假设我们有一个对象数组，你想检查value of name是否像这样定义，

let persons = [ {"name" : "test1"},{"name": "test2"}];

if(persons.some(person => person.name == 'test1')) {
    ... here your code in case person.name is defined and available
}

我认为，这是解决这个问题的最短方法。这里我使用ES6的箭头函数和.filter来检查新添加的用户名是否存在。

var arr = [{
    id: 1,
    username: 'fred'
}, {
    id: 2,
    username: 'bill'
}, {
    id: 3,
    username: 'ted'
}];

function add(name) {
    var id = arr.length + 1;        
            if (arr.filter(item=> item.username == name).length == 0){
            arr.push({ id: id, username: name });
        }
}

add('ted');
console.log(arr);

链接到小提琴

function number_present_or_not() {
  var arr = [2, 5, 9, 67, 78, 8, 454, 4, 6, 79, 64, 688];
  var found = 6;
  var found_two;
  for (i = 0; i < arr.length; i++) {
    if (found == arr[i]) {
      found_two = arr[i];
      break;
    }
  }
  if (found_two == found) {
    console.log('number present in the array');
  } else {
    console.log('number not present in the array');
  }
}

请看下面的例子

$(document).ready(function(){ const arr = document.querySelector(".list"); var abcde = [{ id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 2, username: 'ted' }]; $("#btnCheckUser").click(function() { var tbUsername = $("#tbUsername").val(); if (abcde.some(obj => obj.username === tbUsername)) { alert('existing user ' + tbUsername); return; } else { abcde.push({ id: abcde.length + 1, username: tbUsername }); alert('added new user ' + tbUsername); arr.appendChild(createArray(tbUsername)); return; } }); function createArray(name) { let li = document.createElement("li"); li.textContent = name; return li; } abcde.forEach((x) => arr.appendChild(createArray(x.username))); }); <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script> <p>Add text and click on Check.</p> <input type="text" id="tbUsername" /> <button type="button" id="btnCheckUser">Check</button> <div class="list"> <ul></ul> </div>

我喜欢Andy的回答，但是id不一定是唯一的，所以这里是我想出的创建唯一id的方法。也可以在jsfiddle上查看。请注意arr。如果之前已经删除了任何内容，length + 1可能无法保证唯一的ID。

var array = [ { id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 3, username: 'ted' } ];
var usedname = 'bill';
var newname = 'sam';

// don't add used name
console.log('before usedname: ' + JSON.stringify(array));
tryAdd(usedname, array);
console.log('before newname: ' + JSON.stringify(array));
tryAdd(newname, array);
console.log('after newname: ' + JSON.stringify(array));

function tryAdd(name, array) {
    var found = false;
    var i = 0;
    var maxId = 1;
    for (i in array) {
        // Check max id
        if (maxId <= array[i].id)
            maxId = array[i].id + 1;

        // Don't need to add if we find it
        if (array[i].username === name)
            found = true;
    }

    if (!found)
        array[++i] = { id: maxId, username: name };
}

出于某种原因，我确实尝试了上述步骤，但对我来说似乎并不管用，但这是我对自己问题的最终解决方案，可能对阅读本文的任何人都有帮助:

let pst = post.likes.some( (like) => {  //console.log(like.user, req.user.id);
                                     if(like.user.toString() === req.user.id.toString()){
                                         return true
                                     } } )

这帖子。点赞是喜欢某篇文章的用户的数组。