内联if语句，组

问题描述

| 我这里有2组代码。这两组代码正常工作。但是我面临一个问题，我需要将这两组代码归为一组。请看看，谢谢这是我创建的第一个代码。

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>

之后，我注意到我需要使用另一个逻辑来显示feature_ID的结果。代码是这样的

SELECT r.Restaurant_ID,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

我想将这两个小组结合在一起。第二组必须替换

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

我尝试几种方法来对这2个代码进行分组，但未能成功。我尝试的代码在下面，它会出错。

SELECT r.Name,r.Location_ID
        FROM Restaurants r,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>

解决方法

我仍然不确定你想要什么。这样的事情？

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.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>

, 我自己解决问题。如果有任何方法可以改善，请告诉我，谢谢大家。

SELECT r.Restaurant_ID,f.Feature,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>

内联语句语句

内联if语句，组

问题描述

解决方法

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