Question

I am getting data from the Flickr API with a middle-man file (to avoid crossdomain problems):

<?php

header( Cache-Control: no-cache, must-revalidate );
header( Expires: Mon, 26 Jul 1997 05:00:00 GMT );
header( Content-type: application/json );

die(json_encode( file_get_contents($_REQUEST[ url ]) ) );
?>

This file is fetched by javascript:

//Flickr
var myurl = encodeURIComponent( http://api.flickr.com/services/rest/?method=flickr.photosets.getPhotos&api_key=1408bff5f72a4b84b924d13e8562b6a2&photoset_id=77649470@N03&photoset_id=72157629903184261&format=json );
    $.getJSON( "middle.php?url=" + myurl, function(data){
        console.log(typeof data);
    });

但是,这只是ole。记录显示,结果是一种“扼杀”,而不是“焦耳”。 Ive试图将其改装成一个有:

jQuery.parseJSON(data)

but the console gave me this error:

Uncaught SyntaxError: Unexpected token j

This is the string:

jsonFlickrApi({"photoset":{"id":"72157629903184261", "primary":"7115173307", "owner":"77649470@N03", "ownername":"wedocommunication", "photo":[{"id":"7115173331", "secret":"24900ff306", "server":"5447", "farm":6, "title":"Lounge", "isprimary":"0"}, {"id":"7115173307", "secret":"3435f9a983", "server":"7256", "farm":8, "title":"Hofansicht", "isprimary":"1"}, {"id":"7115173379", "secret":"7747e50597", "server":"7278", "farm":8, "title":"Konfi", "isprimary":"0"}, {"id":"6969093048", "secret":"d4389bc0e4", "server":"7055", "farm":8, "title":"Lounge", "isprimary":"0"}, {"id":"6969093086", "secret":"8e7263005b", "server":"5152", "farm":6, "title":"Eingangsbereich", "isprimary":"0"}], "page":1, "per_page":500, "perpage":500, "pages":1, "total":"5"}, "stat":"ok"})

我可以做些什么来把扼杀变成物体?

Answer 1

有些人把回答说了,但出于某种原因,他们删除了这个职位,因此我在此复制:

$.ajax({
    url:  http://api.flickr.com/services/rest/?method=flickr.photosets.getPhotos&api_key=1408bff5f72a4b84b924d13e8562b6a2&photoset_id=77649470@N03&photoset_id=72157629903184261&format=json ,
    type:  GET ,
    dataType:  jsonp ,
    jsonpCallback:  jsonFlickrApi ,
    success: function(data){
        console.log(data);
    }
});

这完全是工作。感谢任何人:

也许还有其他工作,但我感到更加舒适。谢谢大家。

Answer 2

Flickr正在将答复作为jsonp,即javascript not json,而你应当为此使用jquery jsonp型请求,见:。

in fact that page gives an example for the flickr api-

$.getJSON("http://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?",
  {
    tags: "cat",
    tagmode: "any",
    format: "json"
  },
  function(data) {
    $.each(data.items, function(i,item){
      $("<img/>").attr("src", item.media.m).appendTo("#images");
      if ( i == 3 ) return false;
    });
  });

Answer 3

您无需使用代理来避免交叉主要问题,如。

$.getJSON("http://api.flickr.com/services/rest/?method=flickr.photosets.getPhotos&api_key=1408bff5f72a4b84b924d13e8562b6a2&photoset_id=77649470@N03&photoset_id=72157629903184261&format=json&jsoncallback=?",function(data){
        console.log(data);
    });

DEMO

Answer 4

The JSON starts with the object curly brace { The jsonFlickrApi( is not part of the JSON object. You could string replace data first to remove the jsonFlickrApi( part first if you only need the inner, but it seems like Flickr would have a better built-in solution.

Answer 5

标题9

这样,可以由联合材料作为普通的JSON数据加以分类。

Answer 6

正如上文已经提到的那样,Fakr正在将“JN”的复信发送到“密码>jsonFakrApi >的背靠。

为了归还纯粹的JSON数据,你可以使用nojsonert=1请求参数。

E.g. http://api.flickr.com/services/rest/?method=flickr.photos.getSizes&api_key=_secret_app_key&photo_id=8321754843&format=json&nojsonertback=1

友情链接