我必须遍历一个大型数据集来收集javascript中节点的最后一个子元素。
我需要在子详细信息中找到一组分支代码。 以下是我试图获得的输出：
[1000,1201,1202,2101,3101,3201]
{
"TITLE": {
"FirstLevel": {
"Details": {
"Code": "01",
},
"SecondLevel": [
{
"Details": {
"Description": "{desc}",
},
"ThirdLevel": {
"Details": {
"Code": "01",
},
"FourthLevel": [
{
"Details": {
"Code": "11",
},
"Branch": {
"Details": {
"Code": "1000",
}
}
},
{
"Details": {
"Code": "12",
},
"Branch": [
{
"Details": {
"Code": "1201",
}
},
{
"Details": {
"Code": "1202",
}
}
]
}
]
}
},
{
"Details": {
"Code": "100",
},
"ThirdLevel": [
{
"Details": {
"Code": "02",
},
"FourthLevel": {
"Details": {
"Code": "21"
},
"Branch": {
"Details": {
"Code": "2101",
}
}
}
},
{
"Details": {
"Code": "03",
},
"FourthLevel": [
{
"Details": {
"Code": "31",
},
"Branch": {
"Details": {
"Code": "3101",
}
}
},
{
"Details": {
"Code": "32",
},
"Branch": {
"Details": {
"Code": "3201",
}
}
}
]
}
]
}
]
}
}
}
我已经查看了更多基本问题的答案，并尝试调整解决方案。
一个这样的解决方案接受id传递，并将根据id更新名称。我想我可以在这里使用类似的地图实现。有一个问题，因为阵列不仅仅有孩子。表示将有子节点的位置。
function update(object, passedId) {
object.children.map((element, index) => {
if (element.id === passedId) {
console.log(index)
object.children[index].name = "New Name"
}
if (element.children != null) { // condition for checking Nesting
update(element, passedId)
}
})
console.log(object.children)
}
update(obj, "Branch");
接下来我尝试了一种更简单的方法
function getSubItem(obj) {
for (item in obj) {
for (subItem in obj[item]) {
for (subsubItem in obj[subItem]){
console.log(obj[item][subItem][subsubItem]);
}
}
}
}
getSubItem(obj)
依此类推，在sub子项之后添加sub，但是那种嵌套for循环看起来非常不稳定，而且由于分支可以嵌套在不同的区域，所以它看起来也不可靠。我在这里错过了一个简单的解决方案
答案 0 :(得分：1)
类似的东西：
function update (data, passedId, acc = []) {
if (!_.isObject(data)) {
return acc;
}
return _.reduce(_.values(data), (result, item) => {
result = _.chain(item)
.get(passedId)
.thru(val => [val])
.flatten()
.compact()
.map('Details.Code')
.thru(vals => _.concat(result, vals))
.value();
return update(item, passedId, result);
}, acc);
}
const res = update(data, 'Branch');
答案 1 :(得分：1)
试试这个：
const obj = {
"TITLE": {
"FirstLevel": {
"Details": {
"Code": "01",
},
"SecondLevel": [{
"Details": {
"Description": "{desc}",
},
"ThirdLevel": {
"Details": {
"Code": "01",
},
"FourthLevel": [{
"Details": {
"Code": "11",
},
"Branch": {
"Details": {
"Code": "1000",
}
}
},
{
"Details": {
"Code": "12",
},
"Branch": [{
"Details": {
"Code": "1201",
}
},
{
"Details": {
"Code": "1202",
}
}
]
}
]
}
},
{
"Details": {
"Code": "100",
},
"ThirdLevel": [{
"Details": {
"Code": "02",
},
"FourthLevel": {
"Details": {
"Code": "21"
},
"Branch": {
"Details": {
"Code": "2101",
}
}
}
},
{
"Details": {
"Code": "03",
},
"FourthLevel": [{
"Details": {
"Code": "31",
},
"Branch": {
"Details": {
"Code": "3101",
}
}
},
{
"Details": {
"Code": "32",
},
"Branch": {
"Details": {
"Code": "3201",
}
}
}
]
}
]
}
]
}
}
};
function transData(data, propertyName, path, result = []) {
if (data instanceof Array) {
data.forEach(obj => transData(obj, propertyName, path, result));
} else if (data instanceof Object) {
let existProperty = Object.keys(data).indexOf(propertyName) > -1;
if (existProperty) {
getCode(data[propertyName], path, result);
} else {
Object.keys(data).forEach(key => transData(data[key], propertyName, path, result));
}
}
function getCode(data, path, result) {
if (data instanceof Array) {
data.forEach(obj => getCode(obj, path, result));
} else {
if (typeof path !== 'undefined') {
result.push(path.split(/\./g).reduce((accumulator, val) => accumulator = accumulator[val], data));
} else {
result.push(data);
}
}
}
return result;
}
console.log(transData(obj, 'Branch', 'Details.Code'));
console.log(transData(obj, 'Branch'));
答案 2 :(得分：0)
从Deepdash（Deep Lodash扩展名）尝试_.eachDeep方法。
这是您的情况： https://codepen.io/yurigor/pen/moMpLV
var res = [];
_.eachDeep(data,function(value,key){
if(key=="Branch"){
if(!_.isArray(value)){
value = [value];
}
_.each(value,function(v){
if(v.Details&&v.Details.Code){
res.push(v.Details.Code);
}
});
}
});