我得到了戈登汤普森在类似问题（Combine two tables by joining on the same date or closest prior date (not just exact matches)）上的大力帮助，但现在意识到我的数据并不是我的预期。事实证明，我可以在Product_Interest_Dates之后得到Lead_Dates，这导致之前的SQL代码丢弃这些情况。更具体地说：
我有两张桌子：
和
我希望两个创建一个表，其中，对于每个CustomerID，每个Product_Interest都连接到最接近日期的Lead_Source（在之前或之后）。决赛桌将是：
CustomerID
Product_Interest_Date
Product_Interest
Lead_Date (the closest entry in time to Product_Interest_Date)
Lead_Source (the Lead_Source of the closest Lead_Date)
我研究过戈德的代码，但不能带回家。按照他的例子，我想要图形化：http://i.stack.imgur.com/4ZVDV.jpg
序列的SQL Stack Overflow NEW 1
SELECT
pi.CustomerID,
pi.Product_Interest_Date,
l.Lead_Date,
Abs(pi.Product_Interest_Date-l.Lead_Date) AS Date_Gap
FROM
Test_PI pi
INNER JOIN
Test_Leads l
Stack Overflow NEW 2
SELECT
[Stack Overflow NEW 1].CustomerID,
[Stack Overflow NEW 1].Product_Interest_Date,
Min([Stack Overflow NEW 1].Date_Gap) AS MinOfDate_Gap
FROM [Stack Overflow NEW 1]
GROUP BY [Stack Overflow NEW 1].CustomerID,
[Stack Overflow NEW 1].Product_Interest_Date;
最终
SELECT Test_PI.CustomerID,
Test_PI.Product_Interest_Date,
Test_PI.Product_Interest,
Test_Leads.Lead_Date,
Test_Leads.Lead_Source
FROM (Test_PI INNER JOIN ([Stack Overflow NEW 2]
INNER JOIN [Stack Overflow NEW 1]
ON ([Stack Overflow NEW 2].CustomerID = [Stack Overflow NEW 1].CustomerID)
AND ([Stack Overflow NEW 2].Product_Interest_Date = [Stack Overflow NEW 1].Product_Interest_Date)
AND ([Stack Overflow NEW 2].MinOfDate_Gap = [Stack Overflow NEW 1].Date_Gap))
ON (Test_PI.CustomerID = [Stack Overflow NEW 2].CustomerID)
AND (Test_PI.Product_Interest_Date = [Stack Overflow NEW 2].Product_Interest_Date))
INNER JOIN Test_Leads
ON ([Stack Overflow NEW 1].CustomerID = Test_Leads.CustomerID)
AND ([Stack Overflow NEW 1].Lead_Date = Test_Leads.Lead_Date)
GROUP BY Test_PI.CustomerID, Test_PI.Product_Interest_Date, Test_PI.Product_Interest, Test_Leads.Lead_Date, Test_Leads.Lead_Source;
我尝试将所有这些组合成一个代码，并且无法通过SQL FROM错误！这是我的具体问题，如何在单个SQL代码中编写它？
SELECT
Test_PI.CustomerID,
Test_PI.Product_Interest_Date,
Test_PI.Product_Interest,
Test_Leads.Lead_Date,
Test_Leads.Lead_Source
FROM
(Test_PI
INNER JOIN
( (SELECT
latest.CustomerID,
latest.Product_Interest_Date,
Min(latest.Date_Gap) AS Min_Date_Gap
FROM
latest
) latest1
INNER JOIN
(SELECT
pi.CustomerID,
pi.Product_Interest_Date,
l.Lead_Date,
Abs(pi.Product_Interest_Date - l.Lead_Date) AS Date_Gap
FROM
Test_PI pi
INNER JOIN
Test_Leads l
ON pi.CustomerID = l.CustomerID
) latest
)
ON Test_PI.CustomerID = latest1.CustomerID AND Test_PI.Product_Interest_Date = latest1.Product_Interest_Date
INNER JOIN
Test_Leads
ON Test_Leads.CustomerID = latest1.CustomerID
AND Test_Leads.Lead_Date = latest1.Lead_Date
答案 0 :(得分：6)
既然我们正在考虑过去和未来的[Lead_Date]值，我已经调整了测试数据，涵盖了一个特例
表：Test_PI
CustomerID Product_Interest_Date Product_Interest
---------- --------------------- ----------------
1 2014-09-07 Interest1
1 2014-09-08 Interest2
1 2014-09-15 Interest3
1 2014-09-28 Interest4
表：Test_Leads
CustomerID Lead_Date Lead_Source
---------- ---------- -----------
1 2014-09-07 Source1
1 2014-09-14 Source2
2 2014-09-15 Source3
1 2014-09-16 Source4
我们首先创建一个名为[Date_Gaps]
的已保存Access查询SELECT
pi.CustomerID,
pi.Product_Interest_Date,
l.Lead_Date,
Abs(DateDiff("d", pi.Product_Interest_Date, l.Lead_Date)) AS Date_Gap
FROM
Test_PI pi
INNER JOIN
Test_Leads l
ON pi.CustomerID = l.CustomerID
返回
CustomerID Product_Interest_Date Lead_Date Date_Gap
