您可以为此创建一个原型。只需通过数组元素和您想要从数组元素中删除的值:

Array.prototype.removeItem = function(array,val) {
    array.forEach((arrayItem,index) => {
        if (arrayItem == val) {
            array.splice(index, 1);
        }
    });
    return array;
}
var DummyArray = [1, 2, 3, 4, 5, 6];
console.log(DummyArray.removeItem(DummyArray, 3));

如果元素存在多个实例,您可以进行后回循环,以确保不破坏索引。

var myElement = "chocolate";
var myArray = ['chocolate', 'poptart', 'poptart', 'poptart', 'chocolate', 'poptart', 'poptart', 'chocolate'];

/* Important code */
for (var i = myArray.length - 1; i >= 0; i--) {
  if (myArray[i] == myElement) myArray.splice(i, 1);
}
console.log(myArray);

如果您想要删除删除位置的新数组,您可以总是删除特定元素,并过滤该数组。它可能需要扩展矩阵对象对于不执行过滤法的浏览器来说,但从长远来看,它更容易,因为你所做的只是这个:

var my_array = [1, 2, 3, 4, 5, 6];
delete my_array[4];
console.log(my_array.filter(function(a){return typeof a !== 'undefined';}));

它应显示[1, 2, 3, 4, 6].

使用 JavaScript 原型特性定义列对象上名为删除() 的方法 。

使用使用复数()满足要求的方法。

请看看下面的代码

Array.prototype.remove = function(item) {
    // 'index' will have -1 if 'item' does not exist,
    // else it will have the index of the first item found in the array
    var index = this.indexOf(item);

    if (index > -1) {
        // The splice() method is used to add/remove items(s) in the array
        this.splice(index, 1);
    }
    return index;
}

var arr = [ 11, 22, 67, 45, 61, 89, 34, 12, 7, 8, 3, -1, -4];

// Printing array
// [ 11, 22, 67, 45, 61, 89, 34, 12, 7, 8, 3, -1, -4];
console.log(arr)

// Removing 67 (getting its index, i.e. 2)
console.log("Removing 67")
var index = arr.remove(67)

if (index > 0){
    console.log("Item 67 found at ", index)
} else {
    console.log("Item 67 does not exist in array")
}

// Printing updated array
// [ 11, 22, 45, 61, 89, 34, 12, 7, 8, 3, -1, -4];
console.log(arr)

// ............... Output ................................
// [ 11, 22, 67, 45, 61, 89, 34, 12, 7, 8, 3, -1, -4 ]
// Removing 67
// Item 67 found at  2
// [ 11, 22, 45, 61, 89, 34, 12, 7, 8, 3, -1, -4 ]

注:下面是用完整示例代码执行的完整示例代码。REPL编号js REPL来描述推 ()、 流行 ()、 转移 ()、 非转移 () 和 组合 () 方法的使用 。

> // Defining an array
undefined
> var arr = [12, 45, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34];
undefined
> // Getting length of array
undefined
> arr.length;
16
> // Adding 1 more item at the end i.e. pushing an item
undefined
> arr.push(55);
17
> arr
[ 12, 45, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34, 55 ]
> // Popping item from array (i.e. from end)
undefined
> arr.pop()
55
> arr
[ 12, 45, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
> // Remove item from beginning
undefined
> arr.shift()
12
> arr
[ 45, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
> // Add item(s) at beginning
undefined
> arr.unshift(67); // Add 67 at beginning of the array and return number of items in updated/new array
16
> arr
[ 67, 45, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
> arr.unshift(11, 22); // Adding 2 more items at the beginning of array
18
> arr
[ 11, 22, 67, 45, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
>
> // Define a method on array (temporarily) to remove an item and return the index of removed item; if it is found else return -1
undefined
> Array.prototype.remove = function(item) {
... var index = this.indexOf(item);
... if (index > -1) {
..... this.splice(index, 1); // splice() method is used to add/remove items in array
..... }
... return index;
... }
[Function]
>
> arr
[ 11, 22, 67, 45, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
>
> arr.remove(45);    // Remove 45 (you will get the index of removed item)
3
> arr
[ 11, 22, 67, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
>
> arr.remove(22)    // Remove 22
1
> arr
[ 11, 67, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
> arr.remove(67)    // Remove 67
1
> arr
[ 11, 67, 89, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
>
> arr.remove(89)    // Remove 89
2
> arr
[ 11, 67, 34, 12, 7, 8, 3, -1, -4, -11, 0, 56, 12, 34 ]
>
> arr.remove(100);  // 100 doesn't exist, remove() will return -1
-1
>

这里有很多奇妙的答案, 但对我来说,最有效的答案 不是完全从阵列中移除我的元素, 而是简单地设定它的价值为无效。

这对大多数情况都有效,而且是一个很好的解决方案,因为我稍后会使用变量,而不想让变量消失,只是暂时空的。此外,这个方法完全可以交叉浏览。

array.key = null;

这提供了一个前提值, 而不是一个值 。

注:这将更新给定的数组, 并返回受影响的行 。

用法

var removed = helper.remove(arr, row => row.id === 5 );

var removed = helper.removeAll(arr, row => row.name.startsWith('BMW'));

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var helper = {
 // Remove and return the first occurrence

 remove: function(array, predicate) {
  for (var i = 0; i < array.length; i++) {
   if (predicate(array[i])) {
    return array.splice(i, 1);
   }
  }
 },

 // Remove and return all occurrences

 removeAll: function(array, predicate) {
  var removed = [];

  for (var i = 0; i < array.length; ) {
   if (predicate(array[i])) {
    removed.push(array.splice(i, 1));
    continue;
   }
   i++;
  }
  return removed;
 },
};