Javascript:如何获取嵌套JSON

Javascript:如何获取嵌套JSON,javascript,ajax,json,node.js,Javascript,Ajax,Json,Node.js,} 问题是我正在用nodejs从我当地的电影网站上抓取一个html页面,我有3个数组(ar1,ar2,ar3),它们包含了从现在到未来三天的电影标题 我使用这种方式获取json(这仅限一天),这是函数的“核心”,在这里我生成的数组将是strigify: { "category": [ { "category_id": "1", "category_name": "Editor's Choice ", "cover_url": "http:/

}

问题是我正在用nodejs从我当地的电影网站上抓取一个html页面,我有3个数组(ar1,ar2,ar3),它们包含了从现在到未来三天的电影标题

我使用这种方式获取json(这仅限一天),这是函数的“核心”,在这里我生成的数组将是strigify:

{
"category": [
    {
        "category_id": "1",
        "category_name": "Editor's Choice ",
        "cover_url": "http://www.example.com"
    },
    {
        "category_id": "2",
        "category_name": "New Releases",
        "cover_url": "http://www.example.com"
    }
],
"success": 1
然后在服务器代码的另一点(功能失效!)中,我执行res.send(JSON.stringify(js,null,4))

这就是结果:

            var len = titolo1.length;
            while(j!=len){
                obj1={titolo:titolo1[j],locandina:film1[j]};
                js[j]=obj1;
                j++;
            } 
]

现在,我想谈谈这种类型的json:

[
{
    "titolo": " The Gunman ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/RIBqwqVPwUp6VIWeRZlS-150x214.jpg"
},
{
    "titolo": " Doraemon il film ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/doraemon-il-film-le-avventure-di-nobita-e-dei-cinque-esploratori_notizia-2-150x214.jpg"
},
{
    "titolo": " Child 44 – Il bambino numero 44 ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/Child-44-trailer-italiano-e-locandina-del-thriller-con-Gary-Oldman-e-Tom-Hardy-2-150x214.jpg"
},
{
    "titolo": " Run All Night – Una notte per sopravvivere ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/locandina2-150x214.jpg"
},
{
    "titolo": " Ritorno al Marigold Hotel ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/36366-150x214.jpg"
},
{
    "titolo": " I 7 nani ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/xHIrTDXTLZF1jBd7CQNf-150x214.jpg"
},
{
    "titolo": " Adaline – L’eterna giovinezza ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/adaline-poster-150x214.jpg"
},
{
    "titolo": " Avengers: Age of Ultron – 2D ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/ultron.poster-2-150x214.jpg"
},
{
    "titolo": " Fast and Furious 7 ",
    "locandina": "http://www.metropoliscinemas.it/wp-content/uploads/49946-150x214.jpg"
}
}

我该怎么做? 对不起,我的英语不好。

马可-

如果始终有3个阵列,简单的答案是:

{
"day1": [
    {
        "title": "blablablalbal",
    },
    {
        "title": "blablablalbal",
    }
],
"day2": [
    {
        "title": "blablablalbal",
    },
    {
        "title": "blablablalbal",
    }
],
"day3": [
    {
        "title": "blablablalbal",
    },
    {
        "title": "blablablalbal",
    }
]
您还可以编写一个简单的函数来连接和命名数据。运行下面的示例进行测试

如果这不是您想要的,请提供更多信息

示例:

日历事件
函数main(){ var胶片=联合数据([ar1,ar2,ar3]); document.getElementById('stdout')。value=JSON.stringify(films,null'); } 函数joinData(数据){ 变量i,obj={};
对于(i=0;iMarco-我可以理解你的英语。但是,你需要给出一个你想要的数据结构的示例。我认为你只需要对循环做一个小的更改来添加“film”前缀。这正是我想要的!ty
var films = {day1: ar1, day2: ar2, day3: ar3};
res.send(JSON.stringify( films, null, 4 ));