SQL SERVER - 当不符合条件时,获取0而不是现在的行

Question

我首先要说的是,我不是SQL专家,更不用说SQL SERVER了.

我正在尝试执行以下查询:

基本上它的作用是,在开始日期和结束日期之间的每一天获取一些数据并将其插入表变量中.

BEGIN

declare @sdate datetime = dateadd(day, -10, getdate())
declare @edate datetime = getdate()
declare @max int = datediff(day, @sdate, @edate)
declare @sday int = cast(day(@sdate) as int)
declare @i int = 1;

declare @Days2 table (
    [sending_id] varchar(255) not null,
    [day] datetime not null,
    [conversions] int not null
)

while @i <= @max
BEGIN
    declare @ssdate datetime = cast(dateadd(day, (@max - @i) * -1 , @edate) as date)
    declare @eedate datetime = cast(dateadd(day, ((@max - @i) * -1) + 1 , @edate)as date)

    INSERT into @Days2 ([sending_id], [day], [conversions])
    select CS.send_id, @ssdate as [date], ISNULL(count(*), 0) as day_conversion
    from campaigns_history CH
    LEFT OUTER JOIN campaign_sends CS ON CS.campaign_historyID = CH.id
    LEFT OUTER JOIN c C ON C.subid = convert(varchar(255), CS.id)
    LEFT OUTER JOIN conversions_keygen CK ON CK.campaignID = CS.id AND cast(CK.genkey as varchar(255)) = C.clickid
    LEFT OUTER JOIN lead_feed LF ON LF.leadID = CK.leadID
    WHERE
        CH.id = 19
        and isnumeric(C.subid) <> 0
        and C.tstamp > @ssdate
        and C.tstamp < @eedate
    group by CS.send_id

    SET @i = @i+1
END

select * from @Days2

END

我试图实现的是这样的:

sending_id | days        | conversions
---------------------------------------
send-1     |  2017-08-01 | 1
---------------------------------------
send-1     |  2017-08-02 | 0            -- the problem is here
---------------------------------------
send-2     |  2017-08-01 | 1   
---------------------------------------
send-2     |  2017-08-02 | 4        

但我得到了这个:

sending_id | days        | conversions
---------------------------------------
send-1     |  2017-08-01 | 1
---------------------------------------
send-2     |  2017-08-01 | 1   
---------------------------------------
send-2     |  2017-08-02 | 4  

我知道这是因为这个 - > C.subid = CS.id, C.tstamp >= ...并且C.tstamp <= ...意味着没有符合此标准的行.

但我也需要获得0行!

请帮帮我.

提前谢谢和快乐的编程!

Answer 1

您使用的是正确的轨道left join,但您当前的where子句正在阻止计算空白行.将您的条件移动到连接条件,如下所示:

select CS.send_id, @ssdate as [date], count(c.subid) as day_conversion
from campaigns_history CH
LEFT OUTER JOIN campaign_sends CS ON CS.campaign_historyID = CH.id
LEFT OUTER JOIN c C ON C.subid = convert(varchar(255), CS.id)
    and isnumeric(C.subid) <> 0
    and C.tstamp > @ssdate
    and C.tstamp < @eedate
LEFT OUTER JOIN conversions_keygen CK ON CK.campaignID = CS.id 
  AND cast(CK.genkey as varchar(255)) = C.clickid
LEFT OUTER JOIN lead_feed LF ON LF.leadID = CK.leadID
WHERE
    CH.id = 19
group by CS.send_id

同时转出ISNULL(count(*), 0)的count(c.subid),所以你没有得到的计数1时c.subid是null.