我有以下代码:
$.ajax({
type: "GET",
url: "http://www.flickr.com/services/oembed/?url=http%3A//" + picture_src +"?callback=?&format=json" ,
cache: 'true',
dataType: "jsonp",
success: function(output_data_flickr) {
alert(output_data_flickr.url);
if(output_data_flickr.url != ""){
$('#picture_preview_link').attr('src', output_data_flickr.url);
$('#picture_link_embed').val(output_data_flickr.url);
}
else{
$('#picture_link_embed').val(picture_src);
$('#picture_preview_link').attr('src', picture_src);}
}
});
当我尝试在 firebug 中获取 URL 属性时,为什么会出现这样的错误?
SyntaxError: invalid label
[Break On This Error]
{"type":"photo","title":"Bacon Lollys","author_name":"\u202e\