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Javascript jQuery JSON解析谷歌地图问题_Javascript_Jquery_Json_Api_Google Maps - Fatal编程技术网

Javascript jQuery JSON解析谷歌地图问题

Javascript jQuery JSON解析谷歌地图问题,javascript,jquery,json,api,google-maps,Javascript,Jquery,Json,Api,Google Maps,我正在尝试使用以下代码从google maps获取json响应: var address = 'Sydney,NSW'; $.ajax({ type: "GET", url: "http://maps.googleapis.com/maps/api/geocode/json?sensor=false", dataType: "jsonp", data: ({"address":address}), success: function(json)

我正在尝试使用以下代码从google maps获取json响应:

var address = 'Sydney,NSW';
$.ajax({  
    type: "GET",  
    url: "http://maps.googleapis.com/maps/api/geocode/json?sensor=false",  
    dataType: "jsonp",
    data: ({"address":address}),
    success: function(json){
        if (json.status == "OK") {
           alert(":)");   
        } else {
            alert('wrong!');
        }
   }
});
它工作得很好,但是

因此,我得到了

未捕获的语法错误:意外标记: 铬和 错误:在FireFox中找不到元素

返回的JSON似乎有问题,但看起来不错:

{  
 "status": "OK",
 "results": [ {
 "types": [ "locality", "political" ],
 "formatted_address": "Sydney New South Wales, Australia",
   "address_components": [ {
   "long_name": "Sydney",
   "short_name": "Sydney",
   "types": [ "locality", "political" ]
  }, {
   "long_name": "New South Wales",
   "short_name": "New South Wales",
   "types": [ "administrative_area_level_1", "political" ]
}, {
  "long_name": "Australia",
  "short_name": "AU",
  "types": [ "country", "political" ]
} ],
"geometry": {
  "location": {
    "lat": -33.8689009,
    "lng": 151.2070914
  },
  "location_type": "APPROXIMATE",
  "viewport": {
    "southwest": {
      "lat": -34.1648540,
      "lng": 150.6948538
    },
    "northeast": {
      "lat": -33.5719182,
      "lng": 151.7193290
    }
  },
  "bounds": {
    "southwest": {
      "lat": -34.1692489,
      "lng": 150.5022290
    },
    "northeast": {
      "lat": -33.4245980,
      "lng": 151.3426361
    }
  }
}
}]
}

返回必须是jsonp而不是json(json应该包装到函数调用中)

尝试使用GM2(看起来GM3不支持jsonp)

}
});

返回必须是jsonp而不是json(json应该包装到函数调用中)

尝试使用GM2(看起来GM3不支持jsonp)

}
});

看起来,没有办法通过jsonp实现这一点,也没有办法通过jsonp实现这一点
$.ajax({  
type: "GET",  
url: "http://maps.google.com/maps/geo?sensor=false&output=json&"+$.param({"q":address}),  
dataType: "jsonp",
success: function(json){
    if (json.Status.code == "200") {
       alert(":)");   
    } else {
        alert('wrong!');
    }