是否有从JavaScript数组中删除项的方法?

给定一个数组:

var ary = ['three', 'seven', 'eleven'];

我想做的事情是:

removeItem('seven', ary);

我已经查看了splice(),但它只删除了位置号,而我需要一些东西来删除其值的项目。


当前回答

方法1

var ary = ['three', 'seven', 'eleven'];
var index = ary.indexOf('seven'); // get index if value found otherwise -1

if (index > -1) { //if found
  ary.splice(index, 1);
}

方法2

一条衬垫

var ary = ['three', 'seven', 'eleven'];
filteredArr = ary.filter(function(v) { return v !== 'seven' })


// Or using ECMA6:
filteredArr = ary.filter(v => v !== 'seven')

其他回答

//This function allows remove even array from array
var removeFromArr = function(arr, elem) { 
    var i, len = arr.length, new_arr = [],
    sort_fn = function (a, b) { return a - b; };
    for (i = 0; i < len; i += 1) {
        if (typeof elem === 'object' && typeof arr[i] === 'object') {
            if (arr[i].toString() === elem.toString()) {
                continue;
            } else {                    
                if (arr[i].sort(sort_fn).toString() === elem.sort(sort_fn).toString()) {
                    continue;
                }
            }
        }
        if (arr[i] !== elem) {
            new_arr.push(arr[i]);
        }
    }
    return new_arr;
}

使用实例

var arr = [1, '2', [1 , 1] , 'abc', 1, '1', 1];
removeFromArr(arr, 1);
//["2", [1, 1], "abc", "1"]

var arr = [[1, 2] , 2, 'a', [2, 1], [1, 1, 2]];
removeFromArr(arr, [1,2]);
//[2, "a", [1, 1, 2]]

你可以使用without或pull from Lodash:

const _ = require('lodash');
_.without([1, 2, 3, 2], 2); // -> [1, 3]

在全局函数中,我们不能直接传递自定义值,但有很多方法,如下所示

 var ary = ['three', 'seven', 'eleven'];
 var index = ary.indexOf(item);//item: the value which you want to remove

 //Method 1
 ary.splice(index,1);

 //Method 2
 delete ary[index]; //in this method the deleted element will be undefined

请不要使用带有delete的变体-它会在数组中留下一个洞,因为它不会在删除的项之后重新索引元素。

> Array.prototype.remove=function(v){
...     delete this[this.indexOf(v)]
... };
[Function]
> var myarray=["3","24","55","2"];
undefined
> myarray.remove("55");
undefined
> myarray
[ '3', '24', , '2' ]

从数组中删除所有匹配的元素(而不仅仅是第一个,这似乎是这里最常见的答案):

while ($.inArray(item, array) > -1) {
    array.splice( $.inArray(item, array), 1 );
}

我使用jQuery来完成这些繁重的工作,但是如果您想要本地化,您就可以理解了。