> Jon, John, Ifrim, the == worked ... thank you!!
> gotta read up on it now :)
>
> Cameron, dropping the extra () did not work :(
>
> thanks for the suggestions folks!
> andrew
>
Remember that '=' means assign what is on the right to what is on the left,
whereby '==' means is what is on the left
8 AM
> To: '[EMAIL PROTECTED]'; '[EMAIL PROTECTED]'
> Subject: RE: [PHP] parse error driving me nuts ...
>
>
> Instead of
>
> if ((mysql_num_rows($result)) = 1)
>
> try
>
> if ((mysql_num_rows($result)) == 1)
>
>
> HTH
> Jon
>
> hi, can anyone help me spot the parse error?
> It's throwing up on the first "if" line, and for the life of me I cannot
> find it! :)
>
> tia,
> andrew
>
>
> include("db_connect_params.inc");
> $sql="select path from PHOTO where pid =1";
> $link_id = mysql_connect($host, $usr, $pass);
> $result
Try to use == instead of =
Sorin Ifrim
- Original Message -
From: Andrew <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, February 15, 2001 5:33 PM
Subject: [PHP] parse error driving me nuts ...
> hi, can anyone help me spot the parse error?
> It&#x
Instead of
if ((mysql_num_rows($result)) = 1)
try
if ((mysql_num_rows($result)) == 1)
HTH
Jon
-Original Message-
From: Andrew [mailto:[EMAIL PROTECTED]]
Sent: 15 February 2001 15:34
To: [EMAIL PROTECTED]
Subject: [PHP] parse error driving me nuts ...
hi, can anyone help me spot
hi, can anyone help me spot the parse error?
It's throwing up on the first "if" line, and for the life of me I cannot
find it! :)
tia,
andrew
";
} else {
if (!isset($ii)) $ii = 1;
$i = 1;
while($row=mysql_fetch_array($result))
{
extract($r
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