我有一个包含菜单的对象。
我想输入category ID并获取category name，然后向后移动以找到它parents。在一个物体中这并不容易，所以我想要在此过程中抓住parents。
我遇到的问题是reset parents如果找不到最终的孩子，而且没有别的地方可以去var data = [
{
"tree_id": "10",
"name": "babies & children",
"parent": null,
"position": "1"
}, {
"tree_id": "2",
"name": "clothing",
"parent": null,
"position": "1",
"children": [{
"tree_id": "15",
"name": "kids",
"parent": "2",
"position": "3",
"children": [{
"tree_id": "78",
"name": "fourToTen",
"parent": "15",
"position": "3",
"children": [{
"tree_id": "102",
"name": "fourToSix",
"parent": "78",
"position": "3"
}]
}]
}]
}, {
"tree_id": "55",
"name": "toys",
"parent": null,
"position": "1",
"children": [{
"tree_id": "35",
"name": "lego",
"parent": "55",
"position": "3"
}]
}
];
var crumbs = [];
function getParts(data, elem) {
for(var i = 0; i < data.length; i++) {
var obj = data[i];
if(obj.children !== undefined){
/* push parent into crumbs */
crumbs.push(obj.name);
if(obj.children[0].tree_id === elem){
/* if we've found what we're looking, we're done */
crumbs.push(obj.children[0].name);
console.log(crumbs);
} else {
/* reset parents */
crumbs = []; /* <-- this is wrong here */
/* not found, keep recursing */
getParts(obj.children, elem);
}
}
}
}
/* I want this to return
[
"clothing",
"kids",
"fourToTen",
"fourToSix"
]
but it returns
[
"fourToTen",
"fourToSix"
]
*/
getParts(data, '102');。
这就是我正在尝试的：
parents
问题是，如何保存has数组，直到我在行尾并且找不到子项，然后重置它？
答案 0 :(得分：3)
假设 protected async override void OnActivityResult(int requestCode, Result resultCode, Intent data)
{
base.OnActivityResult(requestCode, resultCode, data);
if (resultCode == Result.Ok)
{
var imageData = GetBytes(ContentResolver.OpenInputStream(data.Data));
}
}
和category id = tree_id
您需要像对待树一样对待category_name = name对象，然后横向移动并跟踪父母。如果找到了某些内容，则转储您需要的信息。
所以data基本上是一个你将要横向移动的对象数组。
示例：
data<强>输出：
"use strict";
var data = [
{
"tree_id": "10",
"name": "babies & children",
"parent": null,
"position": "1"
},
{
"tree_id": "2",
"name": "clothing",
"parent": null,
"position": "1",
"children": [{
"tree_id": "15",
"name": "kids",
"parent": "2",
"position": "3",
"children": [{
"tree_id": "78",
"name": "fourToTen",
"parent": "15",
"position": "3",
"children": [{
"tree_id": "102",
"name": "fourToSix",
"parent": "78",
"position": "3"
}]
}]
}]
},
{
"tree_id": "55",
"name": "toys",
"parent": null,
"position": "1",
"children": [{
"tree_id": "35",
"name": "lego",
"parent": "55",
"position": "3"
}]
}
];
// Solution
function transverse(root, tree, targetId) {
tree.push({
catId : root.tree_id,
catName : root.name
});
/* this if() must come first otherwise fails if you want to stop before end */
if (root.tree_id === targetId) {
console.log("Found id:" + targetId+ ", name=" + root.name);
console.log("Dumping parent info => " + JSON.stringify(tree));
return tree;
}
if (root.hasOwnProperty("children") && root.children instanceof Array)
root.children.forEach(child => {
transverse(child, tree, targetId);
});
}
data.forEach(item => {
transverse(item, [], /*Looking for Id=*/"102");
});
console.log("done");
答案 1 :(得分：1)
这是一种紧凑的功能方式：
data = [{"tree_id":"10","name":"babies & children","parent":null,"position":"1"},{"tree_id":"2","name":"clothing","parent":null,"position":"1","children":[{"tree_id":"15","name":"kids","parent":"2","position":"3","children":[{"tree_id":"78","name":"fourToTen","parent":"15","position":"3","children":[{"tree_id":"102","name":"fourToSix","parent":"78","position":"3"}]}]}]},{"tree_id":"55","name":"toys","parent":null,"position":"1","children":[{"tree_id":"35","name":"lego","parent":"55","position":"3"}]}]
//
first = (ary, fn) => ary.reduce((r, x) => r || fn(x), false);
locate = (data, id) => _locate({children: data}, id, []);
_locate = (node, id, path) => node.tree_id === id ? path
: first(node.children || [], n => _locate(n, id, path.concat(n)));
res = locate(data, '102').map(n => n.name)
console.log(res);