xx_princess_xx Posted May 9, 2009 Share Posted May 9, 2009 I'm really new at the whole PHP thing, i have created a form that allows users to update a record from a database (mysql) at the moment the user has to select the ID they want to update what i am trying to do is create a drop down list with the available ID's for the user to select <?php $con = mysql_connect("localhost", "root", ""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("comreg098hassan", $con); $result = mysql_query("SELECT AID FROM Agents"); while($row_ID = mysql_fetch_array($result)) { echo $row_ID['AID']; echo "<br />"; }mysql_close($con); ?> <form> <select name="aid"><?php echo $row_ID['AID']; ?></select> </form> this code displays all the ID's on the page, but doesnt display in the drop down list, I would really appreciate your help thank-you Link to comment https://forums.phpfreaks.com/topic/157501-solved-selecting-an-option-from-a-drop-down-list-from-mysql-database-using-php/ Share on other sites More sharing options...
wildteen88 Posted May 9, 2009 Share Posted May 9, 2009 The correct syntax for a drop down menu is <select name="aid"> <option value="value1">Value 1</option> <option value="value2">Value 2</option> <option value="value3">Value 3</option> </select> In PHP it'll be // start of form/menu echo '<form> <select name="aid">'; // produce the <option>'s while($row_ID = mysql_fetch_array($result)) { echo '<option value="'.$row_ID['AID'].'">'.$row_ID['AID'].'</option>'; } // end the menu/form echo '</select> </form>'; Link to comment https://forums.phpfreaks.com/topic/157501-solved-selecting-an-option-from-a-drop-down-list-from-mysql-database-using-php/#findComment-830393 Share on other sites More sharing options...
xx_princess_xx Posted May 9, 2009 Author Share Posted May 9, 2009 thanks Link to comment https://forums.phpfreaks.com/topic/157501-solved-selecting-an-option-from-a-drop-down-list-from-mysql-database-using-php/#findComment-830398 Share on other sites More sharing options...
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