array (
array("level" => '1', "caption" => "一级菜单1"),
array("leve1" => '1', "caption" => "一级菜单2"),
array("leve1" => '2', "caption" => "二级菜单1"),
array("leve1" => '2', "caption" => "二级菜单2"),
array("leve1" => '1', "caption" => "一级菜单3"),
)
其中level = 2为其上方最近leve1 = 1 的子菜单,不排除会有level = 3的子菜单(从属于level = 2)现在想弄成下面的形式,想了好久也没什么好方法,同事建议我用递归,可递归小弟不太懂,PHP初接触,给各位大拿添麻烦了
array (
array("level" => '1', "caption" => "一级菜单1"),
array("leve1" => '1', "caption" => "一级菜单2",
"sub"=>array("leve1" => '2', "caption" => "二级菜单1"),
"sub"=>array("leve1" => '2', "caption" => "二级菜单2")
),
array("leve1" => '1', "caption" => "一级菜单3"),
)
php 菜单 递归
array("level" => '1', "caption" => "一级菜单1"),
array("leve1" => '1', "caption" => "一级菜单2"),
array("leve1" => '2', "caption" => "二级菜单1"),
array("leve1" => '2', "caption" => "二级菜单2"),
array("leve1" => '1', "caption" => "一级菜单3"),
)
其中level = 2为其上方最近leve1 = 1 的子菜单,不排除会有level = 3的子菜单(从属于level = 2)现在想弄成下面的形式,想了好久也没什么好方法,同事建议我用递归,可递归小弟不太懂,PHP初接触,给各位大拿添麻烦了
array (
array("level" => '1', "caption" => "一级菜单1"),
array("leve1" => '1', "caption" => "一级菜单2",
"sub"=>array("leve1" => '2', "caption" => "二级菜单1"),
"sub"=>array("leve1" => '2', "caption" => "二级菜单2")
),
array("leve1" => '1', "caption" => "一级菜单3"),
)
php 菜单 递归
搜“php 无限分类”http://bbs.csdn.net/topics/360028778
http://bbs.csdn.net/topics/250025103
http://bbs.csdn.net/topics/320022124
……你的数组里面
level 相当于 pid(把=1改为=0)
caption 拆分为 id 和 name
照着上面抄吧,先完成了,有空自己再理解人家的递归方法偶忙着做其他事情
<?
$arr = array(
array('id'=>1,'city_name'=>'中国','rel_id'=>'1','pid'=>0),
array('id'=>2,'city_name'=>'广东','rel_id'=>'1-2','pid'=>1),
array('id'=>3,'city_name'=>'深圳','rel_id'=>'1-2-3','pid'=>2),
array('id'=>4,'city_name'=>'广州','rel_id'=>'1-2-4','pid'=>2)
);function find_subclass( $pid ){
global $arr;
$__arr = array();
foreach ( $arr as $k=>$v )
{
if( $v['pid']==$pid )$__arr[] = $v;
}
return $__arr;
}function tree_subclass($pid=0){
$__arr = array();
$__arr = find_subclass($pid);
if( !empty($__arr) ){
foreach ( $__arr as $k=>$v )
{ $__arr[$k]['subclass'] = tree_subclass($v['id']);
}
}
return $__arr;
}var_dump(tree_subclass(0));?>运行一下。主要是加了个用id及父id去将节点关系联系起来的。比如 广州的父id是广东id,广东的父id是中国的id。理解下这样的结构就好入手了。
array (
array("level" => '1', "caption" => "一级菜单1"),
array("leve1" => '1', "caption" => "一级菜单2",
"sub"=>array("leve1" => '2', "caption" => "二级菜单1"),
"sub"=>array("leve1" => '2', "caption" => "二级菜单2")
),
array("leve1" => '1', "caption" => "一级菜单3"),
)
任何数组都不会在同一结构中出现同名的键你先调整好需求,再提问
实在对不起楼上几位,刚看到版主的回复,发帖的时候没有检查清楚,确实是数组结构错误了,十分抱歉期望的数组结构应该是下面这样
array (
array("level" => '1', "caption" => "一级菜单1"),
array("leve1" => '1', "caption" => "一级菜单2",
"sub"=>array(
array("leve1" => '2', "caption" => "二级菜单1"),
array("leve1" => '2', "caption" => "二级菜单2")
)
),
array("leve1" => '1', "caption" => "一级菜单3"),
)
大概明白了一点,主要是要在父菜单与子菜单之间添加一个关联用的节点,也就是pid,id,
在递归的时候利用父节点对应的子节点id找到子节点中的菜单名理解的不知道是不是有问题,请指教
$data = array (
array("level" => '1', "caption" => "一级菜单1"),
array("level" => '1', "caption" => "一级菜单2"),
array("level" => '2', "caption" => "二级菜单1"),
array("level" => '2', "caption" => "二级菜单2"),
array("level" => '3', "caption" => "三级菜单1"),
array("level" => '1', "caption" => "一级菜单3"),
);
$last = array(0=>array());
foreach($data as $k=>&$v){
$v['level'] = (int)$v['level'];
$last = array_slice($last, 0, $v['level']);
$last[$v['level']] = &$v;
$last[$v['level']-1]['sub'][] = &$last[$v['level']];
}
var_dump( $last[0]);
这种方法的核心是引用传递,思路还是递归,只是换了种方式实现
感谢snmr_com提供的链接,
anydy2008提供实际代码帮助我理解迭代, dream1206和版主分别提供的两种实现方式,