Question

如何与日期年序列不匹配的记录我试图返回所有未按年订购的客户记录。

例如，

cust1 - 2010
cust1 - 2012
cust1 - 2013
cust1 - 2014
cust2 - 2014

正如您所见，cust1尚未在2011年订购

我想返回显示cust1的行。

我需要能够指定一系列日期，例如最小和最大。

我试过这个：

DECLARE @MINDATE AS INT;
SELECT @MINDATE = '2010';

DECLARE @MAXDATE AS INT;
SELECT @MAXDATE = '2014';

SELECT CUSTID, YEAR(DATE) AS [YEAR]
FROM ORDER
WHERE DATE BETWEEN @MINDATE AND @MAXDATE;

Answer 1

您可以指定如下序列：

DECLARE
    @MINYEAR AS INT = '2010',
    @MAXYEAR AS INT = '2014';

with base as
(
    select @MINYEAR [year]
    union all
    select [year] + 1
    from base
    where [year] < @MAXYEAR
)
select * from [base]

结果：

year
----
2010
2011
2012
2013
2014

Ant然后JOIN与你的表格，例如：

DECLARE
    @MINYEAR AS INT = '2010',
    @MAXYEAR AS INT = '2014';

with base as
(
    select @MINYEAR [year]
    union all
    select [year] + 1
    from base
    where [year] < @MAXYEAR
)
select y.[year], isnull(b.CUSTID, 'No data') [CUSTID]
from base y
left join ORDER b on YEAR(b.DATE) = y.[year]

我不知道如何实现所需的输出，因为您没有提供任何数据架构

Answer 2

SELECT      #sequence.value
FROM        #sequence
LEFT JOIN   MyTable ON #sequence.value = MyTable.value
WHERE       MyTable.value IS NULL

取自answer

Answer 3

我倾向于使用count(distinct)执行此操作：

SELECT CUSTID
FROM ORDER
WHERE DATE BETWEEN @MINDATE AND @MAXDATE
GROUP BY CUSTID
HAVING COUNT(DISTINCT YEAR(DATE)) = (SELECT COUNT(DISTINCT YEAR(DATE))
                                              FROM ORDER
                                              WHERE DATE BETWEEN @MINDATE AND @MAXDATE
                                             );

编辑：这假设所有年份都在数据中表示。

Answer 4

试试这个，它会报告所有年份都没有订单的客户。

它将客户的不同年份数与根据MaxDate和MinDate之间的差异应该存在的年数进行比较。

SELECT CUSTID
FROM ORDER
WHERE DATE BETWEEN @MINDATE AND @MAXDATE
GROUP BY CUSTID
HAVING COUNT(DISTINCT YEAR(DATE)) <> (YEAR(@MAXDATE) - YEAR(@MINDATE)) + 1

显示与日期年序列不匹配的记录

4 个答案: