Posted by Miles Thompson on 10/15/66 11:11

At 12:04 PM 3/21/2005, 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

Steve,

These can be a PITA. Here's how I've done it.

1. Create a function to determine which is selected:

function is_selected( $option_expression )
{
if ($option_expression)
{
$retval = "selected";
}
else
{
$retval = "";
}
return $retval;
} // end of function is_selected

2. Build your list of selection, in this case a list of classifications
fetched from a database;
$option_class is the variable holding the list of options:

$sql = "select * from class";
$result = mysql_query( $sql );
while( $row = mysql_fetch_array( $result ) )
{
$nClassKey = $row["nClassKey"];
$cClass = $row["cClass"];
$selectval = is_selected( $nClassKey == $nClass );
$option_class .= "<option value = \"$nClassKey\"
$selectval >$cClass</option>";
} // end while $row = mysql_fetch...

3. Then display it this way in the form:

<Select name="nClass">
<? echo $option_class ?>
</select>

Hope this is helpful - Miles

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