Question

我试图将json数组数据解析为listview！我搜索了整个互联网并始终在一点上！ Json aray必须有这样的标题

"emp_info":[{"employee name":"Adam","employee no":"101700"},{"employee name":"John","employee no":"101701"},{"employee name":"Paul","employee no":"101702"},{"employee name":"Mark","employee no":"101703"},{"employee name":"Donald","employee no":"101704"},{"employee name":"Brain","employee no":"101705"},{"employee name":"Kevin","employee no":"101706"}]}

根据我的理解，"emp_info"是我必须在android中搜索其中的休息数据的头文件！我的学院假装我可以接受并将相同的数据解析为没有该头名称的listview，但是我在android中搜索解析json的每一段代码都是这样的一行！

JSONObject obj = new JSONObject(jsonString);
JSONArray stations = obj.getJSONArray("emp_inf");

我只需要把jsonarray头文件放在这段代码中就可以了！所以请帮助我是否可以接受没有这段代码的json数组？因为如果我尝试删除这段代码我会得到nullpointer in我的代码！如果你至少可以说是或否，那将非常高兴！发布完整代码！这是获取Json并将其加载到List View

的android类

    private class JsonReadTask extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... params) {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://54.148.41.171/server/index/dompy");
try {
HttpResponse response = httpclient.execute(httppost);
jsonResult = inputStreamToString(
response.getEntity().getContent()).toString();
}

catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return null;
}

private StringBuilder inputStreamToString(InputStream is) {
String rLine = "";
StringBuilder answer = new StringBuilder();
BufferedReader rd = new BufferedReader(new InputStreamReader(is));

try {
while ((rLine = rd.readLine()) != null) {
answer.append(rLine);
}
}

catch (IOException e) {
// e.printStackTrace();
Toast.makeText(getApplicationContext(),
"Error..." + e.toString(), Toast.LENGTH_LONG).show();
}
return answer;
}

@Override
protected void onPostExecute(String result) {
ListDrwaer();
}
}// end async task

public void accessWebService() {
JsonReadTask task = new JsonReadTask();
// passes values for the urls string array
task.execute(new String[] { url });
}

// build hash set for list view
public void ListDrwaer() {
List<Map<String, String>> employeeList = new ArrayList<Map<String, String>>();

try {
JSONObject jsonResponse = new JSONObject(jsonResult);
JSONArray jsonMainNode = jsonResponse.optJSONArray("emp_info");

for (int i = 0; i < jsonMainNode.length(); i++) {
JSONObject jsonChildNode = jsonMainNode.getJSONObject(i);
String name = jsonChildNode.optString("empployee no");
String number = jsonChildNode.optString("etc.");
String outPut = name + "-" + number;
employeeList.add(createEmployee("data", outPut));
}
} catch (JSONException e) {
Toast.makeText(getApplicationContext(), "Error" + e.toString(),
Toast.LENGTH_SHORT).show();
}

SimpleAdapter simpleAdapter = new SimpleAdapter(this, employeeList,
android.R.layout.simple_list_item_1,
new String[] { "employee no" }, new int[] { android.R.id.text1 });
listView.setAdapter(simpleAdapter);
}

private HashMap<String, String> createEmployee(String name, String number) {
HashMap<String, String> employeeNameNo = new HashMap<String, String>();
employeeNameNo.put(name, number);
return employeeNameNo;
}


}

通过这个json数组我成功地能够获取json，其类型是下一个！

 {"emp_info":[{"employee name":"Adam","employee no":"101700"},{"employee name":"John","employee no":"101701"},{"employee name":"Paul","employee no":"101702"},{"employee name":"Mark","employee no":"101703"},{"employee name":"Donald","employee no":"101704"},{"employee name":"Brain","employee no":"101705"},{"employee name":"Kevin","employee no":"101706"}]}

这是我的大学假装我必须接受的json数组！

{"data":"123"}

而且我说它不可能将这个json加载到列表视图中只是因为它没有像emp_info这样的头文件，但是他说它不管我只是必须接受！我们只是在同一个项目上我只是无法理解它甚至可以做他说的话吗？

Answer 1

String name = jsonChildNode.optString("empployee no");
String number = jsonChildNode.optString("etc.");

如何通过“等”获取数据标签我首先无法理解。它应该返回“”而不是员工编号。

Android JsonArray进入ListView

1 个答案: