内连接与if语句，组 [英] Inner join with if statement, group

问题描述

我有2组代码。
这2组代码正常工作没有错误。
但是我面临一个问题，我需要分组这2组代码。
请看看，谢谢
这是我创建的第一个代码。

I have 2 group of code here. This 2 group of code is working without error. But I face a problem , I need to group up this 2 group of code. Please take a look, thank you This is the first code i create.

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 (#ARGUMENTS.Feature_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>

然后，我注意到feature_ID我需要使用另一个逻辑来显示结果。
代码如下：

Later then, I notice the feature_ID i need to use another logic to show the result. The code is like this

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

我想将这两个组合在一起。
第二组必须替换

I want combine this 2 group together. The second group have to replace

<cfif ARGUMENTS.Feature_ID IS NOT "">
    AND f.Feature_ID IN (#ARGUMENTS.Feature_ID#)
</cfif>

我尝试了几种方法来分组这2个代码，但无法做到。代码我试着在下面，它得到错误。

I try few way to group this 2 code, but fail to make it. The code i try is on below, it get error.

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.Location_ID
        FROM Restaurants r, Features f, Cuisine c, State s, Location l
        INNER JOIN Bridge1_Restaurant_Features b ON b.Restaurant_ID = r.Restaurant_ID
        INNER JOIN Features f ON b.Feature_ID = f.Feature_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 "">
        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
        </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>

推荐答案

我自己解决问题。如有任何改进请告诉我，谢谢大家。

I solve the problem by myself. If any way can be improve please let me know, thank everyone.

SELECT r.Restaurant_ID, r.Name, f.Feature, r.Price_Range, r.Cuisine_ID, c.Cuisine,
                l.Location_ID, l.Area, s.State
        FROM   Restaurants r
        <cfif ARGUMENTS.Feature_ID IS NOT "">
        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
        </cfif>
        INNER JOIN Location l ON r.Location_ID = l.Location_ID
        INNER JOIN State s ON l.State_ID = s.State_ID
        INNER JOIN Cuisine c ON r.Cuisine_ID = c.Cuisine_ID 
        INNER JOIN Bridge1_Restaurant_Features b ON b.Restaurant_ID = r.Restaurant_ID
        INNER JOIN Features f ON b.Feature_ID = f.Feature_ID
        <cfif ARGUMENTS.Cuisine IS NOT "">
        AND r.Cuisine_ID = (#ARGUMENTS.Cuisine#)
        </cfif>
        <cfif ARGUMENTS.Price_Range IS NOT "">
        AND r.Price_Range IN (#ARGUMENTS.Price_Range#)
        </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>

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