Javascript 递归地遍历一个对象以找到子对象，一路上接父母、祖父母等
我有一个由菜单组成的对象 我想输入一个Javascript 递归地遍历一个对象以找到子对象，一路上接父母、祖父母等,javascript,Javascript,我有一个由菜单组成的对象 我想输入一个类别ID并获取类别名称，然后向后移动找到它的父项。这在一个对象中并不容易，所以我想在这个过程中捕捉父对象 我遇到的问题是，当找不到最终子项且无处可去时，如何重置父项 这就是我正在尝试的： var数据=[ { “树id”：“10”， “姓名”：“婴儿和儿童”， “父项”：空， “位置”：“1” }, { “树id”：“2”， “姓名”：“服装”， “父项”：空， “职位”：“1”， “儿童”：[{ “树id”：“15”， “姓名”：“孩子们”， “家长”：“
类别ID类别名称父项父对象
我遇到的问题是，当找不到最终子项且无处可去时，如何重置父项
这就是我正在尝试的：
var数据=[
{
“树id”：“10”，
“姓名”：“婴儿和儿童”，
“父项”：空，
“位置”：“1”
}, {
“树id”：“2”，
“姓名”：“服装”，
“父项”：空，
“职位”：“1”，
“儿童”：[{
“树id”：“15”，
“姓名”：“孩子们”，
“家长”：“2”，
“职位”：“3”，
“儿童”：[{
“树id”：“78”，
“名称”：“fourToTen”，
“家长”：“15”，
“职位”：“3”，
“儿童”：[{
“树id”：“102”，
“名称”：“fourToSix”，
“家长”：“78”，
“位置”：“3”
}]
}]
}]
}, {
“树id”：“55”，
“名称”：“玩具”，
“父项”：空，
“职位”：“1”，
“儿童”：[{
“树id”：“35”，
“名称”：“乐高”，
“家长”：“55”，
“位置”：“3”
}]
}
];
var-crumbs=[]；
函数getParts（数据、元素）{
对于（变量i=0；i类别id=tree\u id
和类别名称=name
你需要像对待一棵树一样对待你的数据
对象，然后横穿它，沿途跟踪父对象。如果发现了什么，那么就转储你需要的信息
因此，数据
基本上是一组您将要遍历的对象
示例：
"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");
Found id:102, name=fourToSix
Dumping parent info =>
[
{"catId":"2","catName":"clothing"},
{"catId":"15","catName":"kids"},
{"catId":"78","catName":"fourToTen"},
{"catId":"102","catName":"fourToSix"}]
]
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");
Found id:102, name=fourToSix
Dumping parent info =>
[
{"catId":"2","catName":"clothing"},
{"catId":"15","catName":"kids"},
{"catId":"78","catName":"fourToTen"},
{"catId":"102","catName":"fourToSix"}]
]
假设category id=tree\u id
和category\u name=name
你需要像对待一棵树一样对待你的数据
对象，然后横穿它，沿途跟踪父对象。如果发现了什么，那么就转储你需要的信息
因此，数据
基本上是一组您将要遍历的对象
示例：
"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");
Found id:102, name=fourToSix
Dumping parent info =>
[
{"catId":"2","catName":"clothing"},
{"catId":"15","catName":"kids"},
{"catId":"78","catName":"fourToTen"},
{"catId":"102","catName":"fourToSix"}]
]
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");
Found id:102, name=fourToSix
Dumping parent info =>
[
{"catId":"2","catName":"clothing"},
{"catId":"15","catName":"kids"},
{"catId":"78","catName":"fourToTen"},
{"catId":"102","catName":"fourToSix"}]
]
以下是一种简洁实用的方法：
data=[{“tree\u id”：“10”，“name”：“baby&children”，“parent”：null，“position”：“1”}，{“tree\u id”：“2”，“name”：“1”，“children”：[{“tree\u id”：“15”，“name”：“children”，“parent”：“2”，“position”：“3”，“children”：[{“tree\u id”：“78”，“name”：“fourToTen”，“parent”：“15”，“position”：“3”，“children”：[{“tree\u id”：“102”，“name”：“fourToSix”，“父项”：“78”，“位置”：“3”}]}]}，{“树id”：“55”，“名称”：“玩具”，“父项”：null，“位置”：“1”，“子项”：[{“树id”：“35”，“名称”：“乐高”，“父项”：“55”，“位置”：“3”}]
//
第一个=（ary，fn）=>ary.reduce（（r，x）=>r | | fn（x），false）；
locate=（数据，id）=>u locate（{children:data}，id，[]）；
_locate=（节点、id、路径）=>node.tree\u id==id？路径
：第一个（node.children | |[]，n=>_locate（n，id，path.concat（n））；
res=locate（数据'102'）.map（n=>n.name）
console.log（res）；
这里有一种简洁的功能方式：
data=[{“tree\u id”：“10”，“name”：“baby&children”，“parent”：null，“position”：“1”}，{“tree\u id”：“2”，“name”：“1”，“children”：[{“tree\u id”：“15”，“name”：“children”，“parent”：“2”，“position”：“3”，“children”：“fourToTen”，“parent”：“15”，“position”：“3”，“children”：[{“tree\u id”：“102”，“name”：“fourToSix”父项“：”78“，”位置“：”3“}]}]，{”树id“：”55“，”名称“：”玩具“，”父项“：”空，“位置“：”1“，”子项“：[{”树id“：”35“，”名称“：”乐高“，”父项“：”55“，”位置“：”3“}]]
//
第一个=（ary，fn）=>ary.reduce（（r，x）=>r | | fn（x），false）；
locate=（数据，id）=>u locate（{children:data}，id，[]）；
_locate=（节点、id、路径）=>node.tree\u id==id？路径
：第一个（node.children | |[]，n=>_locate（n，id，path.concat（n））；
res=locate（数据'102'）.map（n=>n.name）
console.log（res）；
Globals不好，请尝试重写您的函数，以便将wantId
作为参数传递。确实，Globals不好。更新后的答案，左acomment@DarrenSweeney我明白你的意思了。谢谢你指出这一点。希望它能帮助你解决你的问题：）全局变量不好，请尝试重写你的函数，以便wantId
是p作为参数assed。确实全局值不好。更新的答案，左acomment@DarrenSweeney我明白你的意思。谢谢你指出。希望它能帮助你解决你的问题：）我可能会遇到浏览器兼容性问题吗？@DarrenSweeney-看看源代码，你会觉得Array.reduce
，Array很好.concat
和Array.map
都是ECMA 5.1语法。@georg的解决方案基于函数式编程概念，仅此而已。我可能会遇到浏览器兼容性问题吗？@DarrenSweeney-查看源代码，您会发现Array.reduce
、Array.concat
和Array.map
都是ECMA5.1语法。@georg的解决方案基于函数式编程概念，仅此而已。