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?
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
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(","));
其他回答
几个简单的方法:
var arr = [1,2,,3,,-3,null,,0,,undefined,4,,4,,5,,6,,,,];
arr.filter(n => n)
// [1, 2, 3, -3, 4, 4, 5, 6]
arr.filter(Number)
// [1, 2, 3, -3, 4, 4, 5, 6]
arr.filter(Boolean)
// [1, 2, 3, -3, 4, 4, 5, 6]
或-(仅适用于“text”类型的单个数组项)
['','1','2',3,,'4',,undefined,,,'5'].join('').split('');
// output: ["1","2","3","4","5"]
或-经典方式:简单迭代
var arr = [1,2,null, undefined,3,,3,,,0,,,[],,{},,5,,6,,,,],
len = arr.length, i;
for(i = 0; i < len; i++ )
arr[i] && arr.push(arr[i]); // copy non-empty values to the end of the array
arr.splice(0 , len); // cut the array and leave only the non-empty values
// [1,2,3,3,[],Object{},5,6]
jQuery:
var arr = [1,2,,3,,3,,,0,,,4,,4,,5,,6,,,,];
arr = $.grep(arr, n => n == 0 || n);
// [1, 2, 3, 3, 0, 4, 4, 5, 6]
带下划线/Loddash:
一般使用情况:
_.without(array, emptyVal, otherEmptyVal);
_.without([1, 2, 1, 0, 3, 1, 4], 0, 1);
有空:
_.without(['foo', 'bar', '', 'baz', '', '', 'foobar'], '');
--> ["foo", "bar", "baz", "foobar"]
无需参阅lodash文档。
我只是在上面的“用全局构造函数调用ES5的数组..filter()”高尔夫技巧中添加了我的声音,但我建议使用Object而不是上面建议的String、Boolean或Number。
具体来说,ES5的filter()已经不会为数组中未定义的元素触发;因此,一个普遍返回true(返回所有元素filter()命中)的函数必然只返回未定义的元素:
> [1,,5,6,772,5,24,5,'abc',function(){},1,5,,3].filter(function(){return true})
[1, 5, 6, 772, 5, 24, 5, 'abc', function (){}, 1, 5, 3]
然而,写出。。。(function(){return true;})的长度大于写入。。。(对象);在任何情况下,Object构造函数的返回值都将是某种类型的对象。与上面建议的基本装箱构造函数不同,没有可能的对象值是假的,因此在布尔设置中,object是function(){return true}的缩写。
> [1,,5,6,772,5,24,5,'abc',function(){},1,5,,3].filter(Object)
[1, 5, 6, 772, 5, 24, 5, 'abc', function (){}, 1, 5, 3]
这是我清理空字段的解决方案。
从费用对象开始:仅获取可用属性(带贴图)筛选空字段(带筛选器)将结果解析为整数(带映射)
fees.map( ( e ) => e.avail ).filter( v => v!== '').map( i => parseInt( i ) );
foo = [0, 1, 2, "", , false, 3, "four", null]
foo.filter(e => e === 0 ? true : e)
回报
[0, 1, 2, 3, "four"]
如果你确定你的数组中没有0,那么它看起来会更好一些:
foo.filter(e => e)