SQL SERVER -不匹配条件时，获取0而不是now行

Question

我首先要说的是，我不是SQL专家，更不用说SQL服务器了。

我正在尝试执行下面的查询：

基本上，它的作用是，对于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到表变量中。

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

我想要实现的东西是这样的：

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        

但我得到的是：

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

我知道发生这种情况是因为这个-> C.subid = CS.id，C.tstamp >= ...和C.tstamp <= ...，这意味着没有符合这个条件的行。

但是我也需要得到那0行！

请帮帮我。

提前感谢并祝您编程愉快！

Answer 1

使用left join是正确的，但是当前的where子句阻止了空行的计数。将条件移动到联接条件，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

另外，为count(c.subid)去掉ISNULL(count(*), 0)，这样当c.subid为null时，您就不会得到1的计数。