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
这是我在@sagar-gavhane的回答之外所做的
const newUser = {_id: 4, name: 'Adam'}
const users = [{_id: 1, name: 'Fred'}, {_id: 2, name: 'Ted'}, {_id: 3, name:'Bill'}]
const userExists = users.some(user => user.name === newUser.name);
if(userExists) {
return new Error({error:'User exists'})
}
users.push(newUser)
其他回答
这个小片段对我有用。
const arrayOfObject = [{ id: 1, name: 'john' }, {id: 2, name: 'max'}];
const checkUsername = obj => obj.name === 'max';
console.log(arrayOfObject.some(checkUsername))
如果你有一个像['john'，'marsh']这样的元素数组，那么我们可以这样做
const checkUsername = element => element == 'john';
console.log(arrayOfObject.some(checkUsername))
const __checkIfElementExists__ = __itemFromArray__ => __itemFromArray__.*sameKey* === __outsideObject__.*samekey*;
if (cartArray.some(checkIfElementExists)) {
console.log('already exists');
} else {
alert('does not exists here')
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 }); }
}
这是我在@sagar-gavhane的回答之外所做的
const newUser = {_id: 4, name: 'Adam'}
const users = [{_id: 1, name: 'Fred'}, {_id: 2, name: 'Ted'}, {_id: 3, name:'Bill'}]
const userExists = users.some(user => user.name === newUser.name);
if(userExists) {
return new Error({error:'User exists'})
}
users.push(newUser)
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');
}
}
