I tried to use the below but it does not work. Please help!
-----Original Message-----
From: Michael Sweeney [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 22, 2003 10:29 AM
To: Ben C.
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] Select value for driopdown box
Assuming that you know which state is selected before you build the
select list:
$sel_state = $_REQUEST['state'];
...
while($row = mysql_fetch_array($result)) {
$buyerid = $row['buyerid'];
$state = $row['state'];
$selected = $state == $sel_state ? "selected=\"selected\"" : "";
$option_block .= "<option value=\"$state\" $selected>".
"$state</option>\n";
}
That will put the string "selected=\"selected\"" (xhtml compliant) into
the option statement for the state that was selected.
..michael..
On Wed, 2003-01-22 at 10:14, Ben C. wrote:
> I am using the query below to edit a record. I want to edit the field
which has a list of states in a dropdown box. I want to have the state
that is in the selected field shown as the selected state. How would I do
this?
>
>
> Query
> -------------------------
> <?
> $sql = "SELECT *
> FROM $table_name
> WHERE buyerid = \"$buyerid\"
> ";
>
> $result = @mysql_query($sql,$connection) or die(mysql_error());
>
>
> while ($row = mysql_fetch_array($result)) {
> $buyerid = $row['buyerid'];
> $state = $row['state'];
>
> $option_block .= "<option value=\"$state\">$state</option>";
> }
>
> $display_block = "
>
> <select name=\"state\" class=Pgtext>
> $option_block
> </select>
> ?>
>
> <tr>
> <td width="258" class="Pgtext" height="21">State</td>
> <td width="259" height="21"><? echo $display_block; ?></td>
> </tr>
>
> --------------------------------
--
Michael Sweeney <[EMAIL PROTECTED]>
Verisity Design, Inc
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