[英]Cannot instantiate the type
我試圖解決此錯誤,即使進行了研究,我的嘗試也沒有成功。 特別是,我收到以下錯誤:無法實例化類型List
下面是代碼:
public class MatchingActivity extends Activity {
protected ParseRelation<ParseUser> mFriendsRelation;
protected ParseUser mCurrentUser;
protected List<ParseUser> mUsers;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
requestWindowFeature(Window.FEATURE_INDETERMINATE_PROGRESS);
setContentView(R.layout.matching);
// Show the Up button in the action bar.
//create list variable
mUsers = new List<ParseUser>();
}
@Override
protected void onResume() {
super.onResume();
mCurrentUser = ParseUser.getCurrentUser();
setProgressBarIndeterminateVisibility(true);
ParseQuery<ParseUser> query = ParseUser.getQuery();
query.findInBackground(new FindCallback<ParseUser>() {
public void done(List<ParseUser> users, ParseException e) {
if (e == null) {
//add all the users to your list variable
mUsers.addAll(users);
} else {
// Something went wrong.
}
}
});
//check the size of your list to see how big it is before accessing it
final int size = mUsers.size();
//or use a loop to loop through each one
for(ParseUser mParseUser : mUsers)
{
//skip over the current user
if(mParseUser == ParseUser.getCurrentUser())
continue;
mParseUser.getString("name");
mParseUser.getNumber("age");
mParseUser.getString("headline");
}
}
}
任何幫助將不勝感激。 提前致謝
更新資料
public class MatchingActivity extends Activity {
protected ParseRelation<ParseUser> mFriendsRelation;
protected ParseUser mCurrentUser;
protected List<ParseUser> mUsers;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
requestWindowFeature(Window.FEATURE_INDETERMINATE_PROGRESS);
setContentView(R.layout.matching);
// Show the Up button in the action bar.
//create list variable
mUsers = (List<ParseUser>) findViewById(R.id.listView1);
}
@Override
protected void onResume() {
super.onResume();
mCurrentUser = ParseUser.getCurrentUser();
setProgressBarIndeterminateVisibility(true);
ParseQuery<ParseUser> query = ParseUser.getQuery();
query.findInBackground(new FindCallback<ParseUser>() {
public void done(List<ParseUser> users, ParseException e) {
if (e == null) {
//add all the users to your list variable
mUsers.addAll(users);
} else {
// Something went wrong.
}
}
});
//check the size of your list to see how big it is before accessing it
final int size = mUsers.size();
//or use a loop to loop through each one
for(ParseUser mParseUser : mUsers)
{
//skip over the current user
if(mParseUser == ParseUser.getCurrentUser())
continue;
mParseUser.getString("name");
mParseUser.getNumber("age");
mParseUser.getString("headline");
ArrayAdapter<String> arrayAdapter = new ArrayAdapter<String>(
this,
android.R.layout.simple_list_item_1, Unsure what to input here,
as I want to return all three items (name, age, headline) from parse into the list);
mUsers.setAdapter(arrayAdapter);
}
}
}
提示錯誤為類型List未定義方法setAdapter(ArrayAdapter)
謝謝你的支持
您不能使用new List()
實例化Interface List
關鍵字new
用於創建(實例化)對象。 在這種情況下,您可以使用implements List
任何類實例化Interface List
implements List
mUsers = new ArrayList<ParseUser>(); //example with ArrayList
查看所有已知實現類:的List
中的Java API 在這里 。
列表是一個接口。 接口無法實例化。
所以試試這個:
mUsers = new ArrayList<ParseUser>();
代替
mUsers = new List<ParseUser>();
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