Question

我希望有一个阵列，其客人没有＆＃34;拒绝＆＃34;作为回应。因此，Tooth Fairy被删除（不应该，她已经接受了＃34;邀请）并且杰克弗罗斯特停留了（不应该，＃34;他拒绝了＃34;邀请）。

function getAttendees(peopleInvited, responses){
    var coming=peopleInvited;
  responses.map(function(cell){
      if (cell.response=='declined') {
        coming.splice(0,1);
         }         
      });
    return coming;  
}

var people = ['Easter Bunny', 'Tooth Fairy', 'Frosty the Snowman', 
              'Jack Frost', 'Cupid', 'Father Time'];
var responses = [ 
     {name: 'Easter Bunny', response: 'declined'}, 
     {name: 'Jack Frost', response: 'declined'}, 
     {name: 'Tooth Fairy', response: 'accepted'} 
   ];

getAttendees(people, responses);

Answer 1

首先，你需要在即将到来的数组中获取人物的索引，然后使用splice根据它的索引删除该人。

function getAttendees(peopleInvited, responses){
  var coming=peopleInvited;
  responses.map(function(cell){
    if (cell.response=='declined') {
      var index = coming.indexOf(cell.name);
      coming.splice(index, 1);
    }         
  });
  return coming;  
}

var people = ['Easter Bunny', 'Tooth Fairy', 'Frosty the Snowman', 
     'Jack Frost', 'Cupid', 'Father Time'];
var responses = [ 
  {name: 'Easter Bunny', response: 'declined'}, 
  {name: 'Jack Frost', response: 'declined'}, 
  {name: 'Tooth Fairy', response: 'accepted'} 
];

getAttendees(people, responses);

Answer 2

您可以使用人员的姓名作为键将响应数组更改为对象。我认为管理起来有点简单。

function getAttendees(people, responses) {
    for (var i = 0, l = people.length, out = []; i < l; i++) {
        var response = responses[people[i]];
        if (!response || response && response === 'accepted') {
            out.push(people[i]);
        }
    }
    return out;
}

var people = ['Easter Bunny', 'Tooth Fairy', 'Frosty the Snowman',
    'Jack Frost', 'Cupid', 'Father Time'];

var responses = {
    'Easter Bunny': 'declined',
    'Jack Frost': 'declined',
    'Tooth Fairy': 'accepted'
}

DEMO

Answer 3

那是因为它们在不同数组中列出的顺序不同。

首先你会找到＃34;复活节兔子＆＃34;在responses数组上，您从coming数组中删除第一个匹配项，而不检查它是什么。在这种情况下，它对应。

然后，你发现＆＃34;拒绝＆＃34;由杰克弗罗斯特＆＃34;并从coming删除第一个（新）事件，现在是＃34; Tooth Fairy＆＃34;。

更改顺序，以便两个数组具有相同的顺序，或者以不同的方式编码以不依赖于顺序（在我看来这更好）。

array.splice删除错误的值并保持错误的值

3 个答案: