如何获取没有数组名称的JSONArray?

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我是JSON新手,尝试这个教程: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。本教程使用我称为命名数组的内容,但是我将在我的Android项目中使用的所有JSON将使用没有命名数组的简单表行。下面是我使用的JSON和本教程中的地震json的示例。

本教程将遍历地震数组,并使用以下代码将其转换为JAVA哈希表列表:

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);            
}

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

我的JSON(Unnamed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@MДΓΓБДLL和@Cody Caughlan的回答,我可以将JSONFunctions.getJSONFromURL格式化为JSONArray而不是JSONObject。这是我修改的工作代码

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;
}
}
提问于
用户回答回答于

你根本不需要调用json.getJSONArray(),因为你正在使用的JSON已经一个数组。所以,不要构建一个实例JSONObject; 使用一个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);            
}

无法使用与本教程中完全相同的方法,因为要处理的JSON需要JSONArray在根目录下解析为a ,而不是解析为a JSONObject

用户回答回答于

JSONArray有一个构造函数需要一个String源(假定是一个数组)。

所以像这样的东西

JSONArray array = new JSONArray(yourJSONArrayAsString);

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