Question

我有一个JSON，如下所示

[
  {
    "name": "Mike",
    "incentives": "23.45",
    "id": "1"
  },
  {
    "name": "Larsen",
    "incentives": "34.78",
    "id": "2"
  },
  {
    "name": "Steve",
    "incentives": "26.78",
    "id": "3"
  }
]

如果JSON对象中不存在ID，我需要将新的JSON对象推送到JSON数组中

以这种方式尝试并工作，这样做的任何更好的方式

$(document).ready(function() {

var idsarray = [];

  var test = [
  {
    "name": "Mike",
    "incentives": "23.45",
    "id": "1"
  },
  {
    "name": "Larsen",
    "incentives": "34.78",
    "id": "2"
  },
  {
    "name": "Steve",
    "incentives": "26.78",
    "id": "3"
  }
];

for(var i=0;i<test.length;i++)
{
idsarray.push(test[i].id)
}

var newobj =  {
    "name": "RAM",
    "incentives": "56.78",
    "id": "4"
  }

  var newid = newobj.id;

if(jQuery.inArray(newid, idsarray) !== -1)
{
test.push(newobj)
}

alert(test.length)

});

Answer 1

inArray很慢。如果您执行一次，这绝不是最糟糕的事情，但是在原始数组上使用find或some会更好：

if (!test.some(o => o.id == newid)) {
  test.push(newobj)
}

但是，如果您经常这样做，那么更好的主意是创建一个集合，因为检查集合中的成员资格非常快：

let idset = new Set();
...
if (!idset.has(newid)) {
  test.push(newobj);
  idset.add(newid);
}

您还可以使用Map（甚至是普通的旧对象）将两者结合起来：

let test = new Map();
...
if (!test.has(newid)) {
    test.set(newobj);
}

您可以使用Map从test.values()获取数组。

Answer 2

您的条件不正确。应该为jQuery.inArray(newid, idsarray) === -1，因为如果jQuery.inArray(newid, idsarray)中没有-1，则操作newid将返回idsarray。因此，您需要检查idsarray中的唯一出现。

您也可以使用indexOf()操作。 idsarray.indexOf(newid) === -1。

OP的代码

$(document).ready(function() {

  var idsarray = [];
  var test = [{
      "name": "Mike",
      "incentives": "23.45",
      "id": "1"
    },
    {
      "name": "Larsen",
      "incentives": "34.78",
      "id": "2"
    },
    {
      "name": "Steve",
      "incentives": "26.78",
      "id": "3"
    }
  ];

  for (var i = 0; i < test.length; i++) {
    idsarray.push(test[i].id)
  }

  var newobj = {
    "name": "RAM",
    "incentives": "56.78",
    "id": "4"
  }

  var newid = newobj.id;
  if (jQuery.inArray(newid, idsarray) === -1) {
    test.push(newobj)
  }
  alert(test.length)

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

使用indexOf()

$(document).ready(function() {

  var idsarray = [];
  var test = [{
      "name": "Mike",
      "incentives": "23.45",
      "id": "1"
    },
    {
      "name": "Larsen",
      "incentives": "34.78",
      "id": "2"
    },
    {
      "name": "Steve",
      "incentives": "26.78",
      "id": "3"
    }
  ];

  for (var i = 0; i < test.length; i++) {
    idsarray.push(test[i].id)
  }

  var newobj = {
    "name": "RAM",
    "incentives": "56.78",
    "id": "4"
  }

  var newid = newobj.id;
  if (idsarray.indexOf(newobj) === -1) {
    test.push(newobj)
  }
  alert(test.length)

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 3

您可以避免使用包含对象array的另一个id。您可以使用本机函数Array.some来检查是否存在并相应地进行推送。

var test = [{"name":"Mike","incentives":"23.45","id":"1"},{"name":"Larsen","incentives":"34.78","id":"2"},{"name":"Steve","incentives":"26.78","id":"3"}];
var newobj = {"name":"RAM","incentives":"56.78","id":"4"};
if(!test.some(({id}) => id == newobj.id)) test.push(newobj);
console.log(test);

此外，如果您的推送操作非常频繁，则每次循环都会影响性能。您可以保留ID的ids或map而不是set的数组，因为搜索的复杂度为O（1）。

var test = [{"name":"Mike","incentives":"23.45","id":"1"},{"name":"Larsen","incentives":"34.78","id":"2"},{"name":"Steve","incentives":"26.78","id":"3"}];
var newobj = {"name":"RAM","incentives":"56.78","id":"4"};
var ids = test.reduce((a, {id}) => Object.assign(a, {[id]:id}), {});
if(!ids[newobj.id]) {
   test.push(newobj);
   ids[newobj.id] = newobj.id;
}
console.log(test);

如果对象不存在，则推送到JSON数组

3 个答案: