通過JSON對象數組迭代

Question

我分析了一個XML文件並檢索了以下JSON對象。 問題是json中有破折號，導致遍歷對象的問題。不幸的是我無法擺脫它們。

$(function() { 
 
    let json = JSON.parse('{"app-app":{"$":{},"notneeded":"123","mapped":{"$":{},"match-match":[{"$":{},"name":"Foo 1","url":"/Bar 1"},{"$":{},"name":"Foo 2","url":"/Bar 2"},{"$":{},"name":"Foo 3","url":"Bar 3"}]},"Nothing":"123"}}'); 
 

 
    var foo = Object.keys(json['app-app']['mapped']['match-match']).length; 
 
    for (var i = 0; i < foo; i++) { 
 
    console.log(json['app-app']['mapped']['match-match'][i].name); 
 
    console.log(json['app-app']['mapped']['match-match'][i].url); 
 
    } 
 
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

我想通過對象進行迭代，每一個 「匹配」 的孩子。類似這樣的：

return [ 
      { 
       name: 'Foo 1', 
       url: 'Bar 1' 
      }, 
      [...] 
     ] 

在此先感謝您。

Answer 1

您可以簡單地使用Array.prototype.map()，遍歷您的JSON數組，並返回您的自定義結構，像這樣：

$(function() { 
 
    let json = JSON.parse('{"app-app":{"$":{},"notneeded":"123","mapped":{"$":{},"match-match":[{"$":{},"name":"Foo 1","url":"/Bar 1"},{"$":{},"name":"Foo 2","url":"/Bar 2"},{"$":{},"name":"Foo 3","url":"Bar 3"}]},"Nothing":"123"}}'); 
 

 

 
    var result = json['app-app']['mapped']['match-match'].map(function(item) { 
 
    return { 
 
     "name": item.name, 
 
     "url": item.url 
 
    }; 
 
    }); 
 
    console.dir(result); 
 

 
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

它會給你預期的結果。

Answer 2

let json = JSON.parse('{"app":{"$":{},"notneeded":"123","mapped":{"$":{},"match":[{"$":{},"name":"Foo 1","url":"/Bar 1"},{"$":{},"name":"Foo 2","url":"/Bar 2"},{"$":{},"regex":"Foo 3","configuration":"Bar 3"}]},"Nothing":"123"}}'); 
 
console.log(json['app']['mapped']['match']);