PHP树更改父结构 [英] PHP Tree Change Parent Structure
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问题描述
我想以其他方式(如Wordpress)更改父级的结构,如果父级是class,则添加类.
当前结构:
< ul>< li>我的清单< ul>< li>菜单1</li>< ul>< li>菜单2</li>< ul>< li>菜单3</li></ul>< li>菜单4</li></ul>< li>菜单5</li></ul></li>
我想要什么:
< ul>< li>我的清单< li>菜单1< ul class ="sub-menu">< li>菜单2< ul class ="sub-menu">< li>菜单3</li></ul></li>< li>菜单4</li></ul></li>< li>菜单5</li></li></ul>
代码:
$ lists = array(大批('id'=>1,'父母'=>0,'名称'=>菜单1",),大批('id'=>2,'父母'=>1,'名称'=>菜单2",),大批('id'=>3,'父母'=>2,'名称'=>菜单3",),大批('id'=>4,'父母'=>1,'名称'=>菜单4",),大批('id'=>5,'父母'=>0,'名称'=>菜单5",));?>< ul>< li>我的清单<?php函数make_list($ lists,$ parent = 0){$ children = array_filter($ lists,function($ v)use($ parent){return $ v ['parent'] == $ parent;});if(!count($ children))返回;回声< ul>";foreach($ children as $ child){回声< li> {$ child ['name']}</li>";make_list($ lists,$ child ['id']);}回声</ul>";}make_list($ lists);?></li></ul>
解决方案
您可以对跟踪菜单级别的功能进行少量更改来做到这一点:
function make_list($ lists,$ parent = 0,$ level = 0){$ children = array_filter($ lists [0],function($ v)use($ parent){return $ v ['parent'] == $ parent;});如果(!count($ children))返回;回声< ul".($ level> 0?'class ="sub-menu"':'').> \ n";foreach($ children as $ child){回声< li> {$ child ['name']}</li> \ n";make_list($ lists,$ child ['id'],++ $ level);}回声</ul> \ n";}make_list($ lists);
I want to change structure of parent other way like Wordpress and add class if is parent.
Current Structure:
<ul>
<li>My List
<ul>
<li>Menu 1</li>
<ul>
<li>Menu 2</li>
<ul>
<li>Menu 3</li>
</ul>
<li>Menu 4</li>
</ul>
<li>Menu 5</li>
</ul>
</li>
What i want:
<ul>
<li>My List
<li>Menu 1
<ul class="sub-menu">
<li>Menu 2
<ul class="sub-menu">
<li>Menu 3</li>
</ul>
</li>
<li>Menu 4</li>
</ul>
</li>
<li>Menu 5</li>
</li>
</ul>
Code:
$lists = array(
array(
'id' => 1,
'parent' => 0,
'name' => 'Menu 1',
),
array(
'id' => 2,
'parent' => 1,
'name' => 'Menu 2',
),
array(
'id' => 3,
'parent' => 2,
'name' => 'Menu 3',
),
array(
'id' => 4,
'parent' => 1,
'name' => 'Menu 4',
),
array(
'id' => 5,
'parent' => 0,
'name' => 'Menu 5',
)
);
?>
<ul>
<li>My List
<?php
function make_list($lists, $parent = 0) {
$children = array_filter($lists, function ($v) use($parent) { return $v['parent'] == $parent; });
if(!count($children)) return;
echo "<ul>";
foreach ($children as $child) {
echo "<li>{$child['name']}</li>";
make_list($lists, $child['id']);
}
echo "</ul>";
}
make_list($lists);
?>
</li>
</ul>
解决方案
You can do that with a minor change to the function to track the menu level:
function make_list($lists, $parent = 0, $level = 0) {
$children = array_filter($lists[0], function ($v) use($parent) { return $v['parent'] == $parent; });
if (!count($children)) return;
echo "<ul" . ($level > 0 ? ' class="sub-menu"' : '') . ">\n";
foreach ($children as $child) {
echo "<li>{$child['name']}</li>\n";
make_list($lists, $child['id'], ++$level);
}
echo "</ul>\n";
}
make_list($lists);
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