How do I then cycle through this array to build the select box? I
currently use this which works fine when I hard code the values..ie.
$yesno=array("Yes"=>"1","No"=>"0") but doesn't seem to work using your
first option.
Here is the form select box code
while($value=each($array))
{
$res .= "<option " .$selected. ($selected == $value['value']
? " selected=\"selected\"" : '') . "
value=".$value['value'].">".$value['key']."</option>\n";
}
Thanks,
Eddie
-----Original Message-----
From: Jeroen Serpieters [mailto:[EMAIL PROTECTED]
Sent: Friday, May 21, 2004 12:16 AM
To: Edward Peloke
Cc: 'php-general'
Subject: Re: [PHP] array_push
On Fri, 21 May 2004, Edward Peloke wrote:
>
> $sql=new Database();
> $sql->query("select fname,lname, id from clients");
> $clients[]=array();
> while($sql->nextRecord()){
> array_push($clients,
$sql->getField('id')=>$sql->getField('fname'));
> } // while
>
> Parse error: parse error, unexpected T_DOUBLE_ARROW in
> C:\obox\Apache2\htdocs\nh_vacdest\admin\test.php on line 38
>
If you want to do it the way you're doing now you should do ik like
this:
array_push($clients,
array($sql->getField('id')=>$sql->getField('fname')));
Another possible solution is like this:
$clients[$sql->getField('id')] = $sql->getField('fname');
--
Jeroen
There are only two kinds of programming languages: those people always
bitch about and those nobody uses.
-- Bjarne Stroustrup
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