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Posted by Andy Hassall on 12/19/05 22:47
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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