如何从多个嵌套对象中计算特定值的对象数？

Question

 var foodwebsites = { "bacon": [{ "url": "stackoverflow.com", }], "icecream": [{ "url": "example.com", }], "cheese": [{ "url": "example.com", }] } var baconfoodwebsites = foodwebsites.bacon.filter(function(elem) { return elem.url == 'example.com'; }).length; var icecreamfoodwebsites = foodwebsites.icecream.filter(function(elem) { return elem.url == 'example.com'; }).length; var cheesefoodwebsites = foodwebsites.cheese.filter(function(elem) { return elem.url == 'example.com'; }).length; var allfoodwebsites = baconfoodwebsites + icecreamfoodwebsites + cheesefoodwebsites; console.log(baconfoodwebsites, icecreamfoodwebsites, cheesefoodwebsites, allfoodwebsites) 

我想做同样的事情，而不必重复所有这些单独的嵌套对象（培根，冰淇淋和奶酪）。

我认为答案将是：

var allfoodwebsites = foodwebsites=.filter(function( elem) {
    return elem.url == 'example.com';
}).length;

附加信息：

如果可能的话，我只想使用jQuery + Pure Javascript。

我想找到所有带有"url": "example.com"嵌套对象"url": "example.com"

Answer 1

最好的选择是reduce计数方法的功能。

 let foodwebsites = {"bacon": [{"url": "stackoverflow.com",}],"icecream": [{"url": "example.com",}],"cheese": [{"url": "example.com",}]}; let allfoodwebsites = Object.values(foodwebsites). reduce((a, array) => a + array. reduce((a, {url}) => a + (url === "example.com"), 0), 0); console.log(allfoodwebsites); 

Answer 2

使用Object.values，Array＃reduce（）和Array＃flat（）。 注意flat（）在某些环境中可能需要polyfill

 let foodwebsites = {"bacon": [{"url": "stackoverflow.com",}],"icecream": [{"url": "example.com",}],"cheese": [{"url": "example.com",}]}; const getUrlCount = (url) => { return Object.values(foodwebsites) .flat() .reduce((a, {url:u})=> a + (url === u) , 0) } console.log(getUrlCount("example.com")) 

Answer 3

解：

您可以创建一个函数，该函数将迭代对象键并在每个访问键的子对象数组中搜索property和value ：

let amount = (p, v, i = 0) => 
(Object.keys(foodwebsites).forEach(k => foodwebsites[k].forEach(o => o[p] === v && i++))
, i);

可以这样使用：

amount("url", "example.com"); // 2

---

工作代码段：

 var foodwebsites = { "bacon": [{ "url": "stackoverflow.com", }], "icecream": [{ "url": "example.com", }], "cheese": [{ "url": "example.com", }] } let amount = (p, v, i = 0) => (Object.keys(foodwebsites).forEach(k => foodwebsites[k].forEach(o => o[p] === v && i++)) , i); console.log( amount("url", "example.com") ); //2 

Answer 4

我实际上建立了所有这些嵌套条目的一个数组：

  const sites = Object.values(foodwebsites).flat();

然后，您可以轻松地对其进行迭代并计算所有键：

 const count = (arr, key, value) => arr.reduce((acc, it) => acc + it[key] === value, 0);

 console.log(
   count(sites, "url", "example.com"),
   count(sites, "url", "stackoverflow.com")
 );