情况是：我有一个对象数组，其中每个对象都有一个对象数组。数组如下所示：
[
{
"dislikes": [
{
"createDate": {
"date": 11,
"day": 0,
"hours": 18,
"minutes": 15,
"month": 10,
"seconds": 11,
"time": 1541956511001,
"timezoneOffset": -60,
"year": 118
},
},
{
"createDate": {
"date": 11,
"day": 0,
"hours": 18,
"minutes": 15,
"month": 10,
"seconds": 11,
"time": 1541956511008,
"timezoneOffset": -60,
"year": 118
},
}
],
},
{
"dislikes": [
{
"createDate": {
"date": 11,
"day": 0,
"hours": 18,
"minutes": 15,
"month": 10,
"seconds": 11,
"time": 1541956511011,
"timezoneOffset": -60,
"year": 118
},
},
{
"createDate": {
"date": 11,
"day": 0,
"hours": 18,
"minutes": 15,
"month": 10,
"seconds": 11,
"time": 1541956511028,
"timezoneOffset": -60,
"year": 118
},
}
],
}
]
因此，我想按time的不喜欢程度对用户和不喜欢的对象进行排序。因此，最早厌恶的用户将是第一个，而最早厌恶的用户将在每个用户的dislikes数组中排名第一。我相信我必须做多种事情，但是我该怎么做呢？
答案 0 :(得分：2)
您可以映射项目并向其中添加包含最早不喜欢的属性，然后对其进行排序：
const data = [{"dislikes":[{"createDate":{"date":11,"day":0,"hours":18,"minutes":15,"month":10,"seconds":11,"time":1541956511001,"timezoneOffset":-60,"year":118}},{"createDate":{"date":11,"day":0,"hours":18,"minutes":15,"month":10,"seconds":11,"time":1541956511008,"timezoneOffset":-60,"year":118}}]},{"dislikes":[{"createDate":{"date":11,"day":0,"hours":18,"minutes":15,"month":10,"seconds":11,"time":1541956511011,"timezoneOffset":-60,"year":118}},{"createDate":{"date":11,"day":0,"hours":18,"minutes":15,"month":10,"seconds":11,"time":1541956511028,"timezoneOffset":-60,"year":118}}]}];
console.log(
data
//map and add newestDislike property
.map((d) => ({
...d,
//reduce and only takes the lowest time value
newestDislike: (d.dislikes || []).reduce(
(result, item) =>
item.createDate.time < result
? item.createDate.time
: result,
Infinity, //defaults to infinity (if no dislikes)
),
}))
.sort((a, b) => a.newestDislike - b.newestDislike),
);
如果已经按照最早的日期对用户的不喜欢进行了排序，那么您可以跳过地图并减少部分。如果用户可能有空的不喜欢或未定义的内容，请确保使用默认的getter函数，这样您的代码就不会崩溃：
//gets a nested prop from object or returns defaultValue
const get = (o = {}, path, defaultValue) => {
const recur = (o, path, defaultValue) => {
if (o === undefined) return defaultValue;
if (path.length === 0) return o;
if (!(path[0] in o)) return defaultValue;
return recur(o[path[0]], path.slice(1), defaultValue);
};
return recur(o, path, defaultValue);
};
console.log(
data.sort(
(a, b) =>
get(
a,
['dislikes', 0, 'createDate', 'time'],
Infinity,
) -
get(
b,
['dislikes', 0, 'createDate', 'time'],
Infinity,
),
),
);
答案 1 :(得分：2)
//Supply the array you've metioned as the argument users to the below method, sortDislikesForAllUsers
let sortDislikesForAllUsers = function(users) {
return users.map(user => {
return {
dislikes: user.dislikes.sort((dislikeA, dislikeB) => ((dislikeA.createDate.time < dislikeB.createDate.time) ? -1 : (dislikeA.createDate.time > dislikeB.createDate.time) ? 1 : 0))
