所以我开始编写 Java Android 并且我正在尝试解析JSON
我创建的字符串。所以,我想将它解析为 aListView
并且我需要人们帮助我。
我的实验 JSON 文件:
[
{
"HoTen":" Nguy\u1ec5n V\u0103n A",
"NamSinh":1999,
"DiaChi":"H\u00e0 N\u1ed9i"
},
{+},
{+},
{+},
{+},
{+},
{+},
{+},
{+}
]
我的代码但它不起作用:
protected void onPostExecute(String s) {
//Toast.makeText(getApplicationContext(),s,Toast.LENGTH_LONG).show();
try {
mangLV = new ArrayList<String>();
JSONArray jsonArray = new JSONArray(s);
JSONObject jsonObject = new JSONObject(s);
for (int i =0;i<=jsonObject.length();i++)
{
JSONObject object = jsonArray.getJSONObject(i);
//HoTen.getString("HoTen");
String HoTen = object.getString("HoTen");
int NamSinh = object.getInt("NamSinh");
String DiaChi = object.getString("DiaChi");
}
ArrayAdapter adapter = new ArrayAdapter(getApplicationContext(),android.R.layout.simple_list_item_1,mangLV);
lvSinhVien.setAdapter(adapter);
} catch (JSONException e) {
e.printStackTrace();
}
}
希望对你有帮助
protected void onPostExecute(String s) {
//Toast.makeText(getApplicationContext(),s,Toast.LENGTH_LONG).show();
try {
ArrayList<String> mangLV = new ArrayList<String>();
JSONArray jsonArray = new JSONArray(s);
for (int i = 0; i < jsonArray.length(); i++) {
JSONObject object = jsonArray.getJSONObject(i);
//HoTen.getString("HoTen");
String HoTen = object.getString("HoTen");
int NamSinh = object.getInt("NamSinh");
String DiaChi = object.getString("DiaChi");
String result = String.format("HoTen: %s, NamSinh: %s, DiaChi: %s",
HoTen, NamSinh, DiaChi);
mangLV.add(result);
}
ArrayAdapter adapter = new ArrayAdapter(getApplicationContext(), android.R.layout.simple_list_item_1, mangLV);
lvSinhVien.setAdapter(adapter);
} catch (JSONException e) {
e.printStackTrace();
}
}
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我来说两句