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Posted by Alp on 02/08/05 23:38
Hi Jarratt,
Thank you for your advice. I will try that and come back with the result.
Alp
"RaTT" <shaggie@gmail.com> wrote in message
news:61f5e49905020802347d67bb4b@mail.gmail.com...
> 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
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
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> >
> >
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