我正在尝试使用此方法GetList()填充ArrayList 我看到我添加了2条记录。 但是当我尝试使用此方法LoadQst()获取这些记录时,在该ArrayList中获得了0条记录 这是我的代码,
public class GameActivity extends AppCompatActivity {
TextView texto1,texto2;
String Option1,Option2;
int CurrentQst,CurrentQstId,int1,int2;
private RequestQueue mQueue;
Question qst = new Question();
public final List<Question> Questions = new ArrayList<Question>();
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_game);
texto1 = (TextView) findViewById(R.id.red);
texto2 = (TextView) findViewById(R.id.blue);
CurrentQst=0;
mQueue = Volley.newRequestQueue(GameActivity.this);
texto1.setText("");
GetList();
LoadQst();
texto1.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
GetList();
}
});
}
private void GetList() {
String url = "*********/api/get_all_products.php";
JsonObjectRequest request = new JsonObjectRequest(Request.Method.GET, url, null,
new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
try {
JSONArray jsonArray = response.getJSONArray("products");
for (int i = 0; i < jsonArray.length(); i++) {
JSONObject employee = jsonArray.getJSONObject(i);
qst = new Question();
String option1 = employee.getString("option1");
qst.id=employee.getInt("id");
qst.option1=employee.getString("option1");
qst.option2=employee.getString("option2");
qst.vote1=employee.getInt("vote1");
qst.vote2=employee.getInt("vote2");
boolean added = Questions.add(qst);
if (added)
{
texto1.append("added\n");
} else
{
texto1.append("not added\n");
}
}
for (int i = 0; i < Questions.size(); i++) {
texto1.append("option " + Questions.get(i).option1 + "\n");
}
texto1.append(" size "+Questions.size()); //here i get size of 2
//Collections.shuffle(Questions);
} catch (JSONException e) {
e.printStackTrace();
texto1.setText(e.getMessage());
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
error.printStackTrace();
}
});
mQueue.add(request);
}
private void LoadQst()
{
try {
Question CurrentQuestion ;
texto2.setText(String.valueOf(Questions.size()));//here i got an error size 0
// CurrentQuestion = (Question)Questions.get(CurrentQst);
//texto1.setText(CurrentQuestion.option1);
//texto2.setText(CurrentQuestion.option1);
// int1=CurrentQuestion.vote1;
//int2=CurrentQuestion.vote2;
//CurrentQstId=CurrentQuestion.id;
}catch (Exception e)
{
texto2.append(e.getMessage());
}
}
}
当我检查我的问题中是否有东西时 我有2个结果
for (int i = 0; i < Questions.size(); i++) {
texto1.append("option " + Questions.get(i).option1 + "\n");
}
但是当我在此方法LoadQst()中调用此Questions ArrayList时,我得到0个结果。 我应该从ArrayList切换到另一个方法吗? 您要提出任何类型的解决方案
请,你可以看看吗?
答案 0 :(得分:1)
这是因为异步中的JsonObjectRequest
onResponse
方法。 LoadQst
方法在onResponse
方法之前执行。
在LoadQst
中调用onResponse
方法,而不是在onCreate
中调用。
new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
try {
JSONArray jsonArray = response.getJSONArray("products");
for (int i = 0; i < jsonArray.length(); i++) {
JSONObject employee = jsonArray.getJSONObject(i);
qst = new Question();
String option1 = employee.getString("option1");
qst.id=employee.getInt("id");
qst.option1=employee.getString("option1");
qst.option2=employee.getString("option2");
qst.vote1=employee.getInt("vote1");
qst.vote2=employee.getInt("vote2");
boolean added = Questions.add(qst);
if (added)
{
texto1.append("added\n");
} else
{
texto1.append("not added\n");
}
}
for (int i = 0; i < Questions.size(); i++) {
texto1.append("option " + Questions.get(i).option1 + "\n");
}
texto1.append(" size "+Questions.size()); //here i get size of 2
//Collections.shuffle(Questions);
LoadQst(); // Call here.
} catch (JSONException e) {
e.printStackTrace();
texto1.setText(e.getMessage());
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
error.printStackTrace();
}
}