Not to worry - manage to solve the problem using the line below
$date=date('H:i');
echo "<input name=\"timeout\" type=\"text\" value=\"$date\">";
Thanks for all the input
-----Original Message-----
From: Baiocchi, Justin (CSIRO IT, Armidale)
Sent: Friday, 6 August 2004 12:50 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Php if statement in a form
Well, this works: so you were all on the right track
if($row['timeout']=="")
{
print "<input name='timeout' type='text' value='"echo
date('H:i')"'>";
}else{
print $row["timeout"];
}
But now I get a parse error on the <input> line
-----Original Message-----
From: Ng Hwee Hwee [mailto:[EMAIL PROTECTED]
Sent: Friday, 6 August 2004 12:30 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Php if statement in a form
what about this?
if(empty($row["timeout"]))
----- Original Message -----
From: "Doug Thompson" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Friday, August 06, 2004 10:18 AM
Subject: Re: [PHP-DB] Php if statement in a form
> if (is_null($row["timeout"]).....)
>
> [EMAIL PROTECTED] wrote:
> > Still no joy. The parse error actually refers to the -- if
> > ($row["timeout"] IS NULL); --- line.
> >
> >
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