[英]How can I return value from onResponse of Volley?
我想在用戶類變量上保存Response的值。 不幸的是,我不能。
請幫幫我。
這是代碼JSONObjectRequest
JsonObjectRequest jsonObjReq = new JsonObjectRequest(Request.Method.GET,
jsonURLFull, new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
Log.d(ActivityLogin.class.getSimpleName(), response.toString());
try {
User userClass = new User();
JSONObject jsonUser = response.getJSONObject("user");
userClass.id = jsonUser.getInt("id");
userClass.name = jsonUser.getString("email");
userClass.email = jsonUser.getString("email");
} catch (JSONException e) {
e.printStackTrace();
Toast.makeText(getApplicationContext(),
"Error: " + e.getMessage(),
Toast.LENGTH_LONG).show();
}
hidepDialog();
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
VolleyLog.d(ActivityLogin.class.getSimpleName(), "Error: " + error.getMessage());
Toast.makeText(getApplicationContext(),
error.getMessage(), Toast.LENGTH_SHORT).show();
hidepDialog();
}
});
AppController.getInstance().addToRequestQueue(jsonObjReq);
這是User類:
public class User {
public int id;
public String name;
public String email;
}
這是API的JSON:
{"user":{"id":12,"name":"test_name","email":"test@email.com",}}
為您提供兩點:
1.您解析email
以name
屬性。
userClass.name = jsonUser.getString("name");
2.您的json數據由於最后一個','
{"user":{"id":12,"name":"test_name","email":"test@email.com"}}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.