如何根据出现在许多地方的另一个属性对一个 JSON 属性进行分组？答案

【问题标题】：How to group one JSON property based on one other property that appears at many places?如何根据出现在许多地方的另一个属性对一个 JSON 属性进行分组？
【发布时间】：2017-05-18 03:16:45
【问题描述】：

我有这个 JSON：

var books = [

    { author: "George Owell", title: "1984", price: 0, page: 328}, 
    { author: "Jane Austen", title: "Pride And Prejudice", price: 20, page: 279}, 
    { author: "J.K. Rowling", title: "Harry Potter and the Sorcerer's Stone", price: 23, page: 320}, 
    { author: "Plato", title: "Apology", price: 0, page: 127}, 
    { author: "Sigmund Freud", title: "The Interpretation of Dreams"},
    { author: "George Owell", title: "Animal Farm", price: 0, page: 100}, 
    { author: "George Owell", title: "Omh", price: 0, page: 100}, 
    { author: "Plato", title: "Aristotle", price: 0, page: 227}

];

现在我想制作一个输出作者和标题的表格。如果作者相同，则标题将合并在一起。例如，“George Owell”的标题为“1984, Animal Farm, Ohm.”，“Plato”的标题为“Apology, Aristotle”，等等。我如何使用 jQuery 或 vanilla JavaScript 来做到这一点？

【问题讨论】：

您的问题中没有 JSON：您有一个对象数组。无论如何，我可能会为此使用数组.reduce() 方法。您是在询问如何生成 HTML 输出以显示结果，还是只是在询问如何合并标题？合并部分可以（或多或少）以与此问题相同的方式完成：stackoverflow.com/questions/33850412/…
如果您不关心“价格”或“页面”，我可以非常快速地编写代码来执行此操作 - 这些是“书籍”，因此想要创建一个名为“作者”，那么我想这是有道理的......但我认为你可能试图以错误的方式解决另一个问题。

标签： javascript jquery

【解决方案1】：

var books = [
    { author: "George Owell", title: "1984", price: 0, page: 328}, 
    { author: "Jane Austen", title: "Pride And Prejudice", price: 20, page: 279}, 
    { author: "J.K. Rowling", title: "Harry Potter and the Sorcerer's Stone", price: 23, page: 320}, 
    { author: "Plato", title: "Apology", price: 0, page: 127}, 
    { author: "Sigmund Freud", title: "The Interpretation of Dreams"},
    { author: "George Owell", title: "Animal Farm", price: 0, page: 100}, 
    { author: "George Owell", title: "Omh", price: 0, page: 100}, 
    { author: "Plato", title: "Aristotle", price: 0, page: 227}
];

var uniqueauthors = [];
$.each(books,function(index,value){
if($.inArray(value.author,uniqueauthors) < 0)
{
uniqueauthors.push(value.author);
}
});

$.each(uniqueauthors,function(index,value){
var bok = $.map(books,function(val,index){
return (val.author == value) ? val.title : null;
});
var $tr = $('<tr>').append('<td>'+value+'</td>').append('<td>'+bok.join(",")+'</td>');
$('table').append($tr);
});

td{
  border: 1px solid black;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
<tr>
<td>Authors</td>
<td>Titles</td>
</tr>
</table>

【讨论】：

【解决方案2】：

以下内容应该可以帮助您入门：

var books = [

    { author: "George Owell", title: "1984", price: 0, page: 328}, 
    { author: "Jane Austen", title: "Pride And Prejudice", price: 20, page: 279}, 
    { author: "J.K. Rowling", title: "Harry Potter and the Sorcerer's Stone", price: 23, page: 320}, 
    { author: "Plato", title: "Apology", price: 0, page: 127}, 
    { author: "Sigmund Freud", title: "The Interpretation of Dreams"},
    { author: "George Owell", title: "Animal Farm", price: 0, page: 100}, 
    { author: "George Owell", title: "Omh", price: 0, page: 100}, 
    { author: "Plato", title: "Aristotle", price: 0, page: 227}
    ];


   var authorBooks = {}

   books.forEach(function(book) {

      if(!authorBooks[book.author]) {
        authorBooks[book.author] = [];
      }

      authorBooks[book.author].push({title: book.title, price: book.price, page: book.page});

    },{});

【讨论】：

【解决方案3】：

昨天在 python 中出现了同样的问题。这是我的回答：https://stackoverflow.com/a/44017627/7633845

这是一个js版本：

var map = {}
for(var i = 0; i < books.length; i++) {
    var book = books[i];
    if(!(book.author in map)) {
        map[book.author] = []
    }
    map[book.author].push(book.title)
}

var out = []
for(var author in map) {
    out.push({
        author: author,
        books: map[author].join(", ")
    })
}

【讨论】：

【解决方案4】：

你可以试试

var books = [

{ author: "George Owell", title: "1984", price: 0, page: 328}, 
{ author: "Jane Austen", title: "Pride And Prejudice", price: 20, page: 279}, 
{ author: "J.K. Rowling", title: "Harry Potter and the Sorcerer's Stone", price: 23, page: 320}, 
{ author: "Plato", title: "Apology", price: 0, page: 127}, 
{ author: "Sigmund Freud", title: "The Interpretation of Dreams"},
{ author: "George Owell", title: "Animal Farm", price: 0, page: 100}, 
{ author: "George Owell", title: "Omh", price: 0, page: 100}, 
{ author: "Plato", title: "Aristotle", price: 0, page: 227}
];

var results = books.reduce((previous, book) => {
(previous[book.author] = previous[book.author] || []).push(book);
    return previous;
}, []);

console.log(results);

【讨论】：