CASE @week WHEN 3 THEN 1 ELSE 0 END) as Week3
  , SUM(CASE @week WHEN 4 THEN 1 ELSE 0 END) as Week4
  , COUNT(*) as Total
from visitors
group by rep

现在我们收到了这些信息，我们将对买家使用相同的信息，然后加入这两个子查询的结果，如下所示：

select v.rep
, v.Week1 , b.Week1
, v.Week2 , b.Week2
, v.Week3 , b.Week3
, v.Week4 , b.Week4
, v.Total, b.Total
from (
     select rep, 
        , SUM(CASE @week WHEN 1 THEN 1 ELSE 0 END) as Week1
        , SUM(CASE @week WHEN 2 THEN 1 ELSE 0 END) as Week2
        , SUM(CASE @week WHEN 3 THEN 1 ELSE 0 END) as Week3
        , SUM(CASE @week WHEN 4 THEN 1 ELSE 0 END) as Week4
        , COUNT(*) as Total
     from visitors
     group by rep) as v
     left join (select rep, 
                 , SUM(CASE @week WHEN 1 THEN 1 ELSE 0 END) as Week1
                 , SUM(CASE @week WHEN 2 THEN 1 ELSE 0 END) as Week2
                 , SUM(CASE @week WHEN 3 THEN 1 ELSE 0 END) as Week3
                 , SUM(CASE @week WHEN 4 THEN 1 ELSE 0 END) as Week4
                 , COUNT(*) as Total
                from visitors
                group by rep) as b on v.rep = b.rep

这应该是你要找的！

以下是针对您的情况更新的相同代码：

select v.rep
    , v.Week1 , b.Week1
    , v.Week2 , b.Week2
    , v.Week3 , b.Week3
    , v.Week4 , b.Week4
    , v.Total, b.Total
from (
        select rep
          , SUM(CASE WEEK(the_date, 5) -
            WEEK(DATE_SUB(the_date, INTERVAL DAYOFMONTH(the_date) - 1 DAY), 5) + 1
            WHEN 1 THEN 1 ELSE 0 END) as Week1
      , SUM(CASE WEEK(the_date, 5) -
            WEEK(DATE_SUB(the_date, INTERVAL DAYOFMONTH(the_date) - 1 DAY), 5) + 1
            WHEN 2 THEN 1 ELSE 0 END) as Week2
      , SUM(CASE WEEK(the_date, 5) -
        WEEK(DATE_SUB(the_date, INTERVAL DAYOFMONTH(the_date) - 1 DAY), 5) + 1
            WHEN 3 THEN 1 ELSE 0 END) as Week3
      , SUM(CASE WEEK(the_date, 5) -
        WEEK(DATE_SUB(the_date, INTERVAL DAYOFMONTH(the_date) - 1 DAY), 5) + 1
            WHEN 4 THEN 1 ELSE 0 END) as Week4
      , COUNT(*) as Total
        from visitors
        where sales_rep = '$sales_rep' AND date('Y-m') = $'chosen_month'
        group by rep ) as v
  left join (
        select rep
          , SUM(CASE WEEK(the_date, 5) -
            WEEK(DATE_SUB(the_date, INTERVAL DAYOFMONTH(the_date) - 1 DAY), 5) + 1
            WHEN 1 THEN 1 ELSE 0 END) as Week1
      , SUM(CASE WEEK(the_date, 5) -
            WEEK(DATE_SUB(the_date, INTERVAL DAYOFMONTH(the_date) - 1 DAY), 5) + 1
            WHEN 2 THEN 1 ELSE 0 END) as Week2
      , SUM(CASE WEEK(the_date, 5) -
        WEEK(DATE_SUB(the_date, INTERVAL DAYOFMONTH(the_date) - 1 DAY), 5) + 1
            WHEN 3 THEN 1 ELSE 0 END) as Week3
      , SUM(CASE WEEK(the_date, 5) -
        WEEK(DATE_SUB(the_date, INTERVAL DAYOFMONTH(the_date) - 1 DAY), 5) + 1
            WHEN 4 THEN 1 ELSE 0 END) as Week4
      , COUNT(*) as Total
        from buyers
        where marketing_type = "URL" AND sales_rep = '$sales_rep' AND date('Y-m') = $'chosen_month'
        group by rep) as b on v.rep = b.rep

挑战MySQL查询html / PHP报告

1 个答案: