Re: Highlighting a stored value as 'selected'
| From: | RaTT | Date: | Tue, 08 Feb 2005 10:34:09 +0000 |
| Subject: | Re: Highlighting a stored value as 'selected' | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-208242@lists.php.net to get a copy of this message | ||
Hello Alp
Try something like,
<?php
//DB fields as an array
$db_fields = array('left' => 'left',
'center' => 'center',
'right' => 'right');
function drop($array,$sel_name='',$sel_field='',$css=''){
$dropdown = "<select name=\"$sel_name\" $css>\n";
foreach($arr as $key => $val){
$sel = ($sel_field == $key)?' selected="selected"':'';
$dropdown .= "\t".'<option
value="'.$key.'"'.$sel.'>'.$val.'</option>'."\n";
}
$dropdown .= "</select>\n";
return $dropdown;
}
usage:
echo drop($db_fields,'test_select','center');
?>
HTH
Jarratt
On Tue, 8 Feb 2005 17:14:24 +0800, Alp <alp.bekisoglu@a-group-int.tv> wrote:
> Is there an easier way to display/highlight the value stored in the database
> for a select option? Such as:
> Stored value is 'center'. The statement is:
> print '<select name="colalign">';
> print '<option value="left">Left';
> print '<option value="center">Center';
> print '<option value="right">Right';
> print '</select>';
>
> I can have 3 sets of the above tied to 'if's but would rather ask for an
> easier or better way.
>
> Thanks in advance.
>
> Alp
>
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