我在我的项目中使用了这个代码

const getValue = (obj, arrPath) => (
  arrPath.reduce((x, y) => {
    if (y in x) return x[y]
    return {}
  }, obj)
)

用法:

const obj = { id: { user: { local: 104 } } }
const path = [ 'id', 'user', 'local' ]
getValue(obj, path) // return 104

这是一个递归的例子。

函数重组(obj，字符串){ Var parts = string.split('.'); var newObj = obj[parts[0]]; If (parts[1]) { 部分。拼接(0,1); var newString = parts.join('.'); return recompose(newObj, newString); ｝ 返回newObj; ｝ var obj ={答:{' 1 ',c:‘2’,d:{答:{b:“胡说”}}}}; console.log(重组(obj, ' a.d.a.b '));/ /等等

其他的建议有点晦涩难懂，所以我想我应该贡献一下:

Object.prop = function(obj, prop, val){
    var props = prop.split('.')
      , final = props.pop(), p 
    while(p = props.shift()){
        if (typeof obj[p] === 'undefined')
            return undefined;
        obj = obj[p]
    }
    return val ? (obj[final] = val) : obj[final]
}

var obj = { a: { b: '1', c: '2' } }

// get
console.log(Object.prop(obj, 'a.c')) // -> 2
// set
Object.prop(obj, 'a.c', function(){})
console.log(obj) // -> { a: { b: '1', c: [Function] } }

几年后，我发现它可以处理范围和数组。如[b] [c] .d.etc

function getScopedObj(scope, str) {
  let obj=scope, arr;

  try {
    arr = str.split(/[\[\]\.]/) // split by [,],.
      .filter(el => el)             // filter out empty one
      .map(el => el.replace(/^['"]+|['"]+$/g, '')); // remove string quotation
    arr.forEach(el => obj = obj[el])
  } catch(e) {
    obj = undefined;
  }

  return obj;
}

window.a = {b: {c: {d: {etc: 'success'}}}}

getScopedObj(window, `a.b.c.d.etc`)             // success
getScopedObj(window, `a['b']["c"].d.etc`)       // success
getScopedObj(window, `a['INVALID']["c"].d.etc`) // undefined

使用数组缩减函数将获得/设置基于提供的路径。

我用a.b.c和a.b.2.c {a:{b:[0,1，{c:7}]}}测试了它，它既可以获取键，也可以将对象更改为设置值

function setOrGet(obj, path=[], newValue){ const l = typeof path === 'string' ? path.split('.') : path; return l.reduce((carry,item, idx)=>{ const leaf = carry[item]; // is this last item in path ? cool lets set/get value if( l.length-idx===1) { // mutate object if newValue is set; carry[item] = newValue===undefined ? leaf : newValue; // return value if its a get/object if it was a set return newValue===undefined ? leaf : obj ; } carry[item] = leaf || {}; // mutate if key not an object; return carry[item]; // return object ref: to continue reduction; }, obj) } console.log( setOrGet({a: {b:1}},'a.b') === 1 || 'Test Case: Direct read failed' ) console.log( setOrGet({a: {b:1}},'a.c',22).a.c===22 || 'Test Case: Direct set failed' ) console.log( setOrGet({a: {b:[1,2]}},'a.b.1',22).a.b[1]===22 || 'Test Case: Direct set on array failed' ) console.log( setOrGet({a: {b:{c: {e:1} }}},'a.b.c.e',22).a.b.c. e===22 || 'Test Case: deep get failed' ) // failed !. Thats your homework :) console.log( setOrGet({a: {b:{c: {e:[1,2,3,4,5]} }}},'a.b.c.e.3 ',22) )

我个人的建议。

除非没有别的办法，否则不要用这种东西!

我看到很多例子，人们用它来翻译json;你会看到locale('app。主页。欢迎')这样的函数。这太糟糕了。如果你已经在object/json中有数据;你知道路径。然后直接使用locale().app.首页示例。欢迎通过改变你的函数返回对象，你得到类型安全，自动完成，不容易出现错别字。

冒着白费口舌的风险…… 我发现这在遍历嵌套对象以引用相对于基对象或具有相同结构的类似对象的位置时非常有用。为此，这对于嵌套的对象遍历函数很有用。注意，我使用了一个数组来保存路径。将其修改为使用字符串路径或数组是很简单的。还要注意，您可以将“undefined”赋值给该值，这与其他一些实现不同。

/* * Traverse each key in a nested object and call fn(curObject, key, value, baseObject, path) * on each. The path is an array of the keys required to get to curObject from * baseObject using objectPath(). If the call to fn() returns falsey, objects below * curObject are not traversed. Should be called as objectTaverse(baseObject, fn). * The third and fourth arguments are only used by recursion. */ function objectTraverse (o, fn, base, path) { path = path || []; base = base || o; Object.keys(o).forEach(function (key) { if (fn(o, key, o[key], base, path) && jQuery.isPlainObject(o[key])) { path.push(key); objectTraverse(o[key], fn, base, path); path.pop(); } }); } /* * Get/set a nested key in an object. Path is an array of the keys to reference each level * of nesting. If value is provided, the nested key is set. * The value of the nested key is returned. */ function objectPath (o, path, value) { var last = path.pop(); while (path.length && o) { o = o[path.shift()]; } if (arguments.length < 3) { return (o? o[last] : o); } return (o[last] = value); }