我建议你使用Set，

集只允许唯一的条目，这将自动解决您的问题。

集合可以这样声明:

const baz = new Set(["Foo","Bar"])

我有这个问题，我做了一个简单的原型，使用它，如果你喜欢它

Array.prototype.findOrPush = function(predicate, fallbackVal) {
    let item = this.find(predicate)
    if(!item){
        item = fallbackVal
        this.push(item)
    }
    return item
}

let arr = [{id: 1}]
let item = arr.findOrPush(e => e.id == 2, {id: 2})
console.log(item) // {id: 2} 

// will not push and just return existing value
arr.findOrPush(e => e.id == 2, {id: 2}) 
conslog.log(arr)  // [{id: 1}, {id: 2}]

我的选择是使用.includes()扩展数组。正如@Darrin Dimitrov所建议的原型:

Array.prototype.pushIfNotIncluded = function (element) {
    if (!this.includes(element)) {
      this.push(element);
    }
}

记住include来自es6，在IE上不起作用: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/includes

我更喜欢使用原生js Array.prototype.some()，即使在jqenv中 文档:w3s some or mdn some

let arr = [
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" }
];
let oneMore = { name: "tom", text: "tasty" };
!arr.some(i => i.name == oneMore.name && i.text == oneMore.text)
  && arr.push(oneMore);

这里你有一种方法可以在一行中为两个数组做这件事:

const startArray = [1,2,3,4]
const newArray = [4,5,6]

const result = [...startArray, ...newArray.filter(a => !startArray.includes(a))]

console.log(result);
//Result: [1,2,3,4,5,6]

推动动态

var a = [
  {name:"bull", text: "sour"},
  {name: "tom", text: "tasty" },
  {name: "Jerry", text: "tasty" }
]

function addItem(item) {
  var index = a.findIndex(x => x.name == item.name)
  if (index === -1) {
    a.push(item);
  }else {
    console.log("object already exists")
  }
}

var item = {name:"bull", text: "sour"};
addItem(item);

用简单的方法

var item = {name:"bull", text: "sour"};
a.findIndex(x => x.name == item.name) == -1 ? a.push(item) : console.log("object already exists")

如果数组只包含基元类型/简单数组

var b = [1, 7, 8, 4, 3];
var newItem = 6;
b.indexOf(newItem) === -1 && b.push(newItem);