var a = [{a1: 1, children: [{a1: 2}, undefined, {a1: 3}]}, undefined, {a1: 5}, undefined, {a1: 6}]
function removeNilItemInArray(arr) {
    if (!arr || !arr.length) return;
    for (let i = 0; i < arr.length; i++) {
        if (!arr[i]) {
            arr.splice(i , 1);
            continue;
        }
        removeNilItemInArray(arr[i].children);
    }
}
var b = a;
removeNilItemInArray(a);
// Always keep this memory zone
console.log(b);

您可能会发现，循环遍历数组并使用要从数组中保留的项构建新数组比按照建议进行循环和拼接更容易，因为在循环遍历时修改数组的长度可能会带来问题。

你可以这样做：

function removeFalsyElementsFromArray(someArray) {
    var newArray = [];
    for(var index = 0; index < someArray.length; index++) {
        if(someArray[index]) {
            newArray.push(someArray[index]);
        }
    }
    return newArray;
}

实际上，这里有一个更通用的解决方案：

function removeElementsFromArray(someArray, filter) {
    var newArray = [];
    for(var index = 0; index < someArray.length; index++) {
        if(filter(someArray[index]) == false) {
            newArray.push(someArray[index]);
        }
    }
    return newArray;
}

// then provide one or more filter functions that will 
// filter out the elements based on some condition:
function isNullOrUndefined(item) {
    return (item == null || typeof(item) == "undefined");
}

// then call the function like this:
var myArray = [1,2,,3,,3,,,,,,4,,4,,5,,6,,,,];
var results = removeElementsFromArray(myArray, isNullOrUndefined);

// results == [1,2,3,3,4,4,5,6]

你明白了——你可以有其他类型的过滤函数。可能比你需要的更多，但我感觉很慷慨…；）

@阿尔尼塔克

实际上，如果您添加一些额外的代码，Array.filter可以在所有浏览器上运行。见下文。

var array = ["","one",0,"",null,0,1,2,4,"two"];

function isempty(x){
if(x!=="")
    return true;
}
var res = array.filter(isempty);
document.writeln(res.toJSONString());
// gives: ["one",0,null,0,1,2,4,"two"]  

这是您需要为IE添加的代码，但过滤器和函数式编程是值得的。

//This prototype is provided by the Mozilla foundation and
//is distributed under the MIT license.
//http://www.ibiblio.org/pub/Linux/LICENSES/mit.license

if (!Array.prototype.filter)
{
  Array.prototype.filter = function(fun /*, thisp*/)
  {
    var len = this.length;
    if (typeof fun != "function")
      throw new TypeError();

    var res = new Array();
    var thisp = arguments[1];
    for (var i = 0; i < len; i++)
    {
      if (i in this)
      {
        var val = this[i]; // in case fun mutates this
        if (fun.call(thisp, val, i, this))
          res.push(val);
      }
    }

    return res;
  };
}

这可能会帮助您：https://lodash.com/docs/4.17.4#remove

var details = [
            {
                reference: 'ref-1',
                description: 'desc-1',
                price: 1
            }, {
                reference: '',
                description: '',
                price: ''
            }, {
                reference: 'ref-2',
                description: 'desc-2',
                price: 200
            }, {
                reference: 'ref-3',
                description: 'desc-3',
                price: 3
            }, {
                reference: '',
                description: '',
                price: ''
            }
        ];

        scope.removeEmptyDetails(details);
        expect(details.length).toEqual(3);

scope.removeEmptyDetails = function(details){
            _.remove(details, function(detail){
                return (_.isEmpty(detail.reference) && _.isEmpty(detail.description) && _.isEmpty(detail.price));
            });
        };

var a = [{a1: 1, children: [{a1: 2}, undefined, {a1: 3}]}, undefined, {a1: 5}, undefined, {a1: 6}]
function removeNilItemInArray(arr) {
    if (!arr || !arr.length) return;
    for (let i = 0; i < arr.length; i++) {
        if (!arr[i]) {
            arr.splice(i , 1);
            continue;
        }
        removeNilItemInArray(arr[i].children);
    }
}
var b = a;
removeNilItemInArray(a);
// Always keep this memory zone
console.log(b);

第一个例子是，当使用上面投票最高的答案时，我得到的字符串长度大于1的单个字符。下面是我解决这个问题的方法。

var stringObject = ["", "some string yay", "", "", "Other string yay"];
stringObject = stringObject.filter(function(n){ return n.length > 0});

如果未定义，则不返回，如果长度大于0，则返回。希望这能帮助一些人。

退换商品

["some string yay", "Other string yay"]