如果我有以下对象数组:

[ { id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 2, username: 'ted' } ]

是否有一种方法通过数组循环检查特定的用户名值是否已经存在,如果它不做任何事情,但如果它没有添加一个新对象到数组的用户名(和新ID)?

谢谢!


当前回答

我假设这里的id是唯一的。Find是一个很棒的数组方法,用于检查数组中是否存在东西:

Const arr = [{id: 1,用户名:'fred'}, {id: 2,用户名:'bill'}, {id: 3,用户名:'ted'}]; 函数add(arr, name) { Const {length} = arr; Const id =长度+ 1; Const found = arr。求(el => el。用户名=== name); 如果(!发现)arr。推送({id,用户名:name}); 返回arr; } console.log(添加(arr“ted”)); console.log(添加(加勒比海盗,“黛西”));

其他回答

极大地简化了我之前的解决方案,并通过在检查指定ID是否存在之前无需遍历整个数组来提供更好的性能。

这应该是最简单的解决方案(我认为):

const users = [{ id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 3, username: 'ted' }];
const addUser = (username) => {
  const user = users.find((user) => user.username === username);
  if (user) return { ...user, new: false };
  const newUser = {
    id: users.length + 1,
    username,
  };
  users.push(newUser);
  return { ...newUser, new: true };
};

下面是一个活生生的例子:

const users = [{ id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 3, username: 'ted' }]; const addUser = (username) => { const user = users.find((user) => user.username === username); if (user) return { ...user, new: false }; const newUser = { id: users.length + 1, username, }; users.push(newUser); return { ...newUser, new: true }; }; // a little extra scripting here to support the input and button in the example const form = document.querySelector('form'); const input = document.querySelector('input'); const span = document.querySelector('span'); const pre = document.querySelector('pre'); const syncDataWithPre = () => { pre.innerHTML = JSON.stringify(users, null, 2); }; form.onsubmit = (e) => { e.preventDefault(); span.textContent = ''; if (input.value) { const user = addUser(input.value); const { new: isNew, ...userDetails } = user; span.classList[isNew ? 'add' : 'remove']('new'); span.textContent = `User ${isNew ? 'added' : 'already exists'}`; } input.value = ''; syncDataWithPre(); }; syncDataWithPre(); body { font-family: arial, sans-serif; } span { display: block; padding-top: 8px; font-weight: 700; color: #777; } span:empty { display: none; } .new { color: #0a0; } .existing: { color: #777; } <form> <input placeholder="New username" /> <button>Add user</button> </form> <span></span> <pre></pre>

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 };
}

你也可以试试这个

 const addUser = (name) => {
    if (arr.filter(a => a.name == name).length <= 0)
        arr.push({
            id: arr.length + 1,
            name: name
        })
}
addUser('Fred')

下面是一个ES6方法链,使用.map()和.includes():

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

const checkForUser = (newUsername) => {
      arr.map(user => {
        return user.username
      }).includes(newUsername)
    }

if (!checkForUser('fred')){
  // add fred
}

映射现有用户以创建用户名字符串数组。 检查该用户名数组是否包含新用户名 如果不存在,则添加新用户

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');
  }
}