Question

我有两个用户对象数组，其中包含_id和lastSeen。

allUsersSeen = [
   {_id: '12csdf', lastSeen: 212512},
   {_id: '232csdf', lastSeen: 34345},
   {_id: '1253csdf', lastSeen: 6346453},
   {_id: '772csfddf', lastSeen: 2345345}
]

currentUsersSeen = [
   {_id: '12csdf', lastSeen: Date.now()},
   {_id: '1253csdf', lastSeen: Date.now()},
   {_id: '1fsdf', lastSeen: Date.now()}
]

我想比较这两个数组，如果allUsersSeen _id与currentUsersSeen匹配，我想将lastSeen更新为Date.now（）。所有其他_id将使其lastSeen保持原样。

来自currentUsersSeen的任何剩余_id应该使用lastSeen of Date.now（）

推送到allUsersSeen上

allUsersSeen的最终数组应如下所示：

allUsersSeen = [
   {_id: '12csdf', lastSeen: Date.now()},
   {_id: '232csdf', lastSeen: 34345},
   {_id: '1253csdf', lastSeen: Date.now()},
   {_id: '772csfddf', lastSeen: 2345345},
   {_id: '1fsdf', lastSeen: Date.now()}
]

显然，Date.now（）将是当前的日期/时间。我该怎么做呢？我尝试过使用嵌套的for循环，并尝试使用forEach（）和map（）。

非常感谢任何帮助。谢谢。

Answer 1

你可能想要那样的东西

allUsersSeen = [
   {_id: '12csdf', lastSeen: 212512},
   {_id: '232csdf', lastSeen: 34345},
   {_id: '1253csdf', lastSeen: 6346453},
   {_id: '772csfddf', lastSeen: 2345345}
]

currentUsersSeen = [
   {_id: '12csdf', lastSeen: Date.now()},
   {_id: '1253csdf', lastSeen: Date.now()},
   {_id: '1fsdf', lastSeen: Date.now()}
]

let newArr = allUsersSeen.map(user => currentUsersSeen.find(curr_user => curr_user._id === user._id) || user);

Answer 2

如果您了解自己要做的事情，可以先从allUsersSeen（如果存在）中删除该元素，然后再从currentUser中再次添加该元素，从而实现这一目标。

allUsersSeen = [
   {_id: '12csdf', lastSeen: 212512},
   {_id: '232csdf', lastSeen: 34345},
   {_id: '1253csdf', lastSeen: 6346453},
   {_id: '772csfddf', lastSeen: 2345345}
]

currentUsersSeen = [
   {_id: '12csdf', lastSeen: Date.now()},
   {_id: '1253csdf', lastSeen: Date.now()},
   {_id: '1fsdf', lastSeen: Date.now()}
]

currentUsersSeen.forEach(user => {
  allUsersSeen = allUsersSeen.filter( obj => obj._id != user.id )
  allUsersSeen.push(user)
})

Answer 3

试试这个，

allUsersSeen = [
        {_id: '12csdf', lastSeen: 212512},
        {_id: '232csdf', lastSeen: 34345},
        {_id: '1253csdf', lastSeen: 6346453},
        {_id: '772csfddf', lastSeen: 2345345}
    ];

    currentUsersSeen = [
        {_id: '12csdf', lastSeen: Date.now()},
        {_id: '1253csdf', lastSeen: Date.now()},
        {_id: '1fsdf', lastSeen: Date.now()}
    ];

    function merge(a, b, prop){
        var reduced =  a.filter( a_item => ! b.find ( b_item => a_item[prop] === b_item[prop]) )
        return reduced.concat(b);
    }
    console.log( merge(allUsersSeen, currentUsersSeen, "_id") );

Answer 4

您可以array#concat两个数组，然后使用array#reduce更新lastSeen。

var allUsersSeen = [{_id: '12csdf', lastSeen: 212512},{_id: '232csdf', lastSeen: 34345},{_id: '1253csdf', lastSeen: 6346453},{_id: '772csfddf', lastSeen: 2345345}],
    currentUsersSeen = [{_id: '12csdf', lastSeen: Date.now()},{_id: '1253csdf', lastSeen: Date.now()},{_id: '1fsdf', lastSeen: Date.now()}],
    
    result = Object.values( allUsersSeen.concat(currentUsersSeen).reduce((r, {_id, lastSeen}) => {
      r[_id] = r[_id] ? {...r[_id], lastSeen} : {_id, lastSeen};
      return r;
    },{}));

console.log(result);

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在Javascript中比较两个对象数组并根据id更新lastSeen

4 个答案: