将变量添加到声音云响应json。

Question

我正在使用声音云api创建应用程序。我在做什么，从

http://api.soundcloud.com/resolve.json?url=http://soundcloud.com/matas/hobnotropic&client_id=YOUR_CLIENT_ID 

得到JSON这工作得很好，并返回所需的东西，但我想怎么过到一个新的变量添加到返回的JSON输出。

目前输出看起来像下面

{"kind":"track","id":49931,"created_at":"2008/10/27 09:56:47 +0000","user_id":1433,"duration":489138,"commentable":true,"state":"finished","original_content_size":130032044,"sharing":"public","tag_list":"dub minimalist hypnotic flanger","permalink":"hobnotropic","streamable":true,"embeddable_by":"all","downloadable":false,"purchase_url":"https://soundcloud.com/matas","label_id":null,"purchase_title":"Get more tracks!","genre":"dub","title":"Hobnotropic","description":"Kinda of an experiment in search for my own sound. I've produced this track from 2 loops I've made using Hobnox Audiotool (http://www.hobnox.com/audiotool.1046.en.html). Imported into Ableton LIve! and tweaked some FX afterwards.","label_name":"","release":"","track_type":"original","key_signature":"","isrc":"","video_url":null,"bpm":84.0,"release_year":null,"release_month":null,"release_day":null,"original_format":"wav","license":"cc-by-nc","uri":"http://api.soundcloud.com/tracks/49931","user":{"id":1433,"kind":"user","permalink":"matas","username":"matas","uri":"http://api.soundcloud.com/users/1433","permalink_url":"http://soundcloud.com/matas","avatar_url":"http://i1.sndcdn.com/avatars-000001548772-zay6dy-large.jpg?16b9957"},"permalink_url":"http://soundcloud.com/matas/hobnotropic","artwork_url":"http://i1.sndcdn.com/artworks-000000103093-941e7e-large.jpg?16b9957","waveform_url":"http://w1.sndcdn.com/IqSLUxN7arjs_m.png","stream_url":"http://api.soundcloud.com/tracks/49931/stream","playback_count":74836,"download_count":280,"favoritings_count":25,"comment_count":23,"attachments_uri":"http://api.soundcloud.com/tracks/49931/attachments"} 

什么我想要的只是添加像"custom_url":"http://example.com/example123"

可变因此，谁能指导我，是否有任何可能的方式来做到这一点。我想知道我是否可以将变量传递给soundcloud，并且soundcloud会自动将该变量追加到响应中。

Answer 1

首先，存储在一个变量的JSON响应：

var obj = response; 

所以，你现在持有JavaScript对象的变量。将属性添加到您的对象，你简单的使用：

obj.custom_url = "Your_url"; 

或

obj["custom_url"] = "your_url"; 

Answer 2

你不需要或想要的jQuery这一点，只是JavaScript的。

采取一个例子

var data = {items: [ 
    {id: "1", name: "Snatch", type: "crime"}, 
    {id: "2", name: "Witches of Eastwick", type: "comedy"}, 
    {id: "3", name: "X-Men", type: "action"}, 
    {id: "4", name: "Ordinary People", type: "drama"}, 
    {id: "5", name: "Billy Elliot", type: "drama"}, 
    {id: "6", name: "Toy Story", type: "children"} 
]}; 

添加项目：

data.items.push(
    {id: "7", name: "Douglas Adams", type: "comedy"} 
); 

，增加了端。见下面添加在中间。

中的项目：

有几种方法。拼接方法是最通用的：

data.items.splice(1, 3); // Removes three items starting with the 2nd, 
         // ("Witches of Eastwick", "X-Men", "Ordinary People") 

拼接修改原始数组，并返回您删除的项目的数组。

添加在中间：

拼接实际上做添加并删除。拼接方法的签名是：

removed_items = arrayObject.splice(index, num_to_remove[, add1[, add2[, ...]]]); 

index - the index at which to start making changes 
num_to_remove - starting with that index, remove this many entries 
addN - ...and then insert these elements 

这样我就可以在第3个位置添加一个项目是这样的：

data.items.splice(2, 0, 
    {id: "7", name: "Douglas Adams", type: "comedy"} 
); 

什么，说的是：在指数2起，除去零个项目，然后插入以下项目。结果是这样的：

var data = {items: [ 
    {id: "1", name: "Snatch", type: "crime"}, 
    {id: "2", name: "Witches of Eastwick", type: "comedy"}, 
    {id: "7", name: "Douglas Adams", type: "comedy"},  // <== The new item 
    {id: "3", name: "X-Men", type: "action"}, 
    {id: "4", name: "Ordinary People", type: "drama"}, 
    {id: "5", name: "Billy Elliot", type: "drama"}, 
    {id: "6", name: "Toy Story", type: "children"} 
]}; 

你可以删除一些，并添加一些在一次：

data.items.splice(1, 3, 
    {id: "7", name: "Douglas Adams", type: "comedy"}, 
    {id: "8", name: "Dick Francis", type: "mystery"} 
); 

...这意味着：在索引1起，删除三个条目，然后添加这两个条目。这会导致：

var data = {items: [ 
    {id: "1", name: "Snatch", type: "crime"}, 
    {id: "7", name: "Douglas Adams", type: "comedy"}, 
    {id: "8", name: "Dick Francis", type: "mystery"} 
    {id: "5", name: "Billy Elliot", type: "drama"}, 
    {id: "6", name: "Toy Story", type: "children"} 
]}; 

Answer 3

我不知道完整的对象结构，但可以说的对象称为result，有一个叫albums属性，它是对象的数组。在这种情况下，你可以这样做：

$.each(result.albums, function() { 
    this.custom_url = 'whatever'; 
}); 

或者香草JS：

for (var i = 0, i < result.albums.length; i++) { 
    result.albums[i].custom_url = 'whatever'; 
}