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
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