Conditional WHERE clause with CASE statement in Oracle

问题

I'm brand-new to the Oracle world so this could be a softball. In working with an SSRS report, I'm passing in a string of states to a view. The twist is that the users could also pick a selection from the state list called "[ No Selection ]" ... (that part was not by doing and I'm stuck with implementing things this way)

If they choose the No Selection option, then I just want to return all states by default, otherwise return just the list of states that are in my comma-separated list.

This really seems like it should be easy but I'm stuck. Here is the code I have so far (just trying to get a sample to work) but my eyes have finally gone crossed trying to get it going.

Could anybody give me some direction on this one please?

begin
  :stateCode:='MO,FL,TX';
  --:stateCode:='[ No Selection ]';
end;
/

select count(*) as StateCount, :stateCode as SelectedVal
from hcp_state vw
where 
  (case 
      when (:stateCode = '') then (1)
      when (:stateCode != '') then (vw.state_cd in (:stateCode))
      (else 0)
  end)
;

回答1:

You can write the where clause as:

where (case when (:stateCode = '') then (1)
            when (:stateCode != '') and (vw.state_cd in (:stateCode)) then 1
            else 0)
       end) = 1;

Alternatively, remove the case entirely:

where (:stateCode = '') or
      ((:stateCode != '') and vw.state_cd in (:stateCode));

Or, even better:

where (:stateCode = '') or vw.state_cd in (:stateCode)

来源：https://stackoverflow.com/questions/18104884/conditional-where-clause-with-case-statement-in-oracle