javascript – 查找每个元素的所有父元素

我正在尝试在不使用url(路由提供程序)的情况下创建面包屑,而不使用jQuery.
我有一棵树

Humans
Trees
Animals
    Cats
    Lions
    Dogs
       Terrier 
       Bulldog
       Cocker
Cars

当我点击Cocker进行显示时,我想要

Animals / Dogs / Cocker

所以,我创建了一个递归函数,以便为我点击的每个元素找到父/父,但它无法正常工作.它发现一个元素有一个父元素,它也找到了元素的第一个父元素,但它没有显示第二个父元素.例如,而不是

Animals / Dogs / Cocker

表明

Dogs / Cocker

那是我的功能

var count = 0;
function iterate(obj) {
    for(var key in obj) {
      var elem = obj[key];
      if(key === "children") {
        count++;
      }
      if(typeof elem === "object") {
        if(elem.children === undefined){
          elem.children = 1;
        }
        if(elem.children.length !==1){
          iterate(elem);
          $scope.showTrail = elem.children;
          $scope.elem = elem;
        }
      }
     }
    if($scope.elem === undefined){
      $scope.elem = {};
      $scope.elem.children = {};
      $scope.elem.roleName = {};
    }

    for (var i = 0; i<$scope.elem.children.length; i++) {
       if($scope.elem.children[i].roleName === selNode.roleName) {
          console.log($scope.elem.roleName + " is a parent of " + selNode.roleName);
       }
    }

  }
iterate($scope.treeData);

这就是JSON

     [
          { "roleName" : "Humans", "roleId" : "role2", "children" : [
           { "roleName" : "", "roleId" : "role11", "children" : [] }
          ]},
          { "roleName" : "Trees", "roleId" : "role2", "children" : [
           { "roleName" : "", "roleId" : "role11", "children" : [] }
          ]},
          { "roleName" : "Animals", "roleId" : "role2", "children" : [
            { "roleName" : "Cats", "roleId" : "role11", "children" : [
             { "roleName" : "", "roleId" : "role11", "children" : [] }
            ]},
            { "roleName" : "Lions", "roleId" : "role11", "children" : [
             { "roleName" : "", "roleId" : "role11", "children" : [] }
            ]},
            { "roleName" : "Dogs", "roleId" : "role11", "children" : [
              { "roleName" : "Terrier", "roleId" : "role11", "children" : [
               { "roleName" : "", "roleId" : "role11", "children" : [] }
              ]},
              { "roleName" : "Bulldog", "roleId" : "role11", "children" : [
               { "roleName" : "", "roleId" : "role11", "children" : [] }
              ]},
              { "roleName" : "Cocker", "roleId" : "role11", "children" : [
               { "roleName" : "", "roleId" : "role11", "children" : [] }
              ]},
            ]}
          ]},
          { "roleName" : "Cars", "roleId" : "role2", "children" : [
           { "roleName" : "", "roleId" : "role11", "children" : [] }
          ]}
     ]

请任何帮助,任何想法.非常感谢你.

解决方法:

您正在迭代树,但如果您不保留某些信息,它将无济于事.解决您问题的最简单方法是构建指向其父节点的所有节点的索引.

如果roleName在整个树中具有唯一值,则此代码将起作用：

var tree = [
    { "roleName" : "Humans", "roleId" : "role2", "children" : []},
    { "roleName" : "Trees", "roleId" : "role2", "children" : []},
    { "roleName" : "Animals", "roleId" : "role2", "children" : [
        { "roleName" : "Cats", "roleId" : "role11", "children" : []},
        { "roleName" : "Lions", "roleId" : "role11", "children" : []},
        { "roleName" : "Dogs", "roleId" : "role11", "children" : [
            { "roleName" : "Terrier", "roleId" : "role11", "children" : []},
            { "roleName" : "Bulldog", "roleId" : "role11", "children" : []},
            { "roleName" : "Cocker", "roleId" : "role11", "children" : []},
        ]}
    ]},
    { "roleName" : "Cars", "roleId" : "role2", "children" : []}
];

var index = {};

function buildIndex(root, children) {
    for(var i in children) {
        index[children[i].roleName] = root;
        buildIndex(children[i].roleName, children[i].children);
    }
}

buildIndex("Root", tree);

function getPath(leaf) {
    return index[leaf] ? getPath(index[leaf]).concat([leaf]) : [leaf];
}

getPath("Bulldog");// returns ["Root", "Animals", "Dogs", "Bulldog"]

JSFiddle：http://jsfiddle.net/E49Ey/

但是它与Angular无关,除了数据驻留在范围内.如果你有一个根据这个数据构建的DOM树,那么你可以通过上升树从DOM获得面包屑.

标签：breadcrumbs,javascript,angularjs
来源： https://codeday.me/bug/20190831/1774887.html