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    <blockquote> <p>If Feature_ID = 1 ,2 , the result must <strong>only</strong> get Hotel_ID = 1</p> </blockquote> <p>Because only Hotel_ID = 1 has both features, correct? Assuming <code>#FORM.Feature_ID#</code> does not contain duplicates, use a HAVING clause to dynamically identify hotels with all of the requested features. </p> <pre><code>SELECT Hotel_ID, COUNT(Feature_ID) AS FeatureCount FROM Bridge_Hotel_Feature &lt;!--- find matching features ---&gt; WHERE Feature_ID IN ( &lt;cfqueryparam value="#FORM.Feature_ID#" cfsqltype="cf_sql_integer" list="true"&gt; ) GROUP BY Hotel_ID &lt;!--- having ALL of the requested features ---&gt; HAVING COUNT(Feature_ID) = &lt;cfqueryparam value="#listLen(FORM.Feature_ID)#" cfsqltype="cf_sql_integer"&gt; </code></pre> <p>You could then join to it as a derived table or possibly a subquery. The sql needs optimization, but conceptually something like</p> <pre><code>SELECT h.Hotel_ID, h.Name, f.Feature FROM Hotel h INNER JOIN Bridge_Hotel_Feature b ON b.Hotel_ID = h.Hotel_ID INNER JOIN Features f ON b.Feature_ID = f.Feature_ID INNER JOIN ( SELECT Hotel_ID, COUNT(Feature_ID) AS FeatureCount FROM Bridge_Hotel_Feature &lt;!--- find matching features ---&gt; WHERE Feature_ID IN ( &lt;cfqueryparam value="#FORM.Feature_ID#" cfsqltype="cf_sql_integer" list="true"&gt; ) GROUP BY Hotel_ID &lt;!--- having ALL of the requested features ---&gt; HAVING COUNT(Feature_ID) = &lt;cfqueryparam value="#listLen(FORM.Feature_ID)#" cfsqltype="cf_sql_integer"&gt; ) ck ON ck.Hotel_ID = h.Hotel_Id </code></pre>
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    1. POHow to get the correct checkbox result from database MySQL?
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    1. COThank you first, I have try the first code u given me. But i fail to get any result even just tick 1 box. By The Way, can i know what the use for COUNT(Feature_ID) AS FeatureCount? because it no connect the following code.
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    2. CO@Raywin - Using `FORM.Feature_ID = "1,2"` the 1st query returns Hotel_ID 1. Can you post your sql? The COUNT(Feature_ID) is used to find only records that have *all* of the selected features.
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    3. COLeigh, thank , if just follow your first code exactly, i can get the answer, when i try just add on restaurant r then i could not get any answer. 2nd code u show me, i keep get error with 'HAVING COUNT(Feature_ID) = 2 ) ck ON ck.Restaurant_ID = r.Restaurant_I'
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