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我正在使用 Google 的地理编码 API 来获取包含地理编码位置信息的 JSON 字符串。这是我从谷歌返回的字符串。

{
  "status": "OK",
  "results": [ {
"types": [ "street_address" ],
"formatted_address": "550 Susong Dr, Morristown, TN 37814, USA",
"address_components": [ {
  "long_name": "550",
  "short_name": "550",
  "types": [ "street_number" ]
}, {
  "long_name": "Susong Dr",
  "short_name": "Susong Dr",
  "types": [ "route" ]
}, {
  "long_name": "Morristown",
  "short_name": "Morristown",
  "types": [ "locality", "political" ]
}, {
  "long_name": "Morristown",
  "short_name": "Morristown",
  "types": [ "administrative_area_level_3", "political" ]
}, {
  "long_name": "Hamblen",
  "short_name": "Hamblen",
  "types": [ "administrative_area_level_2", "political" ]
}, {
  "long_name": "Tennessee",
  "short_name": "TN",
  "types": [ "administrative_area_level_1", "political" ]
}, {
  "long_name": "United States",
  "short_name": "US",
  "types": [ "country", "political" ]
}, {
  "long_name": "37814",
  "short_name": "37814",
  "types": [ "postal_code" ]
} ],
"geometry": {
  "location": {
    "lat": 36.2422740,
    "lng": -83.3219410
  },
  "location_type": "ROOFTOP",
  "viewport": {
    "southwest": {
      "lat": 36.2391264,
      "lng": -83.3250886
    },
    "northeast": {
      "lat": 36.2454216,
      "lng": -83.3187934
    }
  }
}

} ] }

但是，当我在 Java 中运行以下代码时，我收到“java.lang.ClassCastException：java.lang.String 与 net.sf.json.JSONObject 不兼容”错误。

  URL url = new URL(URL + "&address=" + URLEncoder.encode(address, "UTF-8") + "&signature=" + key);
  URLConnection conn = url.openConnection();
  ByteArrayOutputStream output = new ByteArrayOutputStream(1024);
  IOUtils.copy(conn.getInputStream(), output);
  output.close();

  GAddress gaddr = new GAddress();
  JSONObject json = JSONObject.fromObject(output.toString());
  JSONObject placemark = (JSONObject) query(json, "Placemark[0]");

我不确定为什么会收到错误消息。对我来说，Google 响应看起来像是一个有效的 JSON 字符串。其他人有这个问题吗？如果出于某种原因它与 Google 不兼容，我愿意使用 net.sf.json 之外的其他东西。

谢谢，

安德鲁

4

1 回答 1

1

看起来您从原始函数调用中获取了一个字符串。可以肯定的是，您可以添加

System.out.println(query(json, "Placemark[0]").class);

就在最后一行之前。这将为您提供您正在处理的对象的类型。

于 2010-12-30T16:14:04.493 回答