Question

从下面的JSON中，如何使用for循环和ajax从笔记和笔记中检索标题？

{
"infos": {
        "info": [
        {
            "startYear": "1900",
            "endYear": "1930",
            "timeZoneDesc": "daweerrewereopreproewropewredfkfdufssfsfsfsfrerewrBlahhhhh..",
            "timeZoneID": "1",
                            "note": {
                "notes": [
                    {
                        "id": "1",
                        "title": "Mmm"
                    },
                    {
                        "id": "2",
                        "title": "Wmm"
                    },
                    {
                        "id": "3",
                        "title": "Smm"
                    }
                ]
            },
            "links": [
                { "id": "1", "title": "Red House", "url": "http://infopedia.nl.sg/articles/SIP_611_2004-12-24.html" },
                { "id": "2", "title": "Joo Chiat", "url": "http://www.the-inncrowd.com/joochiat.htm" },
                { "id": "3", "title": "Bake", "url": "https://thelongnwindingroad.wordpress.com/tag/red-house-bakery" }
            ]
        }

我尝试了下面的代码，但它不起作用 - 它要么说：

为空

不是对象

长度为空

不是对象

var detail = eval(xmlhttprequest.responseText)
var rss = detail.infos.info
for(var i = 0; i<rss.length; i++)
   startyear += rss[i].startyear

Answer 1

使用

for (i = 0; i < 3; i++) {
    alert(JSON.infos.info[0].note.notes[i].title);
}

在这里尝试：JSFIDDLE WORKING EXAMPLE

BTW您的JSON无效。使用此JSON：

var JSON = {
    "infos": {
        "info": [
            {
                "startYear": "1900",
                "endYear": "1930",
                "timeZoneDesc": "daweerrewereopreproewropewredfkfdufssfsfsfsfrerewrBlahhhhh..",
                "timeZoneID": "1",
                "note": {
                    "notes": [
                        {
                            "id": "1",
                            "title": "Mmm"
                        },
                        {
                            "id": "2",
                            "title": "Wmm"
                        },
                        {
                            "id": "3",
                            "title": "Smm"
                        }
                    ]
                },
                "links": [
                    {
                        "id": "1",
                        "title": "Red House",
                        "url": "http://infopedia.nl.sg/articles/SIP_611_2004-12-24.html"
                    },
                    {
                        "id": "2",
                        "title": "Joo Chiat",
                        "url": "http://www.the-inncrowd.com/joochiat.htm"
                    },
                    {
                        "id": "3",
                        "title": "Bake",
                        "url": "https://thelongnwindingroad.wordpress.com/tag/red-house-bakery"
                    }
                ]
            }
        ]
    }
}

编辑：

这是你想要的：

var infoLength= JSON.infos.info.length;

for (infoIndex = 0; infoIndex < infoLength; infoIndex++) {

    var notesLength= JSON.infos.info[infoIndex].note.notes.length;

    for (noteIndex = 0; noteIndex < notesLength; noteIndex++) {

        alert(JSON.infos.info[infoIndex].note.notes[noteIndex].title);

    }
}

Answer 2

JSON notes类似数组的对象的“路径”是：

json.infos.info[0].note.notes;

所以你可以这样做：

var notes = json.infos.info[0].note.notes;
var titles = [];
for (var i = 0, len = notes.length; i < len; i++)
{
   titles.push(notes[i].title);
}

alert('titles is: ' + titles.join(', '));

小提琴： http://jsfiddle.net/garreh/uDxqD/

您使用的是jQuery吗？ ; - ）

// Assuming your using "success" in ajax response
success: function(json)
{
    var titles = $(json.infos.info[0].note.notes).map(function() {
        return this.title;
    }).get();
    alert(titles.join(', '));
}

Answer 3

将json放入名为obj的var中，使用以下命令：

obj.infos.info[0].note.notes[0].title

http://jsfiddle.net/Znq34/

Answer 4

首先计算音符的长度

var len = jsonobject.infos.info.note.notes.length;

然后循环并获取

var title = jsonobject.infos.info.note.notes[i].title;

如何从数组中选择json项

4 个答案:

编辑：