php - 使用 jQuery 将字符串转换为 JSON

Question

我使用中间人文件从 Flickr API 获取数据(以避免跨域问题):

<?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']) ) );
?>

此文件由 javascript 获取:

//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 console.log 显示结果是一个“字符串”而不是一个 JSON 对象。我尝试将其转换为 JSON 对象:

jQuery.parseJSON(data)

但是控制台给了我这个错误:

Uncaught SyntaxError: Unexpected token j

这是字符串:

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"})

如何将字符串转换为对象？

Answer 1

Flickr 以 jsonp 形式返回响应，这是 javascript 而不是 json，您应该为此使用 jquery jsonp 类型请求，请参阅 - http://api.jquery.com/jQuery.ajax/

事实上，该页面为 flickr api 提供了一个示例-

$.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;
    });
  });