我正在尝试计算重复数据，但我的查询无法正常工作。
每个用户都应该有一个dev_id但是当其他用户拥有相同的dev_id时我想知道这个
表格例如：
dev_id user_id
------------------
111 1
111 1
222 2
111 2
333 3
应该结果：
user_id qu
------------------
1 1
2 1
3 0
这是我的查询
SELECT t1.user_id,
(SELECT Count(DISTINCT t2.dev_id)
FROM reports t2
WHERE t2.user_id != t1.user_id
AND t2.dev_id = t1.dev_id
) AS qu
FROM reports t1
GROUP BY t1.user_id
答案 0 :(得分：1)
您可以通过以下方式获得结果：
select r.user_id, count(*) - 1
from reports r
group by r.user_id;
这是您想要的计算吗？
答案 1 :(得分：0)
好。让我们从简单开始。
首先，您需要获得唯一的user_id / dev id组合
public class EditFragment extends Fragment {
private EditCallback editCallback;
MenuItem save;
MenuItem cancel;
View view;
EditText editText;
int position;
Bundle bundle;
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
setHasOptionsMenu(true);
view = inflater.inflate(R.layout.edit_fragment_layout, container, false);
editText = (EditText) view.findViewById(R.id.editText);
bundle = getArguments();
position = bundle.getInt("position");
editText.setText(bundle.getString("name"));
return view;
}
@Override
public void onAttach(Context context) {
super.onAttach(context);
editCallback = (EditCallback) context;
}
@Override
public void onCreateOptionsMenu(Menu menu, MenuInflater inflater) {
inflater.inflate(R.menu.edit_fragment, menu);
save = menu.findItem(R.id.save);
cancel = menu.findItem(R.id.cancel);
cancel.setOnMenuItemClickListener(new MenuItem.OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {
FragmentTransaction fragmentTransaction = getFragmentManager().beginTransaction();
MyListFragment myListFragment = (MyListFragment) getFragmentManager().findFragmentByTag("MyListFragment");
fragmentTransaction.replace(R.id.fragContainer, myListFragment);
fragmentTransaction.addToBackStack(null);
fragmentTransaction.commit();
return true;
}
});
MenuItem menuItem = save.setOnMenuItemClickListener(new MenuItem.OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {
if (editCallback != null) {
editCallback.updateItem(position, editText.getText().toString());
Toast.makeText(getActivity(), "Yor item was updated", Toast.LENGTH_SHORT).show();
}
return true;
}
});
}
@Override
public void onDetach() {
super.onDetach();
editCallback = null;
}
public interface EditCallback {
void updateItem(int position, String textUpdated);
}
}
结果将是
select distinct dev_id,user_id from reports
之后，您应该获得每个dev_id
的不同user_id的数量 dev_id user_id
------------------
111 1
222 2
111 2
333 3
此类查询的结果将是
select dev_id,c from (
SELECT
dev_id,
count(*)-1 AS c
FROM
(select distinct user_id,dev_id from reports) as fixed_reports
GROUP BY dev_id
) as counts
现在您应该向具有此类dev_id的用户显示。为此你应该加入这样的dev_id列表和step1中的表（显示哪一个user_id，dev_id对存在）
dev_id c
-----------------
111 1
222 0
333 0
＆＃34;鲜明＆＃34;这里需要跳过相同的行。
结果应该是内部查询
select distinct fixed_reports2.user_id,counts.c from (
SELECT
dev_id,
count(*)-1 AS c
FROM
(select distinct user_id,dev_id from reports) as fixed_reports
GROUP BY dev_id
) as counts
join
(select distinct user_id,dev_id from reports) as fixed_reports2
on fixed_reports2.dev_id=counts.dev_id
where counts.c>0 and counts.c is not null
适用于所有查询
dev_id c
-----------------
111 1
如果您确定还需要c = 0的行，则需要执行＆＃34; left join＆＃34;对于fixed_reports2和大型查询，这样你将获得c = null的所有行和行将是0行（可以通过case / when语句更改）
答案 2 :(得分：0)
您的查询已损坏，无法在许多系统上运行。问题是user_id为2的组有两个不同的dev_id。如果您运行下面的“已损坏的查询”，则可以看到min()和max()是不同的，但子查询只能看到随机选择的其中一个值。通过将dev_id添加到分组来更正最后一个查询，该分组显示“缺失”行在计数中的位置。
SELECT -- broken query
t1.user_id, min(t1.dev_id), max(t1.dev_id),
(select distinct t1.dev_id from reports) as should_have_errored,
(SELECT Count(DISTINCT t2.dev_id)
FROM reports t2
WHERE t2.user_id != t1.user_id
AND t2.dev_id = t1.dev_id
) AS qu
FROM reports t1
GROUP BY t1.user_id;
-- On SQL Server that query returns an error
-- Msg 8120, Level 16, State 1, Line 7
-- Column 'reports.dev_id' is invalid in the select list because it is
-- not contained in either an aggregate function or the GROUP BY clause.
SELECT -- query that duplicates your original query
t1.user_id,
(SELECT Count(DISTINCT t2.dev_id)
FROM reports t2
WHERE t2.user_id != t1.user_id
AND t2.dev_id = max(t1.dev_id) /* <-- see here */
) AS qu
FROM reports t1
GROUP BY t1.user_id;
SELECT t1.user_id, t1.dev_id, -- fixed query
(SELECT Count(DISTINCT t2.dev_id)
FROM reports t2
WHERE t2.user_id != t1.user_id
AND t2.dev_id = t1.dev_id
) AS qu
FROM reports t1
GROUP BY t1.user_id, t1.dev_id
http://sqlfiddle.com/#!9/6576e3/20
以下是一些可能有用的查询：
哪个dev_id有多个user_id与之关联？
select dev_id
from reports
group by dev_id
having count(distinct user_id) > 1
哪个user_id与dev_id共享user_id？
select user_id
from reports r1
where exists (
select 1
from reports r2
where r2.dev_id = r1.dev_id and r2.user_id <> ?
)
或者说这实际上只相当于一个内部联接，这也使得一次列出所有人变得容易。请注意，每对将列出两次：
select r1.user_id, r1.dev_id, r2.user_id as common_user_id
from
reports r1 inner join reports r2
on r2.dev_id = r1.dev_id
where
r1.user_id <> r2.user_id
order by
r1.user_id, r1.dev_id, r2.user_id
由于您的表格中有重复的行，因此您需要将其设为select distinct才能获得唯一的行。
答案 3 :(得分：0)
我认为以下sql查询应该可以解决你的问题：
SELECT t1.user_id, t1.dev_id, count(t2.user_id) as qu
FROM (Select Distinct * from reports) t1
Left Join (Select Distinct * from reports) t2
on t1.user_id != t2.user_id and t2.dev_id = t1.dev_id
group by t1.user_Id, t1.dev_id
答案 4 :(得分：0)
SELECT user_id, (COUNT(user_id) -1) as qu
FROM reports
GROUP BY user_id
这样可以在您的情况下获得理想的结果，但是您可以更好地改进它。 干杯,,
答案 5 :(得分：-1)
尝试
SELECT
user_id,
SUM(qu) AS qu
FROM (
SELECT
user_id,
count(*)-1 AS qu
FROM
reports
GROUP BY user_id, dev_id
) AS r
GROUP BY user_id
如果您需要的所有数据都在一个表中，则无需进行连接。
编辑：将组更改为dev_id而不是user_id
Edit2：我认为你需要在group by子句中同时使用dev_id和user_id。
Edit3：添加了一个子查询以获得所需的结果。这可能有点麻烦，也许有人有办法改善这个？