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
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
其他回答
短的例子:
if (typeof(arr[key]) === "undefined") {
arr.push(key);
}
正是出于这些原因，使用像underscore.js这样的js库。union:计算传入数组的并集:在一个或多个数组中出现的唯一项的列表。
_.union([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2, 3, 101, 10]
如果没有结果，可以使用jQuery grep和push: http://api.jquery.com/jQuery.grep/
这基本上是与“扩展原型”解决方案相同的解决方案，但没有扩展(或污染)原型。
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')
这个问题有点老了，但我的选择是
let finalTab = [{id: 1, name: 'dupont'}, {id: 2, name: 'tintin'}, {id: 3, name:'toto'}]; // Your array of object you want to populate with distinct data
const tabToCompare = [{id: 1, name: 'dupont'}, {id: 4, name: 'tata'}]; // A array with 1 new data and 1 is contain into finalTab
finalTab.push(
...tabToCompare.filter(
tabToC => !finalTab.find(
finalT => finalT.id === tabToC.id)
)
); // Just filter the first array, and check if data into tabToCompare is not into finalTab, finally push the result of the filters
console.log(finalTab); // Output : [{id: 1, name: 'dupont'}, {id: 2, name: 'tintin'}, {id: 3, name: 'toto'}, {id: 4, name: 'tata'}];
