Question

这是我的代码，它基于flickr示例，因为我正在尝试学习如何使用JSON。我的问题：

如何让var标题在HTML中正确显示，我仍然是编程尝试自学的新手。

我得到[对象对象]，我认为这意味着它不是一个字符串。不知道如何把它写成一个，可能非常简单但超出我的范围。

谢谢！

$.getJSON("http://api.flickr.com/services/feeds/photos_public.gne?tags=cats&tagmode=any&format=json&thumb_size=s&jsoncallback=?",
                function(data){
                    var title = $(this).children('.title');
              $.each(data.items, function(i,item){
                    var content = '<li>';
                    content += '<p class="title"> <a href="' + item.link + '">' + item.title + '</a></p>'
                    content += '<img width="133" height="100" class="prettyPhoto" src="' +  item.media.m + '"/>';
                    content += '<p class="date">' + title + '</p>'
                    content += '</li>';
                    $(content).appendTo("#recent-photos");
                    if ( i == 15 ) return false;
                  });
                });

更新

我正在使用的JSON示例：我定位所有“item”子项

({
        "title": "Recent Uploads tagged cats",
        "link": "http://www.flickr.com/photos/tags/cats/",
        "description": "",
        "modified": "2009-07-09T16:50:47Z",
        "generator": "http://www.flickr.com/",
        "items": [
       {
            "title": "Beautiful Downtown Gary, Indiana.",
            "link": "http://www.flickr.com/photos/fotomard/3704818882/",
            "media": {"m":"http://farm3.static.flickr.com/2506/3704818882_8a8e793e47_m.jpg"},
            "date_taken": "2009-06-27T17:21:05-08:00",
            "description": "<p><a href=\"http://www.flickr.com/people/fotomard/\">fotomard<\/a> posted a photo:<\/p> <p><a href=\"http://www.flickr.com/photos/fotomard/3704818882/\" title=\"Beautiful Downtown Gary, Indiana.\"><img src=\"http://farm3.static.flickr.com/2506/3704818882_8a8e793e47_m.jpg\" width=\"160\" height=\"240\" alt=\"Beautiful Downtown Gary, Indiana.\" /><\/a><\/p> <p>Backstage at the Palace Theater<\/p>",
            "published": "2009-07-09T16:50:47Z",
            "author": "nobody@flickr.com (fotomard)",
            "author_id": "51533956@N00",
            "tags": "urban cats house church wet beautiful nude theater downtown decay indiana palace crack abandon drugs gary dilapidation trap trespassing dilapitation"
       }

更新2： Ran json2.js

我得到了：这是否意味着我需要对其进行解码，或者我只是将其定位错误？

{"length":0,"prevObject":{"length":1,"0":{"type":"GET","url":"http://api.flickr.com/services/feeds/photos_public.gne?tags=cats&tagmode=any&format=json&thumb_size=s&jsoncallback=jsonp1247160045746&_=1247160045957","data":null,"dataType":"script","global":true,"contentType":"application/x-www-form-urlencoded","processData":true,"async":true,"accepts":{"xml":"application/xml, text/xml","html":"text/html","script":"text/javascript, application/javascript","json":"application/json, text/javascript","text":"text/plain","_default":"*/*"},"cache":false}},"selector":".children(undefined)"}

Answer 1

如果你使用Firefox和Firebug，你可以这样做：

console.log(data);

在每个（）调用中

以查看数据的结构。而你的title var应该在each（）调用中是本地的。

var title = item.title;

或

var title = $(item).children().attr('title'); // depens on how the data is structured

Answer 2

您最好的选择是包含json2.js文件并调用stringify（）：

alert(JSON.stringify(data));

这样你就可以看到ajax调用返回的内容。

Answer 3

我想通了把var放在每个内部，我能够得到它。工作

如何使用json和jquery显示var？

3 个答案: