SQL函数问题 [英] SQL Function problem

问题描述

declare @diff float
set @diff=0.8
if (@diff>=0.0) and (@diff<0.3) 
begin
 	set @diff=0.0
print 'a'
end
else 
	if (@diff>=0.3) and (@diff<0.8)
begin
 	set @diff=0.5
print 'b'
end
	if @diff >= 0.8 
begin
 	set @diff=1.0
print 'c'
end
print @diff

-------------------------------------------------- -

----------------------------------------------------

declare @monthdays float
declare @lgwdays int --legalworkingdays
declare @lgpdays float --legalpresentworkingdays
declare @diff float
set @diff=0
set @monthdays=day(dateadd(mm,datediff(mm,-1,'30-09-2011'),-1))
set @lgwdays=@monthdays-4
set @lgpdays=(9/@monthdays)*@lgwdays
set @diff=round(@lgpdays,2)-floor(@lgpdays)
print @diff
if (@diff>=0.0) and (@diff<0.3) 
begin
 	set @diff=0.0
print 'a'
end
if (@diff>=0.3) and (@diff<0.8)
begin
 	set @diff=0.5
print 'b'
end
if (@diff>=0.8) 
begin
 	set @diff=1.0
print 'c'
end
print @diff

------------为什么上述两个函数的输出具有相同的值

------------why the output is different for above 2 functions with same values

推荐答案

可能是因为它们包含不同的逻辑.
顶部的条件是else ，底部的条件是else .

Probably because they contain different logic.
The top one has an else condition the bottom one lacks.

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