---------- --------------------- ---------- --------
1 2014-09-07 2014-09-07 0
1 2014-09-08 2014-09-07 1
1 2014-09-15 2014-09-07 8
1 2014-09-28 2014-09-07 21
1 2014-09-07 2014-09-14 7
1 2014-09-08 2014-09-14 6
1 2014-09-15 2014-09-14 1
1 2014-09-28 2014-09-14 14
1 2014-09-07 2014-09-16 9
1 2014-09-08 2014-09-16 8
1 2014-09-15 2014-09-16 1
1 2014-09-28 2014-09-16 12
现在查询
SELECT
CustomerID,
Product_Interest_Date,
Min(Date_Gap) AS MinOfDate_Gap
FROM Date_Gaps
GROUP BY
CustomerID,
Product_Interest_Date
返回
CustomerID Product_Interest_Date MinOfDate_Gap
---------- --------------------- -------------
1 2014-09-07 0
1 2014-09-08 1
1 2014-09-15 1
1 2014-09-28 12
所以如果我们只是加入[Date_Gaps]查询以获取[Lead_Date]
SELECT
mingap.CustomerID,
mingap.Product_Interest_Date,
Date_Gaps.Lead_Date
FROM
Date_Gaps
INNER JOIN
(
SELECT
CustomerID,
Product_Interest_Date,
Min(Date_Gap) AS MinOfDate_Gap
FROM Date_Gaps
GROUP BY
CustomerID,
Product_Interest_Date
) mingap
ON Date_Gaps.CustomerID = mingap.CustomerID
AND Date_Gaps.Product_Interest_Date = mingap.Product_Interest_Date
AND Date_Gaps.Date_Gap = mingap.MinOfDate_Gap
我们得到了
CustomerID Product_Interest_Date Lead_Date
---------- --------------------- ----------
1 2014-09-07 2014-09-07
1 2014-09-08 2014-09-07
1 2014-09-15 2014-09-14
1 2014-09-15 2014-09-16
1 2014-09-28 2014-09-16
请注意，我们在09-15获得两次点击，因为他们都有1天的差距（之前和之后）。因此，我们需要通过使用Min(Lead_Date)（或Max(Lead_Date)，您的选择）将上述查询包装在聚合查询中来打破这种关系
SELECT
CustomerID,
Product_Interest_Date,
Min(Lead_Date) AS MinOfLead_Date
FROM
(
SELECT
mingap.CustomerID,
mingap.Product_Interest_Date,
Date_Gaps.Lead_Date
FROM
Date_Gaps
INNER JOIN
(
SELECT
CustomerID,
Product_Interest_Date,
Min(Date_Gap) AS MinOfDate_Gap
FROM Date_Gaps
GROUP BY
CustomerID,
Product_Interest_Date
) mingap
ON Date_Gaps.CustomerID = mingap.CustomerID
AND Date_Gaps.Product_Interest_Date = mingap.Product_Interest_Date
AND Date_Gaps.Date_Gap = mingap.MinOfDate_Gap
)
GROUP BY
CustomerID,
Product_Interest_Date
给我们
CustomerID Product_Interest_Date MinOfLead_Date
---------- --------------------- --------------
1 2014-09-07 2014-09-07
1 2014-09-08 2014-09-07
1 2014-09-15 2014-09-14
1 2014-09-28 2014-09-16
所以现在我们已经准备好加入原始表
了SELECT
Test_PI.CustomerID,
Test_PI.Product_Interest_Date,
Test_PI.Product_Interest,
Test_Leads.Lead_Date,
Test_Leads.Lead_Source
FROM
(
Test_PI
INNER JOIN
(
SELECT
CustomerID,
Product_Interest_Date,
Min(Lead_Date) AS MinOfLead_Date
FROM
(
SELECT
mingap.CustomerID,
mingap.Product_Interest_Date,
Date_Gaps.Lead_Date
FROM
Date_Gaps
INNER JOIN
(
SELECT
CustomerID,
Product_Interest_Date,
Min(Date_Gap) AS MinOfDate_Gap
FROM Date_Gaps
GROUP BY
CustomerID,
Product_Interest_Date
) mingap
ON Date_Gaps.CustomerID = mingap.CustomerID
AND Date_Gaps.Product_Interest_Date = mingap.Product_Interest_Date
AND Date_Gaps.Date_Gap = mingap.MinOfDate_Gap
)
GROUP BY
CustomerID,
Product_Interest_Date
) closest
ON Test_PI.CustomerID = closest.CustomerID
AND Test_PI.Product_Interest_Date = closest.Product_Interest_Date
)
INNER JOIN
Test_Leads
ON Test_Leads.CustomerID = closest.CustomerID
AND Test_Leads.Lead_Date = closest.MinOfLead_Date
返回
CustomerID Product_Interest_Date Product_Interest Lead_Date Lead_Source
---------- --------------------- ---------------- ---------- -----------
1 2014-09-07 Interest1 2014-09-07 Source1
1 2014-09-08 Interest2 2014-09-07 Source1
1 2014-09-15 Interest3 2014-09-14 Source2
1 2014-09-28 Interest4 2014-09-16 Source4