我是JSON的新手并尝试本教程： 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将使用没有命名数组的简单表行。我正在使用的JSON和教程中的地震json的例子如下。
本教程遍历地震数组并使用以下代码转换为JAVA哈希地图列表：
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，如果我的JSON如下所示，我该如何使用json.getJSONArray("")？我可以转换剩下的代码，如果我没有getJSONArray("strJsonArrayName")，我只需要知道如何使用strJsonArrayName加载JSON。
我的JSON（UnNamed Array）
[
{
"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，基于@MДΓΓБДLL和@Cody Caughlan的答案，我能够将JSONFunctions.getJSONFromURL重新格式化为JSONArray而不是JSONObject。这是我修改过的工作代码，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;
}
}
答案 0 :(得分：134)
您根本不需要调用json.getJSONArray()，因为您已经使用的JSON是数组。所以，不要构造JSONObject的实例;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需要在根目录下解析为JSONArray，而不是JSONObject。
答案 1 :(得分：5)
JSONArray有一个构造函数，它接受String源（假定为数组）。
这样的事情
JSONArray array = new JSONArray(yourJSONArrayAsString);
答案 2 :(得分：0)
我假设一个名为JSONArray的是一个JSONObject，并从服务器访问数据以填充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;
}
答案 3 :(得分：0)
以下是19API lvl下的解决方案：
首先。做一个Gson obj。 - ＆GT; Gson gson = new Gson();
第二步是使用StringRequest将您的jsonObj作为String（而不是 JsonObjectRequest）
YoursObjArray[] yoursObjArray = gson.fromJson(response, YoursObjArray[].class);