Gabung dalam dengan pernyataan if, grup

saya masih tidak yakin apa yang Anda inginkan.

sesuatu seperti ini?

WITH group2 AS (

    SELECT r.Restaurant_ID, r.Name, f.Feature
    FROM   Restaurants r
    INNER JOIN Bridge1_Restaurant_Features b ON b.Restaurant_ID = r.Restaurant_ID
    INNER JOIN Features f ON b.Feature_ID = f.Feature_ID
    INNER JOIN
    (
         SELECT Restaurant_ID, COUNT(Feature_ID) AS FeatureCount
         FROM   Bridge1_Restaurant_Features
         <!--- find matching features --->
         WHERE  Feature_ID IN ( <cfqueryparam value="#ARGUMENTS.Feature_ID#"   cfsqltype="cf_sql_integer" list="true"> )
        GROUP BY Restaurant_ID
        <!--- having ALL of the requested features --->
         HAVING COUNT(Feature_ID) = <cfqueryparam value="#listLen(ARGUMENTS.Feature_ID)#" cfsqltype="cf_sql_integer">
        ) ck ON ck.Restaurant_ID = r.Restaurant_Id

)

SELECT r.Name , r.Restaurant_ID, f.feature, r.Price_Range, r.Cuisine_ID, c.Cuisine,
    s.State_ID, s.State, l.Location_ID, l.Area, l.State_ID, r.Name, r.Location_ID
FROM Restaurants r, Bridge1_Restaurant_Features b, Features f, Cuisine c, State s, Location l
    where 0=0
    AND b.Feature_ID = f.Feature_ID 
    AND b.Restaurant_ID = r.Restaurant_ID
    AND r.Cuisine_ID = c.Cuisine_ID 
    AND r.Location_ID = l.Location_ID
    AND l.State_ID = s.State_ID
    <cfif ARGUMENTS.Feature_ID IS NOT "">
    AND f.Feature_ID IN (group2.Restaurant_ID)
    </cfif>
    <cfif ARGUMENTS.Price_Range IS NOT "">
    AND r.Price_Range IN (#ARGUMENTS.Price_Range#)
    </cfif>
    <cfif ARGUMENTS.Cuisine IS NOT "">
    AND r.Cuisine_ID = (#ARGUMENTS.Cuisine#)
    </cfif>
    <cfif val(ARGUMENTS.LocationID2) IS #val(ARGUMENTS.StateID)#>
    AND l.State_ID = #val(ARGUMENTS.LocationID2)#
    <cfelse>
        AND l.Location_ID = #val(ARGUMENTS.LocationID2)#
    </cfif>