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

someArray = [{a: 'a1 value', b: {c: "c1 value"},
             {a: 'a2 value', b: {c: "c2 value"}]
newObject = {a: 'a2 value', b: {c: "c2 value"}}

//New object which needs check for duplicity

let isExists = checkForExists(newObject) {
    return someArray.some(function(el) {
        return el.a === newObject.a && el.b.c === newObject.b.c;
    });
}
// write your logic here 
// if isExists is true then already object in an array else you can add

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])]
}

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)

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}]

短的例子:

if (typeof(arr[key]) === "undefined") {
  arr.push(key);
}

像这样的吗?

var item = "Hello World";
var array = [];
if (array.indexOf(item) === -1) array.push(item);

与对象

var item = {name: "tom", text: "tasty"}
var array = [{}]
if (!array.find(o => o.name === 'tom' && o.text === 'tasty'))
    array.push(item)