给定一个JavaScript对象,

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

和字符串

"a.b"

我怎么把字符串转换成点符号呢

var val = obj.a.b

如果字符串只是'a',我可以使用obj[a]。但这个更复杂。我想应该有什么简单的方法,但现在想不起来了。


当前回答

这是ninjagecko提出的我的扩展方案。

对我来说,简单的字符串表示法是不够的,所以下面的版本支持如下内容:

index(obj, 'data.accounts[0].address[0].postcode');

 

/**
 * Get object by index
 * @supported
 * - arrays supported
 * - array indexes supported
 * @not-supported
 * - multiple arrays
 * @issues:
 *  index(myAccount, 'accounts[0].address[0].id') - works fine
 *  index(myAccount, 'accounts[].address[0].id') - doesnt work
 * @Example:
 * index(obj, 'data.accounts[].id') => returns array of id's
 * index(obj, 'data.accounts[0].id') => returns id of 0 element from array
 * index(obj, 'data.accounts[0].addresses.list[0].id') => error
 * @param obj
 * @param path
 * @returns {any}
 */
var index = function(obj, path, isArray?, arrIndex?){

    // is an array
    if(typeof isArray === 'undefined') isArray = false;
    // array index,
    // if null, will take all indexes
    if(typeof arrIndex === 'undefined') arrIndex = null;

    var _arrIndex = null;

    var reduceArrayTag = function(i, subArrIndex){
        return i.replace(/(\[)([\d]{0,})(\])/, (i) => {
            var tmp = i.match(/(\[)([\d]{0,})(\])/);
            isArray = true;
            if(subArrIndex){
                _arrIndex =  (tmp[2] !== '') ? tmp[2] : null;
            }else{
                arrIndex =  (tmp[2] !== '') ? tmp[2] : null;
            }
            return '';
        });
    }

    function byIndex(obj, i) {
        // if is an array
        if(isArray){
            isArray = false;
            i = reduceArrayTag(i, true);
            // if array index is null,
            // return an array of with values from every index
            if(!arrIndex){
                var arrValues = [];
                _.forEach(obj, (el) => {
                    arrValues.push(index(el, i, isArray, arrIndex));
                })
                return arrValues;
            }
            // if array index is specified
            var value = obj[arrIndex][i];
            if(isArray){
                arrIndex = _arrIndex;
            }else{
                arrIndex = null;
            }
            return value;
        }else{
            // remove [] from notation,
            // if [] has been removed, check the index of array
            i = reduceArrayTag(i, false);
            return obj[i]
        }
    }

    // reduce with the byIndex method
    return path.split('.').reduce(byIndex, obj)
}

其他回答

你可以通过点表示法获得对象成员的值,只需一行代码:

new Function('_', 'return _.' + path)(obj);

对你来说:

var obj = { a: { b: '1', c: '2' } }
var val = new Function('_', 'return _.a.b')(obj);

为了简化,你可以这样写一个函数:

function objGet(obj, path){
    return new Function('_', 'return _.' + path)(obj);
}

解释:

Function构造函数创建一个新的Function对象。在JavaScript中,每个函数实际上都是一个function对象。使用function构造函数显式创建函数的语法如下:

new Function ([arg1[, arg2[, ...argN]],] functionBody)

其中arguments(arg1到argN)必须是一个对应于有效javaScript标识符的字符串,functionBody是一个包含包含函数定义的javaScript语句的字符串。

在我们的例子中,我们利用字符串函数体来检索点表示法的对象成员。

希望能有所帮助。

我通过ninjagecko扩展了这个优雅的答案,这样函数就可以处理点和/或数组样式的引用,并且空字符串会导致父对象返回。

给你:

string_to_ref = function (object, reference) {
    function arr_deref(o, ref, i) { return !ref ? o : (o[ref.slice(0, i ? -1 : ref.length)]) }
    function dot_deref(o, ref) { return ref.split('[').reduce(arr_deref, o); }
    return !reference ? object : reference.split('.').reduce(dot_deref, object);
};

查看我的jsFiddle工作示例:http://jsfiddle.net/sc0ttyd/q7zyd/

解决方案:

function deepFind(key, data){
  return key.split('.').reduce((ob,i)=> ob?.[i], data)
}

用法:

const obj = {
   company: "Pet Shop",
   person: {
      name: "John"
   },
   animal: {
      name: "Lucky"
   }
}

const company = deepFind("company", obj) 
const personName = deepFind("person.name", obj) 
const animalName = deepFind("animal.name", obj) 

这是ninjagecko提出的我的扩展方案。

对我来说,简单的字符串表示法是不够的,所以下面的版本支持如下内容:

index(obj, 'data.accounts[0].address[0].postcode');

 

/**
 * Get object by index
 * @supported
 * - arrays supported
 * - array indexes supported
 * @not-supported
 * - multiple arrays
 * @issues:
 *  index(myAccount, 'accounts[0].address[0].id') - works fine
 *  index(myAccount, 'accounts[].address[0].id') - doesnt work
 * @Example:
 * index(obj, 'data.accounts[].id') => returns array of id's
 * index(obj, 'data.accounts[0].id') => returns id of 0 element from array
 * index(obj, 'data.accounts[0].addresses.list[0].id') => error
 * @param obj
 * @param path
 * @returns {any}
 */
var index = function(obj, path, isArray?, arrIndex?){

    // is an array
    if(typeof isArray === 'undefined') isArray = false;
    // array index,
    // if null, will take all indexes
    if(typeof arrIndex === 'undefined') arrIndex = null;

    var _arrIndex = null;

    var reduceArrayTag = function(i, subArrIndex){
        return i.replace(/(\[)([\d]{0,})(\])/, (i) => {
            var tmp = i.match(/(\[)([\d]{0,})(\])/);
            isArray = true;
            if(subArrIndex){
                _arrIndex =  (tmp[2] !== '') ? tmp[2] : null;
            }else{
                arrIndex =  (tmp[2] !== '') ? tmp[2] : null;
            }
            return '';
        });
    }

    function byIndex(obj, i) {
        // if is an array
        if(isArray){
            isArray = false;
            i = reduceArrayTag(i, true);
            // if array index is null,
            // return an array of with values from every index
            if(!arrIndex){
                var arrValues = [];
                _.forEach(obj, (el) => {
                    arrValues.push(index(el, i, isArray, arrIndex));
                })
                return arrValues;
            }
            // if array index is specified
            var value = obj[arrIndex][i];
            if(isArray){
                arrIndex = _arrIndex;
            }else{
                arrIndex = null;
            }
            return value;
        }else{
            // remove [] from notation,
            // if [] has been removed, check the index of array
            i = reduceArrayTag(i, false);
            return obj[i]
        }
    }

    // reduce with the byIndex method
    return path.split('.').reduce(byIndex, obj)
}

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] } }