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

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]

其他回答

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);

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'}];

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]

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]

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。