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    <p>I would just make put the variables in javascript with php and then use javascript functions.. no jquery or AJAX needed.</p> <p>However you need to have a foreign key for subcategories no matter what.. ie - For every record in subcat table you need to give it a catid so for referencing...</p> <pre><code>&lt;?php $db = new mysqli('localhost','user','password','dbname');//set your database handler $query = "SELECT id,cat FROM cat"; $result = $db-&gt;query($query); while($row = $result-&gt;fetch_assoc()){ $categories[] = array("id" =&gt; $row['id'], "val" =&gt; $row['cat']); } $query = "SELECT id, catid, subcat FROM subcat"; $result = $db-&gt;query($query); while($row = $result-&gt;fetch_assoc()){ $subcats[$row['catid']][] = array("id" =&gt; $row['id'], "val" =&gt; $row['subcat']); } $jsonCats = json_encode($categories); $jsonSubCats = json_encode($subcats); ?&gt; &lt;!docytpe html&gt; &lt;html&gt; &lt;head&gt; &lt;script type='text/javascript'&gt; &lt;?php echo "var categories = $jsonCats; \n"; echo "var subcats = $jsonSubCats; \n"; ?&gt; function loadCategories(){ var select = document.getElementById("categoriesSelect"); select.onchange = updateSubCats; for(var i = 0; i &lt; categories.length; i++){ select.options[i] = new Option(categories[i].val,categories[i].id); } } function updateSubCats(){ var catSelect = this; var catid = this.value; var subcatSelect = document.getElementById("subcatsSelect"); subcatSelect.options.length = 0; //delete all options if any present for(var i = 0; i &lt; subcats[catid].length; i++){ subcatSelect.options[i] = new Option(subcats[catid][i].val,subcats[catid][i].id); } } &lt;/script&gt; &lt;/head&gt; &lt;body onload='loadCategories()'&gt; &lt;select id='categoriesSelect'&gt; &lt;/select&gt; &lt;select id='subcatsSelect'&gt; &lt;/select&gt; &lt;/body&gt; &lt;/html&gt; </code></pre>
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    1. COyou will notice i selected catid from subcat as well so that you can relate subcats to cat.. this is necessary
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    2. COhey thanks for that! Worked great! How could I get it to do that for three sets of dropdowns just like that?
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    3. COYea no problem.. the third dropdown based on the second drop down I assume?
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