}
})
}
//Supply the array returned in the above method as input to the below method, sortUsers
let sortUsers = function(arrayOfSortedDislikesPerUser) {
return arrayOfSortedDislikesPerUser.sort((userA, userB) => ((userA.dislikes[0].createDate.time < userB.dislikes[0].createDate.time) ? -1 : (userA.dislikes[0].createDate.time > userB.dislikes[0].createDate.time) ? 1 : 0))
}
let arrayOfSortedDislikesPerUser = sortDislikesForAllUsers(users);
let finalSortedArray = sortUsers(arrayOfSortedDislikesPerUser);
console.log(finalSortedArray);
在以下代码段中，
sortDislikesForAllUsers 这种方法可以对个人的不喜欢进行排序 用户sortUsers 该方法根据用户的第一次不喜欢时间对用户进行排序 以上方法获得的排序的不喜欢数组的元素
简单：）
运行以下代码段。您可以直接复制粘贴到您的代码中！
let users = [{
"dislikes": [
{
"createDate": {
"date": 11,
"day": 0,
"hours": 18,
"minutes": 15,
"month": 10,
"seconds": 11,
"time": 1541956511001,
"timezoneOffset": -60,
"year": 118
},
},
{
"createDate": {
"date": 11,
"day": 0,
"hours": 18,
"minutes": 15,
"month": 10,
"seconds": 11,
"time": 1541956511008,
"timezoneOffset": -60,
"year": 118
},
}
],
},
{
"dislikes": [
{
"createDate": {
"date": 11,
"day": 0,
"hours": 18,
"minutes": 15,
"month": 10,
"seconds": 11,
"time": 1541956511011,
"timezoneOffset": -60,
"year": 118
},
},
{
"createDate": {
"date": 11,
"day": 0,
"hours": 18,
"minutes": 15,
"month": 10,
"seconds": 11,
"time": 1541956511028,
"timezoneOffset": -60,
"year": 118
},
}
],
}]
let sortDislikesForAllUsers = function(users) {
return users.map(user => {
return {
dislikes: user.dislikes.sort((dislikeA, dislikeB) => ((dislikeA.createDate.time < dislikeB.createDate.time) ? -1 : (dislikeA.createDate.time > dislikeB.createDate.time) ? 1 : 0))
}
})
}
let sortUsers = function(arrayOfSortedDislikesPerUser) {
return arrayOfSortedDislikesPerUser.sort((userA, userB) => ((userA.dislikes[0].createDate.time < userB.dislikes[0].createDate.time) ? -1 : (userA.dislikes[0].createDate.time > userB.dislikes[0].createDate.time) ? 1 : 0))
}
let arrayOfSortedDislikesPerUser = sortDislikesForAllUsers(users);
let finalSortedArray = sortUsers(arrayOfSortedDislikesPerUser);
console.log(finalSortedArray);
编辑：@HMR对评论进行WRT ：
1。它会改变原始数组。是。如果要避免突变，则必须创建已发送数组的副本。
let noRefCopy = new Array()
noRefCopy = noRefCopy.concat(originalArr)
现在，对副本进行排序并返回副本。
2。如果您想检查未定义的等，请确定。
以上答案试图解决逻辑问题。如果问题确实是针对他们的，我们当然可以解决以上两个问题。
干杯，
克鲁西卡
答案 2 :(得分：1)
如下所示（使用lodash.js）
_.each(users, (u) => { u.dislikes = _.sortBy(u.dislikes, 'createdDate.time'); });
users = _.sortBy(users, 'dislikes[0].createdDate.time');
答案 3 :(得分：1)
检查下面的代码。这将使您可以根据时间进行排序：
function sortByTime(obj1, obj2){
return obj1.time - obj2.time;
}
array.sort((obj1, obj2)=>{
obj1.dislikes.sort(sortByTime);
obj2.dislikes.sort(sortByTime);
return obj1.dislikes[0].time - obj2.dislikes[0].time;
});
最早的时间我没听懂你的意思。上面的代码按时间升序排序。
注意：：上面的代码无法处理缺少财产之夜的极端情况