Show lines according to the date range

RATE     EFF_FROM      EFF_TILL         SO_NO        SO_DATE
24.5      03-07-2008     26-11-2010      001            20-03-2010
47         27-11-2010     31-03-2012      001            20-03-2010
The above is the result of a query. I want to display only the line where is located between EFF_FROM and the EFF_TILL SO_DATE.
In above case 1st row lies between the range of dates, 07/03/2008 and 11/26/2010. Therefore, only the first line should appear.

Note: even if the the SO_DATE lies between 27/11/2010 and 31/03/2012, that the opener should appear.

Please suggest how can this be achieved?

Sanjay

Why not use this

select * from (
select 24.5 rate,'03-jul-2008' eff_from,'26-dec-2010' eff_till,001 so_no,'20-mar-2010' so_date from dual union all
select 47 rate,'27-nov-2010' eff_from,'31-mar-20102' eff_till,001 so_no,'20-mar-2010' so_date from dual)
where so_date between eff_from and eff_till

Iqbal

Tags: Database

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    Published by: 1004407 on June 7, 2013 11:06

    Hello

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    Post edited by: FrankKulash

    Sorry; I wanted to respond to OP, not Greg

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