[英]Is there a more efficient way to sort this array?
So I have an array of a certain user's invites (invites he has received from other users).所以我有一系列特定用户的邀请(他从其他用户那里收到的邀请)。 And I want the frontend to sort these invites by their status.
我希望前端按状态对这些邀请进行排序。 4 possible categories, expired, completed, accepted and pending.
4 个可能的类别,已过期、已完成、已接受和待处理。 The invites are fetched as objects from the DB.
邀请作为对象从数据库中获取。 And this is how I sort them.
这就是我对它们进行排序的方式。 The array
partArray
includes all of the invites and needs to be sorted into acceptedArr
, expiredArr
, completedArr
and pendingArr
.数组
partArray
包括所有邀请,需要分类为acceptedArr
、 expiredArr
、 completedArr
和pendingArr
。
for(let j = partarray.length-1;j>=0;j--){
if(partarray[j].userStats[auth.uid]==1 &&partarray[j].inviteStat==1){
acceptedArr.push(partarray[j].id)
partarray.splice(j, 1);
}
else if(partarray[j].userStats[auth.uid]==-1||partarray[j].inviteStat==0){
expiredArr.push(partarray[j].id)
partarray.splice(j, 1);
}
else if(partarray[j].userStats[auth.uid]==0&&partarray[j].inviteStat==1){
pendingArr.push(partarray[j].id)
partarray.splice(j, 1);
}else if(partarray[j].inviteStat==2){
completedArr.push(partarray[j].id)
partarray.splice(j, 1);
}
}
}
As you can see, in order for the invite to be sorted into the acceptedArr it needs to fulfil 2 conditions partarray[j].userStats[auth.uid]==1 && partarray[j].inviteStat==1
there are different conditions for each category.如您所见,为了将邀请分类到 acceptedArr 中,它需要满足 2 个条件
partarray[j].userStats[auth.uid]==1 && partarray[j].inviteStat==1
有不同的条件对于每个类别。 Is there a way to achieve what I'm doing in a more efficient manner?有没有办法以更有效的方式实现我正在做的事情?
Why not just filter them like this?为什么不像这样过滤它们呢?
const acceptedArr = partarray.filter(invite => invite.userStats[auth.uid]===1 && invite.inviteStat===1).map(invite => invite.id); const completedArr = partarray.filter(invite => invite.inviteStat === 2).map(invite => invite.id);
I would avoid splitting array into 4 categories and use array.sort
instead.我会避免将数组分成 4 个类别,而是使用
array.sort
。
Here's the implementation with sample data.这是示例数据的实现。
const auth = { uid: 'test' } const partarray = [{ userStats: { test: 1 }, inviteStat: 1 }, { userStats: { test: -1 }, inviteStat: 0 }, { userStats: { test: 0 }, inviteStat: 1 }, { userStats: { test: 1 }, inviteStat: 2 } ] const evaluateOrder = ({ userStats, inviteStat }) => { // lower value - higher ranking if (userStats[auth.uid] === 1 && inviteStat === 1) return 1; if (userStats[auth.uid] === -1 && inviteStat === 0) return 2; if (userStats[auth.uid] === 0 && inviteStat === 1) return 3; if (inviteStat === 2) return 4; } const sortUsers = (array) => { array.sort((prev, next) => { const prevOrder = evaluateOrder(prev); const nextOrder = evaluateOrder(next); return prevOrder - nextOrder; }) return array; } console.log(sortUsers(partarray))
Not sure if I understand what you're trying to do.不确定我是否理解你想要做什么。 You can eliminate a lot of that code with a switch statement though, and by moving the common code to the bottom:
不过,您可以使用 switch 语句并通过将公共代码移至底部来消除大量此类代码:
for (let j = partarray.length - 1; j >= 0; j--) {
let el = partarray[j]
let userStats = el.userStats[auth.uid]
let inviteStat = el.inviteStat
switch(inviteStat){
case 0:
//do something
break;
case 1:
if(userStats == 0){
//do something
} else if(userStats == 1){
//do something
}
break;
case 2:
//do something
break;
}
if (userStats == -1 ){
//do something
}
pendingArr.push(el.id)
partarray.splice(j, 1);
}
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