簡體   English   中英

如何排序對象數組?

[英]How to sort an array of objects?

我有一個對象數組:

var data= [{
    "title": "All pages",
    "page": "all",

}, {
    "title": "Post with builder",
    "page": "pt_post_6188",

}, {
    "title": "Blog Categories",
    "page": "tx_category",

}, {
    "title": "Single Blog Posts",
    "page": "pt_post",

}];

以及由對象項標題構成的排序順序數組。

var order = ["Post with builder", "All pages", "Blog Categories", "Single Blog Posts"];

我如何按順序數組對第一個數組進行排序,以便新數據像這樣

var newdata= [{
    "title": "Post with builder",
    "page": "pt_post_6188",

}, {
    "title": "All pages",
    "page": "all",

}, {
    "title": "Blog Categories",
    "page": "tx_category",

}, {
    "title": "Single Blog Posts",
    "page": "pt_post",

}];

與參考文章不同。 我按沒有任何特定邏輯的特定順序數組進行排序。

您可以將對象用作哈希表

{
    "Post with builder": 0,
    "All pages": 1,
    "Blog Categories": 2,
    "Single Blog Posts": 3
}

索引並對其進行排序。

 var data = [{ "title": "All pages", "page": "all", }, { "title": "Post with builder", "page": "pt_post_6188", }, { "title": "Blog Categories", "page": "tx_category", }, { "title": "Single Blog Posts", "page": "pt_post", }], order = ["Post with builder", "All pages", "Blog Categories", "Single Blog Posts"], orderObject = {}; order.forEach(function (a, i) { orderObject[a] = i; }) data.sort(function (a, b) { return orderObject[a.title] - orderObject[b.title]; }); document.write('<pre>' + JSON.stringify(data, 0, 4) + '</pre>'); 

由於您已經將訂購數組索引作為一種哈希值用於訂購參考,因此我們可以像這樣簡單地使用它們:

sorted = Array(order.length);
data.forEach((c,i) => sorted[order.indexOf(c.title)] = c);

//使用loadash

var data= [{
"title": "All pages",
"page": "all",
}, {
"title": "Post with builder",
"page": "pt_post_6188",
}, {
"title": "Blog Categories",
"page": "tx_category",
}, {
"title": "Single Blog Posts",
"page": "pt_post",
}];

var order = [“帶有構建器的帖子”,“所有頁面”,“博客類別”,“單個博客帖子”];

function (data, order){
 return _.reduce(order, function (output, v) {
  output.push(_.filter(data,{title:v})[0]);
  return output;
 },[]);
}

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM