0

我有一个查询，它返回有关工作委员会的分析风格信息以及申请人来自以下结构的位置：

查询数据格式

要访问的查询是：

SELECT   g.name AS [Source]
        ,SUM(CASE WHEN v.cost = 0 THEN 1 ELSE 0 END) AS [Organic Clicks]
        ,SUM(CASE WHEN v.cost <> 0 THEN 1 ELSE 0 END) AS [Paid Clicks]
        ,COUNT(v.id) AS [Total Clicks]
        ,SUM(CASE WHEN a.applicant = 1 AND v.cost = 0 THEN 1 ELSE 0 END) AS [Organic Applicants]
        ,SUM(CASE WHEN a.applicant = 1 AND v.cost <> 0 THEN 1 ELSE 0 END) AS [Paid Applicants]
        ,SUM(CASE WHEN a.applicant = 1 THEN 1 ELSE 0 END) AS [Total Applicants]
        ,SUM(v.cost/100.0) AS [Spend]
FROM a_views v
LEFT OUTER JOIN a_views a
    ON v.viewerid = a.viewerid
    AND v.sessionsourceid = a.sessionsourceid
    AND a.applicant = 1
JOIN a_sources s
    ON v.sourceid = s.id
JOIN a_sourcegroups g ON s.fk_sourcegroup = g.id
--JOIN jobs j ON v.jobid = j.anal_id AND j.featured = 1
WHERE v.hostName = @jobboard
    AND v.viewed_at >= @start AND v.viewed_at <= @end
GROUP BY g.name

唯一的问题是在LEFT OUTER JOIN a_views块中可能返回多条记录。我需要做的是只在Click总和中跟踪一次记录，但每次都为Applicant总和找到它。

我确实在这个问题上发现了一个类似的问题，但回答者实际上并没有提供太多信息。

为了恢复我需要的内容，连接右侧的每个记录实例，但左侧只有一个记录实例。

4

2 回答 2

0

出于某种原因，我认为您不希望通过自我加入来回答这个问题。也许只是聚合 a_views 一次就​​可以解决问题，首先是申请人，然后是游戏。

我发现很难从查询中确定您的意图，但是作为第一个聚合，如下所示：

select g.name, valuationDups.viweerid, v.sessionsourceid,
       SUM(CASE WHEN v.cost = 0 THEN 1 ELSE 0 END) AS [Organic Clicks],
       SUM(CASE WHEN v.cost <> 0 THEN 1 ELSE 0 END) AS [Paid Clicks],
       COUNT(v.id) AS [Total Clicks],
       SUM(CASE WHEN a.applicant = 1 AND v.cost = 0 THEN 1 ELSE 0 END) AS [Organic Applicants],
       SUM(CASE WHEN a.applicant = 1 AND v.cost <> 0 THEN 1 ELSE 0 END) AS [Paid Applicants],
       SUM(CASE WHEN a.applicant = 1 THEN 1 ELSE 0 END) AS [Total Applicants],
       SUM(v.cost/100.0) AS [Spend]       
FROM a_views v JOIN
     a_sources s
     ON v.sourceid = s.id JOIN
     a_sourcegroups g
     ON s.fk_sourcegroup = g.id
WHERE v.hostName = @jobboard and
      v.viewed_at >= @start AND
      v.viewed_at <= @end
group by g.name, valuationDups.viweerid, v.sessionsourceid

然后通过 g.name 再次聚合。

于 2012-12-11T18:51:22.920 回答
0

您可以在不完全更改查询的情况下执行此操作的一种方法是将点击总和和申请人总和分成两个单独的查询，UNION将它们放在一起，然后再次对它们进行分组/求和。

像这样的东西（伪代码）：

SELECT
        Source
        ,SUM([Organic Clicks])
        ,.....
FROM
(
SELECT   g.name AS [Source]
        ,SUM(CASE WHEN v.cost = 0 THEN 1 ELSE 0 END) AS [Organic Clicks]
        ,SUM(CASE WHEN v.cost <> 0 THEN 1 ELSE 0 END) AS [Paid Clicks]
        ,COUNT(v.id) AS [Total Clicks]
        ,0 AS [Organic Applicants]
        ,0 AS [Paid Applicants]
        ,0 AS [Total Applicants]
        ,SUM(v.cost/100.0) AS [Spend]
FROM
        ClickOnlyTables

UNION ALL

SELECT   g.name AS [Source]
        ,0 AS [Organic Clicks]
        ,0 AS [Paid Clicks]
        ,0 AS [Total Clicks]
        ,SUM(CASE WHEN a.applicant = 1 AND v.cost = 0 THEN 1 ELSE 0 END) AS [Organic Applicants]
        ,SUM(CASE WHEN a.applicant = 1 AND v.cost <> 0 THEN 1 ELSE 0 END) AS [Paid Applicants]
        ,SUM(CASE WHEN a.applicant = 1 THEN 1 ELSE 0 END) AS [Total Applicants]
        ,0 AS [Spend]
FROM
        ApplicantTables
)
GROUP BY
        Source
于 2012-12-11T17:55:48.783 回答