我正在尝试为 CMS 系统创建一个动态菜单这是我的SQL表,它有一个额外的字段,称为内容
parent_id order id titel
0 1 2 Home
0 2 4 Contact
0 3 9 stuff
2 1 1 more stuff
2 2 3 way more stuff
9 1 10 all stuff
我想要这个结果:
home
more stuff
way more stuff
contact
stuff
all stuff
这就是我如何称呼菜单:
$resultPaginas = mysqli_query($mysqli, "select id, titel, parent_id, volgorde from TABLE order by order");
if (!$resultPaginas) {
die(mysqli_error($mysqli));
}
这将显示按parent_id顺序排序的表。
一切都完成了,我知道我需要做一个 while 循环,但是我如何执行正确的 while 循环以获得我想要的结果?
附言这将是一个菜单
使用此递归函数构建菜单:
function buildMenu($parentId = 0) {
$menu = array();
$result = mysqli_query($mysqli, "SELECT `id`, `titel`, `parent_id`, `volgorde` FROM [TABLE] WHERE `parent_id` = $parentId ORDER BY `order`");
if ($result !== false) {
while ($page = $result->fetch_assoc()) {
$page["children"] = buildMenu($page["id"]);
$menu[] = $page;
}
}
return $menu;
}
像这样调用递归函数
$resultPaginas = buildMenu();
$resultPaginas的示例内容:
Array
(
[0] => Array
(
[id] => 1
[parent_id] => 0
[titel] => home
[children] => Array
(
[0] => Array
(
[id] => 2
[parent_id] => 1
[titel] => more stuff
)
[1] => Array
(
[id] => 3
[parent_id] => 1
[titel] => way more stuff
)
)
)
[1] => Array
(
[id] => 4
[parent_id] => 0
[titel] => contact
)
)
首先尝试从父 id 为 0 的表中全部选择
并循环播放
样本:
foreach($parent as $par){
//Then you need to write another query to take all submenus for current parentid menu
$submenus= mysqli_query($mysqli, "select id, titel, parent_id, volgorde from TABLE where parent_id = $par.id order by order");
//Then again loop trough
foreach($submenus as $sub){
// and add existing submenus to parent menu
}
}
注意:如果语法不正确,这只是概念抱歉。