Re: A real problem
| From: | John W. Holmes | Date: | Mon, 26 Apr 2004 20:05:50 +0000 |
| Subject: | Re: A real problem | ||
| References: | 1 2 3 4 5 6 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-184696@lists.php.net to get a copy of this message | ||
From: "Daniel Clark" <dclark@nwlink.com>
> If I was displaying it. :-)
> Here is a "ruff" guess.
>
> <?
> $query = "SELECT id, name FROM shop_brands";
> $result = mysql_query($query) or die("error in query - $query
> ".mysql_error());
>
> while($row = mysql_fetch_row($result)) {
>
> list($brand_id, $brand_name) = $row;
>
> $query2 = "SELECT * FROM shop_models WHERE brand_id = $brand_id";
Not sure what you're trying to accomplish here, but anytime you have nested
SELECT queries like this, you're begging for it to be replaced with a single
SELECT query with a JOIN.
SELECT sb.id, sb.name, sm.* FROM shop_brands sb, shop_models sm WHERE sb.id
= sm.brand_id
Now you have everything you need in a single, more efficienct, query.
---John Holmes...