我已经尝试解决这个错误,即使经过研究,我的尝试也没有成功。特别是,我收到以下错误: 无法实例化类型列表
代码如下:
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");
}
}
}
如有任何帮助,我们将不胜感激。 提前致谢
更新
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);
}
}
}
错误提示 方法 setAdapter(ArrayAdapter) 未定义类型 List
感谢您的支持
最佳答案
您不能使用 new List()
实例化 Interface List
关键字new
用于创建(实例化)对象。在这种情况下,您可以使用任何实现 List
Interface List
mUsers = new ArrayList<ParseUser>(); //example with ArrayList
参见 All Known Implementing Classes: of List
in Java api here .
关于java - 无法实例化类型,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25111867/