如何在JavaScript中删除数组中的空元素?

是否有一种简单的方法,或者我需要循环并手动删除它们?


当前回答

使用正则表达式筛选出无效条目

array = array.filter(/\w/);
filter + regexp

其他回答

我需要完成同样的任务,并遇到了这个线程。我最终使用数组“join”使用“_”分隔符创建字符串,然后使用正则表达式:-

1. replace "__" or more with just one "_",
2. replace preceding "_" with nothing "" and similarly 
3. replace and ending "_" with nothing ""

…然后使用数组“拆分”生成一个已清理的数组:-

var myArr = new Array("","","a","b","","c","","","","","","","","","e","");
var myStr = "";

myStr = myArr.join("_");

myStr = myStr.replace(new RegExp(/__*/g),"_");
myStr = myStr.replace(new RegExp(/^_/i),"");
myStr = myStr.replace(new RegExp(/_$/i),"");
myArr = myStr.split("_");

alert("myArr=" + myArr.join(","));

…或1行代码:-

var myArr = new Array("","","a","b","","c","","","","","","","","","e","");

myArr = myArr.join("_").replace(new RegExp(/__*/g),"_").replace(new RegExp(/^_/i),"").replace(new RegExp(/_$/i),"").split("_");

alert("myArr=" + myArr.join(","));

…或,扩展Array对象:-

Array.prototype.clean = function() {
  return this.join("_").replace(new RegExp(/__*/g),"_").replace(new RegExp(/^_/i),"").replace(new RegExp(/_$/i),"").split("_");
};

var myArr = new Array("","","a","b","","c","","","","","","","","","e","");

alert("myArr=" + myArr.clean().join(","));

@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;
  };
}

实际上,您可以使用ES6+方法,假设数组如下:

const arr = [1,2,3,undefined,4,5,6,undefined,7,8,undefined,undefined,0,9];

答案可以是以下两种方式之一:

第一种方式:常量clearArray=arr.filter(i=>i);//[ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]第二种方式:const clearArray=arr.filter(布尔值);//[ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]

2022年10月14日更新:

这两个答案并不完全正确,即使在给定的例子中,是的,它是有效的,但请注意给定数组中的数字0,通过这两种方式,数字0都会消失,这显然与使用布尔强制检查项目有关。

完全正确的方法是检查nulish并删除它们:

const notNil = (i) => !(typeof i === 'undefined' || i === null);

const clearArray = arr.filter(i => isNil(i));

常量arr=[1,2,3,未定义,4,5,6,未定义、7,8,未定义;未定义,0,9];常量notNil=(i)=>!(类型i===“未定义”||i===null);console.log(“非nil:”,arr.filter(非nil));

以上答案都不适用于所有类型。下面的解决方案将删除null、undefined、{}[]、NaN,并保留日期字符串,最好的是它甚至从嵌套对象中删除。

function removeNil(obj) {
    // recursively remove null and undefined from nested object too.
    return JSON.parse(JSON.stringify(obj), (k,v) => {
      if(v === null || v === '') return undefined;
      // convert date string to date.
      if (typeof v === "string" && /^\d\d\d\d-\d\d-\d\dT\d\d:\d\d:\d\d.\d\d\dZ$/.test(v))
        return new Date(v);
      // remove empty array and object.
      if(typeof v === 'object' && !Object.keys(v).length) return undefined;
      return v;
    });
  }

函数removeNil(obj){//递归地从嵌套对象中删除null和undefined。返回JSON.parse(JSON.stringify(obj),(k,v)=>{如果(v===null||v==='')返回undefined;//将日期字符串转换为日期。if(typeof v==“string”&&/^\d\d\d\d-\d\dT\dd:\d\d:\d\d.d.d\dZ$/.test(v))返回新日期(v);//删除空数组和对象。if(typeof v=='object'&&!object.keys(v).length)返回undefined;返回v;});}常量ob={s: “a”,b: 43中,国家:['a','b','c'],l: 空,n: {ks:“a”,efe:null,ce:“”},d: new Date(),nan:nan,k: 未定义,emptyO:{},emptyArr:[],}常量输出=removeNil(ob);console.log(输出);console.log('测试:',ob.countries.length,typeof(ob.d))

简单ES6

['a','b','',,,'w','b'].filter(v => v);