Re: Forms

The way I did: # taking values from DB $query = mysql_query("select * from VALUES") $result = mysql_fet_array($query);

#    create an array of all values of dropdow menu

$values = arra('value1', 'value2', 'value3', 'value4'); # create SELECT form using for loop echo '<select name="category" id="category"><option selected="true" value="Option Value">Option Value</option><option value="line">---------------------</option>'; for($i=0; $i<count($values); $i++) { # if value from DB is equal to value in SELECT form set it as SELECTED $selected = ($result['value_from_db'] == $values[$i]) ? 'SELECTED' : ''; echo '<option value='.$values[$i].' '.$selected.'> '.$values[$i].'</option>'; } echo '</select>'; and it work just fine for me :) -afan Marquez Design wrote:

Greetings, Does anyone know how to get a particular option to display in a drop menu? <select name="category" id="category"> <option selected="true" value="Option Value">Option Value</option> <option value="line">---------------------</option> <option value="value1 ">value1</option> <option value="value2 ">value2 </option> <option value="value3 ">value3</option> <option value="value4 ">value4</option> </select> The user has previously selected a category. That information is in the database. Here they are editing the record. What I would like is for the option that was selected and is in the database to be displayed as the selectd option. Does anyone know how I can do this, or can you point me in the right direction? Thank you, -- Steve Marquez Marquez Design