script blocks on $news=mysql_fetch_array($result);

[Georges.Kuntz]

Hi,
Why my script blocks on : $news=mysql_fetch_array($result); ?
It seems to work properly in other sites (emplacements), The table
(lagreze_frontpage) existst and also in the right base...
************************************************** ********
<?
$query2="SELECT * FROM lagreze_frontpage WHERE id=$n";
$result=mysql_query($query2, $connect);
$news=mysql_fetch_array($result);

if ($lFormatDate=="fr")
{
list($y,$m,$d) = explode("-",$news['date']);
$sep= "-";
$news['date'] = $d.$sep.$m.$sep.$y;
}

?>

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in c:\program files\easyphp\www\lagreze\site\index.php on line 52

Georges.kuntz

[Alvaro G Vicario]

*** Georges.Kuntz wrote/escribió (Fri, 8 Oct 2004 11:49:02 +0200):

> $result=mysql_query($query2, $connect);
> $news=mysql_fetch_array($result);

You take it for granted that mysql_query() always returns a valid result.
This is what man says:

"Only for SELECT,SHOW,EXPLAIN or DESCRIBE statements mysql_query() returns
a resource identifier or FALSE if the query was not executed correctly. For
other type of SQL statements, mysql_query() returns TRUE on success and
FALSE on error. A non-FALSE return value means that the query was legal and
could be executed by the server."

What if your query fails? $result will be FALSE and then:

"Usage: array mysql_fetch_array ( resource result [, int result_type ] )"

The first argument for mysql_fetch_array() must a resource result, not a
boolean value. Thus the error.


To sum up: do error checking.


--
-- Álvaro G. Vicario - Burgos, Spain
-- Thank you for not e-mailing me your questions
--

[Georges.Kuntz]

I think I have understood.
Thanks for the answer.
Georges

"Alvaro G Vicario" <alvaro_QUITAR_REMOVE@telecomputeronline.com> a écrit
dans le message de news:1geaevevux0zb$.z5o19kyavswt$.dlg@40tude.net.. .

> *** Georges.Kuntz wrote/escribió (Fri, 8 Oct 2004 11:49:02 +0200):

> > $result=mysql_query($query2, $connect);
> > $news=mysql_fetch_array($result);

>
> You take it for granted that mysql_query() always returns a valid result.
> This is what man says:
>
> "Only for SELECT,SHOW,EXPLAIN or DESCRIBE statements mysql_query() returns
> a resource identifier or FALSE if the query was not executed correctly.

For

> other type of SQL statements, mysql_query() returns TRUE on success and
> FALSE on error. A non-FALSE return value means that the query was legal

and

> could be executed by the server."
>
> What if your query fails? $result will be FALSE and then:
>
> "Usage: array mysql_fetch_array ( resource result [, int result_type ] )"
>
> The first argument for mysql_fetch_array() must a resource result, not a
> boolean value. Thus the error.
>
>
> To sum up: do error checking.
>
>
> --
> -- Álvaro G. Vicario - Burgos, Spain
> -- Thank you for not e-mailing me your questions
> --