php - 为什么这个 MySQL 连接语句不起作用？

Question

这是我的代码:

$query2 = mysql_query("SELECT * FROM categories WHERE parent = $id JOIN SELECT * FROM posts WHERE main_nav_page = '$idTwo'");

                    while ($row2 = mysql_fetch_assoc($query2)) {
                        $id   = $row2['id'];
                        $name = $row2['name'];
                        $slug = $row2['slug'];
                        $subMenuOrder = $row2['sub_menu_order'];

                        echo "<tr>\n";
                        echo "<td>&nbsp; -- $name</td>\n";
                        echo "</tr>\n";
                    }

我的语法有误吗？

编辑:

错误信息是:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/studentw/public_html/new_nav.php on line 30

Answer 1

取而代之的是你可能想要更像的东西:

SELECT * 
FROM 
       categories c 
    INNER JOIN 
       posts p ON c.categoryid = p.categoryid 
WHERE 
       c.parent = $id 
       AND p.main_nav_page = '$idTwo'; 

请注意，表是连接的，而不是选择语句。此外，连接在 FROM 子句中指定。

试试这个:

$results = mysql_query("query here") or die(mysql_error());