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

如果你的项目包含lodash,使用unionBy方法会很简单

import {unionBy} from "lodash";

let arrayObj = [
    { name: "jhon", text: "guitar"},
    { name: "paul", text: "bass" },
    { name: "george", text: "guitar" }
];

// this object will be added to the array
arrayObj = unionBy(arrayObj, [{name: 'ringo', text: 'drums'}], 'name')

// this object will be ignored because already exists  
arrayObj = unionBy(arrayObj, [{name: "jhon", text: "guitar"}], 'name')

其他回答

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);
}

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);
}

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])]
}

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]