迭代“子对象”的javascript对象属性答案

【问题标题】：Iterating javascript Object properties of "subobjects"迭代“子对象”的javascript对象属性
【发布时间】：2016-02-26 20:52:58
【问题描述】：

我有一个这样的对象：

customers:
{
  239:
  {
    firstname : "Peter",
    lastname  : "Johnson"
  },
  242:
  {
    firstname : "Peter",
    lastname  : "Johnson"
  },
  etc...
}

如何迭代以获取所有名字，最好使用 jQuery？编号的 ID 不一定按顺序排列。

【问题讨论】：

$.each(customers, function(index, item) { console.log(item.firstname);});
@fuyushimoya .map 与 array 合作
谢谢指出。
@fuyushimoya，我喜欢你的解决方案。简单易懂。请把它作为一个答案，这样我就可以把它变成“接受的答案”:)

标签： javascript jquery object iteration

【解决方案1】：

要遍历object，并获取每个子对象的项目，您可以使用$.each 来实现，例如：

var customers = {
  239:
  {
    firstname : "Peter",
    lastname  : "Johnson"
  },
  242:
  {
    firstname : "Peter",
    lastname  : "Johnson"
  }
};

var firstnamesInArray = [],
        firstnamesInObject = [];

// The $.each requires an object/array as first param, and a function as second param.
// And it'll iterate through the object and call the passed in function with (key, value).
$.each(customers, function(index, item) {
  // So at here, you'll get, for example: index as 239, and item as {firstname: "Peter", lastname: "Johnson"}.
  
  // Do anything with the sub item.
  console.log(item.firstname);
  
  // You can also get the targets to arrayform or object form.
  firstnamesInArray.push(item.firstname);
  firstnamesInObject[index] = item.firstname;
});

console.log("Get into array:", firstnamesInArray);
console.log("Get into object:", firstnamesInObject);

&lt;script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"&gt;&lt;/script&gt;

【讨论】：

【解决方案2】：

var customers = {
    239: {
        firstname: "Peter",
        lastname: "Johnson"
    },
    242: {
        firstname: "Paul",
        lastname: "Johnson"
    },
};

var firstnames = Object.keys(customers)
    .reduce((p, c) => 
        p.push(customers[c].firstname) && p, []);

document.write(firstnames.join(', ')); // "Peter, Paul"

【讨论】：

【解决方案3】：

var data = {customers:
{
  239:
  {
    firstname : "Peter",
    lastname  : "Johnson"
  },
  242:
  {
    firstname : "Peter",
    lastname  : "Johnson"
  }
}};


$.each(data,function(k,v){
  $.each(v,function(key,value){
    console.log(key+"    "+value.firstname);
  });
});

&lt;script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"&gt;&lt;/script&gt;

你可以在 jquery 中这样做

【讨论】：