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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