wright67uk Posted April 20, 2011 Share Posted April 20, 2011 Ive been scanning over and over this code and I cant work out why my 2nd drop down menu doesnt have unique values. Please if anyone can give me guidance, you never know I may be able to get rid of my headache! <body> <p> <form action="" method="post"> <select name="drop_1" id="drop_1"> <option value="" selected="selected" disabled="disabled">Select a Category</option> <?php getTierOne(); ?> </select> <span id="wait_1" style="display: none;"> <img alt="Please Wait" src="ajax-loader.gif"/> </span> <span id="result_1" style="display: none;"></span> </form> </p> <p> <?php if(isset($_POST['submit'])){ $drop = $_POST['drop_1']; $tier_two = $_POST['Subtype']; echo "You selected "; echo $drop." & ".$tier_two; } ?> </body> <?php function getTierOne() { $result = mysql_query("SELECT DISTINCT Type FROM business") or die(mysql_error()); while($tier = mysql_fetch_array( $result )) { echo '<option value="'.$tier['Type'].'">'.$tier['Type'].'</option>'; } } if($_GET['func'] == "drop_1" && isset($_GET['func'])) { drop_1($_GET['drop_var']); } function drop_1($drop_var) { include_once('db.php'); $result = mysql_query("SELECT DISTINCT Subtype FROM business WHERE Type='$drop_var'") or die(mysql_error()); echo '<select name="Subtype" id="Subtype"> <option value=" " disabled="disabled" selected="selected">Choose one</option>'; while($drop_2 = mysql_fetch_array( $result )) { echo '<option value="'.$drop_2['Subtype'].'">'.$drop_2['Subtype'].'</option>'; } echo '</select> '; echo '<input type="submit" name="submit" value="Submit" />'; } ?> Quote Link to comment https://forums.phpfreaks.com/topic/234294-why-doesnt-my-2nd-drop-down-have-distinct-values/ Share on other sites More sharing options...
btherl Posted April 20, 2011 Share Posted April 20, 2011 Can you post the raw output of that script? Quote Link to comment https://forums.phpfreaks.com/topic/234294-why-doesnt-my-2nd-drop-down-have-distinct-values/#findComment-1204262 Share on other sites More sharing options...
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