oracle Where子句是否有整数范围？
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Is there integer ranges for Where Clause?
提问by zmische
I need to check that result of expression in where clause is in range of Integers.
我需要检查 where 子句中的表达式结果是否在整数范围内。
something lke this:
像这样的东西：
select * from table where (col1 / col2 ) in (1..8).
With (1..8)representing a range of integers.
用(1..8)代表一个整数范围。
I mean that it must be integer, not float. So that I cant use between 1 and 8, because 1.2 will be correct.
我的意思是它必须是整数，而不是浮点数。所以我不能使用between 1 and 8，因为 1.2 是正确的。
回答by Tony Andrews
You can of course do this:
你当然可以这样做：
select * from table where (col1 / col2 ) in (1,2,3,4,5,6,7,8);
or
或者
select * from table where (col1 / col2 ) between 1 and 8
and mod (col1 , col2 ) = 0;
回答by Martin B?gelund
How about
怎么样
select *
from table
where (col1 / col2 ) BETWEEN 1 AND 8
and (col1 / col2 ) = FLOOR(col1 / col2 )
This simply checks if the fraction is in the interval, andinteger.
这只是检查分数是否在区间和整数中。
回答by Nelson
You could cast it from float to int and use between. (You might want to make a virutal/computed column depending on the query's performance.)
您可以将其从 float 转换为 int 并使用between. （您可能希望根据查询的性能创建虚拟/计算列。）
回答by Bill
To test col1/col2 is an integer, you could Mod them together...
要测试 col1/col2 是一个整数，您可以将它们一起修改...
Where (col1/col2) = (col1\col2) and (col1/col2) between 1 and 8
其中 (col1/col2) = (col1\col2) 和 (col1/col2) 介于 1 和 8 之间
回答by jva
You can use pipeline function to generate integer range.
您可以使用管道函数生成整数范围。
create or replace type my_number_collection is table of number;
create or replace function
gen_range (p_from in number, p_to in number)
return my_number_collection
PIPELINED
as
begin
for i in p_from..p_to loop
pipe row(i);
end loop;
return;
end;
/
select *
from my_table
where col1/col2 in (select column_value from table(gen_range(1,8)));
