简单解决方案(ES6)

如果你没有重复元素

Array.prototype.remove = function(elem) {
  var indexElement = this.findIndex(el => el === elem);
  if (indexElement != -1)
    this.splice(indexElement, 1);
  return this;
};   

在线演示(小提琴)

DEMO

你需要用.indexOf()找到你要找的位置，然后用.splice()删除它

function remove(arr, what) {
    var found = arr.indexOf(what);

    while (found !== -1) {
        arr.splice(found, 1);
        found = arr.indexOf(what);
    }
}

var array = new Array();
array.push("A");
array.push("B");
array.push("C");
 ​   
remove(array, 'B');
alert(array)​​​​;

这将解决所有问题。

这只对str list有效，查一下

myStrList.filter(item=> !["deletedValue","deletedValue2"].includes(item))

使用array.splice

/*array.splice(index , howMany[, element1[, ...[, elementN]]])

array.splice(index) // SpiderMonkey/Firefox extension*/

array.splice(1,1)

来源: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice

简单解决方案(ES6)

如果你没有重复元素

Array.prototype.remove = function(elem) {
  var indexElement = this.findIndex(el => el === elem);
  if (indexElement != -1)
    this.splice(indexElement, 1);
  return this;
};   

在线演示(小提琴)

如果想要从字符串数组中移除字符串数组:

const names = ['1','2','3','4']
const excludeNames = ['2','3']
const filteredNames = names.filter((name) => !excludeNames.includes(name));
// ['1','4']