如何根据 javascript 中的数组中的路径遍历嵌套的 object？

Question

const obj = { 
    first: { second: { third: 'done'} },
    hello: { world: { foo: { bar: 'wrong' } } },
    second: { third: 'wrong'}
};

const arr = [ 'first', 'second', 'third' ];

function traverse(obj, arr) {

}
// output = 'done'

Given a first input as a nested object, and a second input as an array containing strings. What is the best way to traverse the nested object based on the path set by the array to output 'done'?

Answer 1

You can reduce the array arr , changing the accumulator to a deeper object at each step.

 const obj = { first: { second: { third: 'done'} }, hello: { world: { foo: { bar: 'wrong' } } }, second: { third: 'wrong'} }; const arr = [ 'first', 'second', 'third' ]; function traverse(obj, arr) { return arr.reduce((acc, curr) => acc? acc[curr]: undefined, obj); } console.log(traverse(obj, arr)); console.log(traverse(obj, ['hello', 'world', 'foo'])); console.log(traverse(obj, ['first', 'hello', 'world']));

Answer 2

You can use references

• Loop over the array untill the last second last element
• If the ref[key] is an object, change ref to ref[key]
• else return Not Found
• Check if the ref has property name same as last variable then return ref[last] else return Not Found

 const obj = { first: { second: { third: 'done'} },hello: { world: { foo: { bar: 'wrong' } } },second: { third: 'wrong'}}; const arr = [ 'first', 'second', 'third' ]; function traverse(obj, arr) { let ref = obj let last = arr[arr.length-1] for(let i=0; i<arr.length-1; i++){ let key = arr[i] if(typeof ref[key] === 'object'){ ref = ref[key] } else{ return "Not found" } } return ref.hasOwnProperty(last)? ref[last]: "Not found" } console.log(traverse(obj,arr)) console.log(traverse(obj,['first','third'])) console.log(traverse(obj,['hello','world']))

Answer 3

if you want it short and to handle system file path like:

it returns the object if exist and -1 if not

 const obj = { first: { second: { third: 'done'} }, hello: { world: { foo: { bar: 'wrong' } } }, second: { third: 'wrong'} }; function traverse(obj, path) { let seq = path.split('/').filter(x => x;= ''). seq?map( s => obj =:obj[s]; -1; obj[s] ). return obj, } console;log(traverse(obj. '/hello/notexist/')), console;log(traverse(obj, '/hello/world/'));

or even shorter with a reduce() instead of a map() :

function traverse(obj, path) {
    let seq = path.split('/').filter(x => x != '');
    return seq.reduce((acc, curr) => !acc[curr] ? -1 : acc[curr], obj);
}

or a even even shorter:

 function traverse(obj, path) { return path.split('/').filter(Boolean).reduce((acc, curr) => acc[curr]? acc[curr]: -1, obj); } const obj = { first: { second: { third: 'done'} }, hello: { world: { foo: { bar: 'wrong' } } }, second: { third: 'wrong'} }; const res = traverse(obj, '/hello/world/'); // return {foo:...} console.log(res);