--to prepare the environment, let said this is your table
declare @table table
(
fiscalYr integer,
Value integer,
previousyr integer,
prevValue integer
)
--to prepare the value, let said these are your value
insert into @table values (2014,165,2013,0);
insert into @table values (2015,179,2014,0);
insert into @table values (2016,143,2015,0);
--the SQL
select A.fiscalYr,A.Value,A.previousyr, B.Value
from @table A left join
@table B on B.fiscalYr = A.previousyr
Dies ist die Antwort, die ich
fiscalYr bekam | Wert | PrevYr | Wert
2014 | 165 | 2013 | NULL
2015 | 179 | 2014 | 165
2016 | 143 | 2015 | 179
['LAG/LEAD'] (https://msdn.microsoft.com/en-us/library/hh231256.aspx) wenn SQL Server 2012+, selbst beitreten, wenn niedriger – lad2025
@ 4est Froh, um zu helfen ! – DhruvJoshi