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

点击这里查看:

https://stackoverflow.com/a/53644664/1084987

你可以在后面创建if条件,比如

if(!contains(array, obj)) add();

其他回答

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
}

你可以建立你的数组原型,使它更模块化,尝试这样的东西

    Array.prototype.hasElement = function(element) {
        var i;
        for (i = 0; i < this.length; i++) {
            if (this[i] === element) {
                return i; //Returns element position, so it exists
            }
        }

        return -1; //The element isn't in your array
    };

你可以这样使用它:

 yourArray.hasElement(yourArrayElement)

Lodash中的xorWith可以用来实现这一点

let objects = [ { id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 2, username: 'ted' } ]
let existingObject = { id: 1, username: 'fred' };
let newObject = { id: 1729, username: 'Ramanujan' }

_.xorWith(objects, [existingObject], _.isEqual)
// returns [ { id: 2, username: 'bill' }, { id: 2, username: 'ted' } ]

_.xorWith(objects, [newObject], _.isEqual)
// returns [ { id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 2, username: 'ted' } ,{ id: 1729, username: 'Ramanujan' } ]
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');
  }
}

接受的答案也可以这样写,在。some上使用箭头函数

 function checkAndAdd(name) {
     var id = arr.length + 1;
     var found = arr.some((el) => {
           return el.username === name;
     });
     if (!found) { arr.push({ id: id, username: name }); }
 }