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> -- $name</td>\n";
echo "</tr>\n";
}
我的语法错了吗?
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
答案 0 :(得分:4)
而不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select语句。此外,联接在FROM
子句中指定。
试试这个:
$results = mysql_query("query here") or die(mysql_error());
答案 1 :(得分:1)
我认为语法错误很小。看起来你错过了WHERE parent
- 子句中“$ id”周围的单引号('')。应该是这样的,我猜:
SELECT * FROM categories WHERE parent = '$id' ...