从嵌套属性数组中获取对象的嵌套值 [英] Getting an Objects nested value from an array of nested properties

问题描述

我遇到一种情况，需要将两个对象压缩在一起，同时保留两个值.我可以遍历两个对象并构建所有键的数组.

I have a situation where I need to zip two Objects together retaining both of the values. I can iterate through both the objects and build an array of all the keys.

    var traverse = function (obj, chain) {

        for (var prop in obj) {
            if (obj.hasOwnProperty(prop)) {

                var tempChain = [].concat(chain || []);
                tempChain.push(prop);

                if (typeof obj[prop] === 'object') {
                    traverse(obj[prop], tempChain);
                }
                console.log(tempChain);
            }
        }
    };

传递以下内容:

    traverse({
        'a': {
            'b': 'hello world',
            'b1': 'hello world 1',
            'c': {
                'd': 'dello'
            }
        }
    })

将退还给我:

[a]
[a,b]
[a,b1]
[a,c]
[a, c, d]

所以我现在在对象中有一个嵌套属性数组.我如何才能访问obj [[a，c，d]]?我知道我可以通过评估解决问题，但是我不相信内容.

So I now have an array of nested properties in an object. How can I access essentially obj[[a,c,d]]? I know I can solve the problem through eval but I can't trust the content.

eval('window.' + ['a','c','d'].join('.')); 

如果我可以遍历该数组并检查两个属性中是否都存在该属性，则构建一个合并了压缩"值的新对象.

If I can loop through that array and check to see if the property exists in both of them, then build a new object of the combined 'zipped' values.

推荐答案

也许是这样的事情?

function getValueAt(obj, keyPathArray) {
    var emptyObj = {};

    return keyPathArray.reduce(function (o, key) {
        return (o || emptyObj)[key];
    }, obj);
}

然后您可以像使用它一样

Then you can use it like:

var o = { a: { c: { d: 1 } } };

getValueAt(o, ['a', 'c', 'd']); //1

但是，对于不存在的属性来说效率不高，因为它不会短路. 这是不使用reduce的另一种方法:

However it's not efficient for non-existing properties, since it will not short-circuit. Here's another approach without using reduce:

function getValueAt(o, keyPathArray) {
    var i = 0, 
        len = keyPathArray.length;

    while (o != null && i < len) o = o[keyPathArray[i++]];

    return o;    
}

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