Question

我在java中解析一个简单的json时遇到了问题。这是json的样本。

[
  {
    "politics": [
      {
        "type": "admin2",
        "friendly_type": "country",
        "name": "United States",
        "code": "usa"
      },
      {
        "type": "admin6",
        "friendly_type": "county",
        "name": "Gratiot",
        "code": "26_057"
      },
      {
        "type": "constituency",
        "friendly_type": "constituency",
        "name": "Eighth district, MI",
        "code": "26_08"
      },
      {
        "type": "admin6",
        "friendly_type": "county",
        "name": "Clinton",
        "code": "26_037"
      },
      {
        "type": "admin4",
        "friendly_type": "state",
        "name": "Michigan",
        "code": "us26"
      },
      {
        "type": "constituency",
        "friendly_type": "constituency",
        "name": "Fourth district, MI",
        "code": "26_04"
      }
    ],
    "location": {
      "latitude": 43.111976,
      "longitude": -84.71275
    }
  }
]

现在这给了我正确的json索引。

JSONParser parser = new JSONParser();
Object obj = parser.parse(output);
JSONArray array = (JSONArray)obj;
String jsonobj = array.get(0).toString();

{"politics":[{"code":"usa","name":"United States","type":"admin2","friendly_type":"country"},{"code":"26_057","name":"Gratiot","type":"admin6","friendly_type":"county"},{"code":"26_08","name":"Eighth district, MI","type":"constituency","friendly_type":"constituency"},{"code":"26_037","name":"Clinton","type":"admin6","friendly_type":"county"},{"code":"us26","name":"Michigan","type":"admin4","friendly_type":"state"},{"code":"26_04","name":"Fourth district, MI","type":"constituency","friendly_type":"constituency"}],"location":{"latitude":43.111976,"longitude":-84.71275}}

但我似乎无法从中得到我想要的属性。

JSONObject obj1 = new JSONObject(jsonobj);
String n = obj1.getString("admin4");
System.out.println(n);

我所需要的就是这个json就是密歇根州。我哪里错了？非常感谢帮助。

Answer 1

简化json对象上的搜索和其他操作的最佳解决方案是将json字符串转换为java对象并进行操作。

将json字符串转换为java对象使用跟随代码：

  import org.codehaus.jackson.map.ObjectMapper;
  import org.json.JSONException;
  import org.json.JSONObject; 

YourObject myObject;
ObjectMapper mapper = new ObjectMapper();
try{
    myObject= mapper.readValue(jsonData, myObject.class);
}
catch (Exception e) {
    e.printStackTrace();
}

例如定义你的班级屁股跟随：

public class myObject{

      private List<Politics> politics;
      private Location location;
      // define getters and setters
}

define Politics and Location class:

public class Politics
{
      String  type;
      String  friendly_type;
      String  name;
      String  code;
      // define getters and setters
}

public class Location
{
  String latitude;
  String longitude;
   // define getters and setters
}

Answer 2

首先，array.get(0)将为您提供主数组中的第一个元素。第一个元素是一个JSON对象，它有两个属性politics和location。您似乎对politics属性的数组值内的值感兴趣。您必须使用此((JSONArray)((JSONObject)array.get(0)).get("politics"))来获取该数组。

其次，admin4不是属性，它实际上是type属性的值。你必须遍历数组才能找到它。

这是一个完整的例子：

JSONParser parser = new JSONParser();
Object obj = parser.parse(output);
JSONArray array = (JSONArray)obj;
JSONArray politics = ((JSONObject)array.get(0)).get("politics"));
JSONObject obj = null;
for(int i = 0; i < politics.size(); i++){
    if(((JSONObject)politics.get(i)).getString("type").equals("admin4")){
        obj = ((JSONObject)politics.get(i));
    }
}
if(obj != null){
    // Do something with the object.
}

您似乎正在使用simple json库。我不记得确切的是.get("politics")还是.getJSONObject("politics")。在我的例子中，方法名称可能存在其他错误。

Answer 3

这是因为您正在尝试获取JSON对象的内部元素。尝试

JSONObject obj1 = new JSONObject(jsonobj);
JSONArray arr = (JSONArray) obj1.getObject("politics");

您将获得一个JSONArray对象，该对象进一步构成JSON对象。现在，为了使用键获取值，您必须迭代数组，如下所示：

for(int i=0; i<arr.size(); i++){
    JSONObject obj = arr.getJSONArray(i);
    System.out.println(obj.getString("type"));
}

现在将为您提供输出：

admin2
admin6
constituency
admin6
admin4
constituency

无法解析json数组中的json数组

3 个答案: