Posted by briansmccabe on 12/20/05 00:02

Andy Hassall wrote:
> On 19 Dec 2005 12:24:35 -0800, briansmccabe@gmail.com wrote:
>
> Kimmo originally wrote:
>
> >>$result = mysql_query('SELECT COUNT(*) AS foo FROM table') or
> >>die(mysql_error());
>
> But you've used:
>
> >$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
> >AND format = 'reg'" or die(mysql_error());
>
> You are missing the call to mysql_query().
>
> >$result6 = mysql_fetch_array($query6);
>
> ... so $query6 just contains the SQL string, and not a MySQL result set
> resource identifier. You're not actually executing the SQL anywhere, or if you
> are in the rest of the code, you haven't done the error checking there.
>
> >later on the page:
> >
> >echo $result6[quant]
>
> This should be: $result6['quant'].
>
> See:
>
> http://www.php.net/manual/en/language.types.array.php#language.types.array.donts
>
> p.s. You've got the posting style nearly right - however you should put your
> new message _under_ the old one, not above, so the whole message makes sense
> read on its own. You have "top posted".
>
> --
> Andy Hassall :: andy@andyh.co.uk :: http://www.andyh.co.uk
> http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

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