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!
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 };
}
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');
}
}
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))
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)