如果两个值都不存在,我如何推入数组?这是我的数组:

[
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" }
]

如果我试图再次推入数组的名字:“tom”或文本:“tasty”,我不希望发生任何事情…但如果这两个都不存在那么我就输入。push()

我该怎么做呢?


当前回答

推送后删除重复项

如果你已经有一个包含重复项的数组,将对象数组转换为字符串数组,然后使用Set()函数消除重复项:

let arr_obj = [
    { name: "tom", text: "tasty" }, 
    { name: "tom", text: "tasty" }
]

let arr_str = arr_obj.map(JSON.stringify)

let arr_unique = [...new Set(arr_str)].map(JSON.parse) 

推前检查

如果你到目前为止没有重复的元素,你想在推入一个新元素之前检查重复:

let arr_obj = [
    { name: "tom", text: "tasty" },
    { name: "tim", text: "tusty" }
]

let new_obj = { name: "tom", text: "tasty" }

let arr_str = arr_obj.map(JSON.stringify)

!arr_str.includes(JSON.stringify(new_obj)) && arr_obj.push(new_obj)

其他回答

如果不在列表中,则添加

对于一个简单值的列表,它是一行程序…

[...new Set([...someArray, someElement])]

JavaScript的用法:

var myArray = ['bill','bob']
var alreadyIn = [...new Set([...myArray, 'bob'])] // ['bill','bob']
var notAlreadyIn = [...new Set([...myArray, 'peter'])] // ['bill','bob','peter']

TypeScript文本(注意include vs includes):

interface Array<T> {
  include(element: T): Array<T>
}
Array.prototype.include = function (element: any): any[] {
  return [...new Set([...this, obj])]
}

...但对于对象来说,情况就复杂多了

[...new Set([...someArray.map((o) => JSON.stringify(o)),
    JSON.stringify(someElement)]).map((o) => JSON.parse(o))

TypeScript文本处理任何事情:

Array.prototype.include = function (element: any): any[] {
  if (element && typeof element === 'object')
    return [
      ...new Set([
        ...this.map((o) => JSON.stringify(o)),
        JSON.stringify(element),
      ]),
    ].map((o) => JSON.parse(o))
  else return [...new Set([...this, element])]
}

使用数组是很容易做到的。函数findIndex,它以函数作为参数:

var arrayObj = [{name:"bull", text: "sour"},
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" }
]
var index = arrayObj.findIndex(x => x.name=="bob"); 
// here you can check specific property for an object whether it exist in your array or not

index === -1 ? arrayObj.push({your_object}) : console.log("object already exists")
 

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

如果有人有不太复杂的要求,这里是我对一个简单字符串数组的答案的改编:

Array.prototype.pushIfNotExist = function(val) {
    if (typeof(val) == 'undefined' || val == '') { return; }
    val = $.trim(val);
    if ($.inArray(val, this) == -1) {
        this.push(val);
    }
};

更新:替换indexOf和trim与jQuery的IE8兼容性的替代品

不确定速度,但stringification + indexOf是一个简单的方法。首先将数组转换为字符串:

let strMyArray = JSON.stringify(myArray);

然后,对于一系列属性-值对,您可以使用:

if (strMyArray.indexOf('"name":"tom"') === -1 && strMyArray.indexOf('"text":"tasty"') === -1) {
   myArray.push({ name: "tom", text: "tasty" });
}

查找整个对象更简单:

if (strMyArray.indexOf(JSON.stringify(objAddMe) === -1) { 
   myArray.push(objAddMe);
}