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

试试这个

第一种方法使用一些

  let arr = [{ id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 3, username: 'ted' }];
    let found = arr.some(ele => ele.username === 'bill');
    console.log(found)

第二种方法使用包括、映射

   let arr = [{ id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 3, username: 'ted' }];
    let mapped = arr.map(ele => ele.username);
    let found = mapped.includes('bill');
    console.log(found)

其他回答

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)

试试这个

第一种方法使用一些

  let arr = [{ id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 3, username: 'ted' }];
    let found = arr.some(ele => ele.username === 'bill');
    console.log(found)

第二种方法使用包括、映射

   let arr = [{ id: 1, username: 'fred' }, { id: 2, username: 'bill' }, { id: 3, username: 'ted' }];
    let mapped = arr.map(ele => ele.username);
    let found = mapped.includes('bill');
    console.log(found)

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')
const __checkIfElementExists__ = __itemFromArray__ => __itemFromArray__.*sameKey* === __outsideObject__.*samekey*;

    if (cartArray.some(checkIfElementExists)) {
        console.log('already exists');
    } else {
        alert('does not exists here')

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
}