Question

我想计算发送的邮件（主表：ex_deliverylog）＆amp;他们的收件人（详细信息表：ex_deliverylog）来自日志。以下查询返回[session]和[recipients]的相同值。总之，我无法组织和计数[会话]。

    Select 
        deliveryaccount,
        DATEDIFF(d,deliverytime, getdate()) AS ago
        ,COUNT(ex_deliverylog.deliveryid) as session
        ,COUNT(ex_deliverylog_recipients.deliveryid) as recipients

        --,( select count(*) from ex_deliverylog_recipients where ex_deliverylog.deliveryid = ex_deliverylog_recipients.deliveryid )  

    from ex_deliverylog

    left join ex_deliverylog_recipients 
           on ex_deliverylog_recipients.deliveryid = ex_deliverylog.deliveryid


    group by
        deliveryaccount,
        DATEDIFF(d,deliverytime, getdate())


    order by ago, session desc

查询＆amp;结果：

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表＆amp;字段：

enter image description here

我如何计算两个会话和＆amp;他们的总收件人？

Answer 1

现在你得到两个相同的值，因为你的查询为每个GROUP BY返回了一定数量的行，并且对于每个COUNT()语句，所有这些行都被填充，所以你在这些字段中计算的结果中收到相同的值。您需要计算这些ID的唯一值。所以改变

Select 
    deliveryaccount,
    DATEDIFF(d,deliverytime, getdate()) AS ago
    ,COUNT(ex_deliverylog.deliveryid) as session
    ,COUNT(ex_deliverylog_recipients.deliveryid) as recipients

到

Select 
    deliveryaccount,
    DATEDIFF(d,deliverytime, getdate()) AS ago
    ,COUNT(distinct ex_deliverylog.deliveryid) as session
    ,COUNT(distinct ex_deliverylog_recipients.deliveryid) as recipients

如果这不能满足您的需求，我建议您将会话数和收件人数分成您认为可以控制的单独查询。

Answer 2

如果我了解您要执行的操作，我认为您需要在会话数上使用COUNT DISTINCT而不是COUNT，默认为COUNT ALL：

SELECT 
        deliveryaccount,
        DATEDIFF(d,deliverytime, getdate()) AS ago
        ,COUNT(DISTINCT ex_deliverylog.deliveryid) as session
        ,COUNT(ex_deliverylog_recipients.deliveryid) as recipients
FROM ex_deliverylog
LEFT JOIN ex_deliverylog_recipients 
       ON ex_deliverylog_recipients.deliveryid = ex_deliverylog.deliveryid
GROUP BY
    deliveryaccount,
    DATEDIFF(d,deliverytime, getdate())
ORDER BY ago, session desc

这样，会话计数将反映不同会话的数量，收件人计数将反映不同收件人的数量。如果既未指定ALL也未指定DISTINCT，则COUNT默认为ALL，您将获得您遇到的行为（即两者的计数相同）。

Answer 3

我如何计算两个会话和＆amp;他们的总收件人？
需要计算ex_deliverylog_recipients.deliveryrecipientid

SELECT 
        deliveryaccount
       ,DATEDIFF(d,deliverytime, getdate()) AS ago
       ,COUNT(DISTINCT ex_deliverylog.deliveryid) as session
       ,COUNT(DISTINCT ex_deliverylog_recipients.deliveryrecipientid) as recipients
FROM ex_deliverylog
LEFT JOIN ex_deliverylog_recipients 
       ON ex_deliverylog_recipients.deliveryid = ex_deliverylog.deliveryid
GROUP BY
    deliveryaccount,
    DATEDIFF(d,deliverytime, getdate())
ORDER BY ago, session desc

一个人可以多次收到同一封电子邮件例如。发送到两个小组，每个小组都有人如果您需要一个唯一收件人的计数，那么deliveryrecipientaddress（不是deliveryrecipientid）：

SELECT 
        deliveryaccount
       ,DATEDIFF(d,deliverytime, getdate()) AS ago
       ,COUNT(DISTINCT ex_deliverylog.deliveryid) as session
       ,COUNT(DISTINCT ex_deliverylog_recipients.deliveryrecipientaddress) as recipients
FROM ex_deliverylog
LEFT JOIN ex_deliverylog_recipients 
       ON ex_deliverylog_recipients.deliveryid = ex_deliverylog.deliveryid
GROUP BY
    deliveryaccount,
    DATEDIFF(d,deliverytime, getdate())
ORDER BY ago, session desc

Answer 4

SQL Server有一个很好的SQL扩展来执行此操作。

添加＆＃34; WITH ROLLUP＆＃34;在你的GROUP BY子句之后，它将在每个GROUP BY级别产生小计。

SQL：对主人和主人进行分组和统计细节

4 个答案: