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我是 JSON 新手，正在尝试本教程： http: //p-xr.com/android-tutorial-how-to-parse-read-json-data-into-a-android-listview/#comments

我是 JSON、C 语言、Java 和 Android 的新手，但我正在学习。本教程使用我所谓的命名数组，但我将在我的 android 项目中使用的所有 JSON 都将使用没有命名数组的简单表行。我正在使用的 JSON 示例和教程中的地震 json 示例如下。

本教程遍历地震数组并使用以下代码转换为 JAVA hashmap 列表：

JSONArray  earthquakes = json.getJSONArray("earthquakes");
    for(int i=0;i<earthquakes.length();i++){                        
        HashMap<String, String> map = new HashMap<String, String>();    
        JSONObject e = earthquakes.getJSONObject(i);

        map.put("id",  String.valueOf(i));
        map.put("name", "Earthquake name:" + e.getString("eqid"));
        map.put("magnitude", "Magnitude: " +  e.getString("magnitude"));
        mylist.add(map);            
}

json.getJSONArray("")我的问题是，如果我的 JSON 如下所示，我该如何使用？我可以转换其余的代码，我只需要知道如何使用getJSONArray("strJsonArrayName")如果我没有strJsonArrayName.

我的 JSON（未命名数组）

[
  {
    "cnt":1,
    "name":"American",
    "pk":7
  },
  {
    "cnt":2,
    "name":"Celebrities",
    "pk":3
  },
  {
    "cnt":1,
    "name":"Female",
    "pk":2
  },
  {
    "cnt":1,
    "name":"Language",
    "pk":8
  },
  {
    "cnt":1,
    "name":"Male",
    "pk":1
  },
  {
    "cnt":1,
    "name":"Region",
    "pk":9
  }
]

教程的 JSON（命名数组）

{
  "earthquakes":[
    {
      "eqid":"c0001xgp",
      "magnitude":8.8,
      "lng":142.369,
      "src":"us",
      "datetime":"2011-03-11 04:46:23",
      "depth":24.4,
      "lat":38.322
    },
    {
      "eqid":"c000905e",
      "magnitude":8.6,
      "lng":93.0632,
      "src":"us",
      "datetime":"2012-04-11 06:38:37",
      "depth":22.9,
      "lat":2.311
    },
    {
      "eqid":"2007hear",
      "magnitude":8.4,
      "lng":101.3815,
      "src":"us",
      "datetime":"2007-09-12 09:10:26",
      "depth":30,
      "lat":-4.5172
    },
    {
      "eqid":"c00090da",
      "magnitude":8.2,
      "lng":92.4522,
      "src":"us",
      "datetime":"2012-04-11 08:43:09",
      "depth":16.4,
      "lat":0.7731
    },
    {
      "eqid":"2007aqbk",
      "magnitude":8,
      "lng":156.9567,
      "src":"us",
      "datetime":"2007-04-01 18:39:56",
      "depth":10,
      "lat":-8.4528
    },
    {
      "eqid":"2007hec6",
      "magnitude":7.8,
      "lng":100.9638,
      "src":"us",
      "datetime":"2007-09-12 21:49:01",
      "depth":10,
      "lat":-2.5265
    },
    {
      "eqid":"a00043nx",
      "magnitude":7.7,
      "lng":100.1139,
      "src":"us",
      "datetime":"2010-10-25 12:42:22",
      "depth":20.6,
      "lat":-3.4841
    },
    {
      "eqid":"2010utc5",
      "magnitude":7.7,
      "lng":97.1315,
      "src":"us",
      "datetime":"2010-04-06 20:15:02",
      "depth":31,
      "lat":2.3602
    },
    {
      "eqid":"2009mebz",
      "magnitude":7.6,
      "lng":99.9606,
      "src":"us",
      "datetime":"2009-09-30 08:16:09",
      "depth":80,
      "lat":-0.7889
    },
    {
      "eqid":"2009kdb2",
      "magnitude":7.6,
      "lng":92.9226,
      "src":"us",
      "datetime":"2009-08-10 17:55:39",
      "depth":33.1,
      "lat":14.0129
    }
  ]
}

在本教程中，根据 @MДΓΓ БДLL 和 @Cody Caughlan 的回答，我能够将 JSONFunctions.getJSONFromURL 重新格式化为 JSONArray 而不是 JSONObject。这是我修改后的工作代码，谢谢！

public class JSONfunctions {
public static JSONArray getJSONfromURL(String url){
    InputStream is = null;
    String result = "";
    JSONArray jArray = null;

            HttpClient httpclient = new DefaultHttpClient();
            HttpGet httpget = new HttpGet(url);
            HttpResponse response = httpclient.execute(httpget);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();

            BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
            }
            is.close();
            result=sb.toString();

        jArray = new JSONArray(result);            
    return jArray;
}
}
4

4 回答 4

148

您根本不需要调用json.getJSONArray()，因为您正在使用的 JSON 已经是一个数组。所以，不要构造 ; 的实例JSONObject。使用JSONArray. 这应该足够了：

// ...
JSONArray json = new JSONArray(result);
// ...

for(int i=0;i<json.length();i++){                        
    HashMap<String, String> map = new HashMap<String, String>();    
    JSONObject e = json.getJSONObject(i);

    map.put("id",  String.valueOf(i));
    map.put("name", "Earthquake name:" + e.getString("eqid"));
    map.put("magnitude", "Magnitude: " +  e.getString("magnitude"));
    mylist.add(map);            
}

您不能使用与教程中完全相同的方法，因为您正在处理的 JSON 需要JSONArray在根目录中解析为 a，而不是JSONObject.

于 2012-04-15T18:25:14.650 回答
7

JSONArray有一个构造函数，它接受一个String源（假定是一个数组）。

所以像这样

JSONArray array = new JSONArray(yourJSONArrayAsString);
于 2012-04-15T18:24:05.930 回答
0

我假设一个命名的 JSONArray 是一个 JSONObject，并从服务器访问数据以填充一个 Android GridView。值得我的方法是：

private String[] fillTable( JSONObject jsonObject ) {
   String[] dummyData = new String[] {"1", "2", "3", "4", "5", "6", "7","1", "2", "3", "4", "5", "6", "7","1", "2", "3", "4", "5", "6", "7", };
  if( jsonObject != null ) {
      ArrayList<String> data = new ArrayList<String>();
      try {
          // jsonArray looks like { "everything" : [{}, {},] }
          JSONArray jsonArray = jsonObject.getJSONArray( "everything" );
          int number = jsonArray.length(); //How many rows have got from the database?
          Log.i( Constants.INFORMATION, "Number of ows returned:  " + Integer.toString( number ) );
                  // Array elements look like this
          //{"success":1,"error":0,"name":"English One","owner":"Tutor","description":"Initial Alert","posted":"2013-08-09 15:35:40"}
          for( int element = 0; element < number; element++ ) { //visit each element
             JSONObject jsonObject_local = jsonArray.getJSONObject( element );
             //  Overkill on the error/success checking
             Log.e("JSON SUCCESS", Integer.toString( jsonObject_local.getInt(Constants.KEY_SUCCESS) ) );
             Log.e("JSON ERROR", Integer.toString( jsonObject_local.getInt(Constants.KEY_ERROR) ) );
                if ( jsonObject_local.getInt( Constants.KEY_SUCCESS) == Constants.JSON_SUCCESS ) {
                   String name = jsonObject_local.getString( Constants.KEY_NAME );
                   data.add( name );
                   String owner = jsonObject_local.getString( Constants.KEY_OWNER );
                   data.add( owner );
                   String description = jsonObject_local.getString( Constants.KEY_DESCRIPTION );
                   Log.i( "DESCRIPTION", description );
                   data.add( description ); 
                   String date = jsonObject_local.getString( Constants.KEY_DATE );
                   data.add( date );
                }
                else {
                    for( int i = 0; i < 4; i++ ) {
                        data.add( "ERROR" );
                    }
                }
          }
  }  //JSON object is null
  catch ( JSONException jsone) {
      Log.e( "JSON EXCEPTION", jsone.getMessage() );
  }
      dummyData = data.toArray( dummyData );
  }
  return dummyData;

}

于 2013-08-15T10:39:46.370 回答
0

这是19API lvl下的解决方案：

  • 首先。制作一个 Gson 对象。-->Gson gson = new Gson();

  • 第二步是使用 StringRequest（而不是 JsonObjectRequest）将您的 jsonObj 作为字符串

  • 获取 JsonArray 的最后一步...

YoursObjArray[] yoursObjArray = gson.fromJson(response, YoursObjArray[].class);

于 2015-10-02T15:14:25.137 回答