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Posted by bettina on 11/29/05 23:25
I've tried the following:
$search_breweries = mysql_query("SELECT b.BREWERY_CODE, b.BREWERY,
COUNT(c.ID) FROM breweries as b JOIN coasters as c ON
b.BREWERY_CODE = c.BREWERY_CODE Group by b.BREWERY_CODE order by
COUNT(c.ID) DESC ");
$i = 0;
while ($row = mysql_fetch_array($search_breweries, MYSQL_NUM))
{
$mybreweries[$i][0] = $row[0];
$mybreweries[$i][1] = $row[1];
$mybreweries[$i][2] = $row[2];
$i = $i + 1;
}
......
And I get the following message:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result resource in....
Sean schrieb:
> Follow this up in the sql forums, but you want something like this:
>
> SELECT b.BREWERY_CODE, b.BREWERY, COUNT(c.id) as num_coaster
> FROM breweries b LEFT JOIN coasters c ON b.BREWERY_CODE =
> c.BREWERY_CODE
> Group by b.BREWERY_CODE
> order by COUNT(c.id) DESC
>
> UNTESTED!
>
> This will get you what you want in one pass. (i think)
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