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HomeBackend DevelopmentPHP Tutorial用 list 处理树状数据(邻接列表)

现有一个数组

$d = array(  array( '公告', 1, 0 ),  array( '文章', 2, 0 ),  array( '文章1', 3, 2 ),  array( '文章2', 4, 2),  array( '文章1评论', 5, 3 ),  array( '文章2评论', 6, 4 ),  array( '文章1评论1', 7, 3 ),  array( '文章1评论评论', 8, 5 ),);
期望如下输出
公告文章  文章1    文章1评论      文章1评论评论    文章1评论1  文章2    文章2评论

于是可以
foreach($d as $t) list($a[$pid][$id], $id, $pid) = $t;
得到
Array(    [0] => Array        (            [1] => 公告            [2] => 文章        )    [2] => Array        (            [3] => 文章1            [4] => 文章2        )    [3] => Array        (            [5] => 文章1评论            [7] => 文章1评论1        )    [4] => Array        (            [6] => 文章2评论        )    [5] => Array        (            [8] => 文章1评论评论        ))
可以看到,数据按第3列聚类了
于是再用一个递归函数就可实现数据的展示了
function foo($ar, $pid=0, $deep=0) {  foreach($ar[$pid] as $k=>$v) {    printf("%s%s\n", str_repeat(' ', $deep), $v);    if(isset($ar[$k])) foo($ar, $k, $deep+2);  }}
调用 foo($a);


回复讨论(解决方案)

版主是个大好人


斑竹对无限级树情有独钟。
每次看都有新收获。

前排 学习!

学习了。呵呵

原来是这样表现的。

真简洁,学习了。

写的不错啊,学习了

  static void Main(string[] args)        {            double a, b, c, p, h, area;            Console.Write("请输入三角形的边A: ");            string s = Console.ReadLine();            a = double.Parse(s);            Console.Write("请输入三角形的边B: ");            s = Console.ReadLine();            b = double.Parse(s);            Console.Write("请输入三角形的边C: ");            s = Console.ReadLine();            c = double.Parse(s);            if (a > 0 && b > 0 && c > 0 && a + b > c && a + c > b && b + c > a)            {                Console.WriteLine("三角形的三边分别为:a={0},b={1},c={2}", a, b, c);                p = a + b + c;                h = p / 2;                area = Math.Sqrt(h * (h - a) * (h - b) * (h - c));                Console.WriteLine("三角形的周长={0},面积={1}",p,area);            }            else Console.WriteLine("无法构成三角形!");            Console.ReadKey();        }

不好意思,上面那个发错了。。。我不是故意的。。我是想试一下这个编辑器的功能

学习了

牛X,学习了








http://www.javadad.com

学习了  版主

不错,学习学习了

很不错,学习了

支持一下

hao......

观摩,学习,支持,接分

学习了。

谢谢楼主分享

谢谢楼主分享,学习啦!

学习了

眩技那  

简洁明了 树 总是用递归方便 

真的很不错,很有用,可以学习学习

LZ的思路和我用的思路是一样的,但代码比我的要少很多,比我高明多了。

这种算法的思路就是建立一人以parent_id为键名的二级数组,用递归调用这个数组。

array(2978) {  [0]=>  object(stdClass)#1 (4) {    ["id"]=>    string(8) "50094064"    ["subject"]=>    string(22) "在线影视/电子书"    ["parent_id"]=>    int(0)    ["type_id"]=>    int(0)  }.........

上面是组织形式,个人习惯把这分类生成对象保存到txt文本里面,生成的有自定义键名,不能直接list(),用array_values取出后,再用list还是不正常,又换了$t的元素键名,还是不正常。
//源数组 转化了数组 下面是foreach里面的内容 这个list形式不能用$t=array_values($t);list($id,$a[$pid][$t[$id],$pid,$tid) = $t;

于是又改了一下,能用了。
//源数据转换成了对象foreach($object_tmp as $t) {	$t=array($t->subject,$t->id,$t->parent_id);	list($a[$pid][$id],$id,$pid) = $t;}


不过,换了一位置。
//源数据转换成了对象foreach($object_tmp as $t) {	$t=array($t->id,$t->subject,$t->parent_id);	list($id,$a[$pid][$id],$pid) = $t;}

又不能用了,看了半天list的说明,也没能找到答案

附上本人原来的做法:
对于list转化二维数组的,个人采取比较笨拙的方法,源数组与上面的类似

//以父级id为键名的  更多一维的(一般3维)数组		$tmp=array();		foreach ($item_category as $it){			if( count($tmp[$it["parent_id"]]) ){				$tmp[$it["parent_id"]][count($tmp[$it["parent_id"]])]=$it;			}else{				$tmp[$it["parent_id"]][0]=$it;			}		}

$tmp就是相当于LZ方法的数组$a

从项目文件里面拿出来的,写在控制器里面的,就不改了
//用于存放数据(整理好的)的公共变量	var $sorta=array();	//树形排序核心部分  pid:父级起始  tmp以父级id为第一维的数组	public function get_all_($pid,$tmp){			$tt=$tmp[$pid];			foreach($tt  as $ttt){				$this->sorta[count($this->sorta)]=$ttt;				$this->get_all_($ttt["id"],$tmp);			}	}

换成javascript的就最好

学习了,不错

不错不错,这个必须收藏一下。

好   很简单  方便了

错不错,这个必须收藏一下。

学习了    方法很简便

学习了    方法很简便

学习了

已阅..

前面也做过这个,不过是直接写递归函数从数据库取的。

很?大,又?到一招

观摩,学习,支持,接分

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