blocks|key|4387933|text|我认为一个简单的group+by就可以做到这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4387934|SELECT
++++name,
++++MAX(cardId)
FROM+table
GROUP+BY+name|code-block|syntax|javascript|4387935|编辑由于问题已更新，我认为您要查找的内容如下所示：|offset|length|style|BOLD|4387936|SELECT
++++MIN(name),
++++MAX(name),
++++cardId,
++++COUNT(name)+name_count
FROM+table
GROUP+BY+cardId
HAVING+COUNT(name)+>=+2|4387937|您也可以通过将表连接到自身来实现这一点，但是使用HAVING的版本的优点是，当您有3个或更多具有相同cardId的人时，它不会给出重复的结果。|CODE|4387938|entityMap^0|0|0|0|2|0|0|O|6|1E|6|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@$I|W|J|X|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Z|8|@$I|10|J|11|K|Q]|$I|12|J|13|K|Q]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|S|$]]

I think a simple group by would do the trick:
<pre><code>SELECT
 name,
 MAX(cardId)
FROM table
GROUP BY name
</code></pre>
Edit
Since the question has been updated, I think you're looking for something like this:
<pre><code>SELECT
 MIN(name),
 MAX(name),
 cardId,
 COUNT(name) name_count
FROM table
GROUP BY cardId
HAVING COUNT(name) &gt;= 2
</code></pre>
You could also do this by joining the table to itself, but the advantage of the version with <code>HAVING</code> is that it doesn't give duplicate results when you have 3 or more people with the same <code>cardId</code>.

blocks|key|2071532|text|使用GROUP+BY可以解决您的问题|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2071533|select+name,+max(cardid)+as+cardid+from+table_1
group+by+name+
order+by+cardid+asc|code-block|syntax|javascript|2071534|entityMap^0|2|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

USING <code>GROUP BY</code> will solve you problem
<pre><code>select name, max(cardid) as cardid from table_1
group by name 
order by cardid asc
</code></pre>

blocks|key|5856361|text|with+t+as+(
++++select+1+Id+,++'John'+name,+123+CardId+union+all
++++select+2+,++'Mark'+,+123++++union+all
++++select+3+,++'Tom'+,+231+union+all
++++select+4+,++'Debra'+,+321+++union+all
++++select+5+,++'Debra'+,+321+++union+all
++++select+6+,++'Debra'+,+321+++union+all
++++select+7+,++'Tom'+,+321+union+all
++++select+8+,++'Tom'+,+322
)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5856362|我认为测试数据需要Tom的另一行，通过max()/group+by解决方案，Tom+322将被返回，但根据规范不应该返回，因为它是单个CardId|unstyled|offset|length|style|CODE|5856363|++select+distinct+name,+CardId+from+(
++++++++select+name,+CardId,+Count(*)+over(partition+by+CardId)+c
++++++++from+t
++++)x
++++where+c>1|5856364|您需要计算distinct+CardIds的数量并过滤单行|5856365|entityMap^0|0|J|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@$G|S|H|T|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|W|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>with t as (
 select 1 Id , 'John' name, 123 CardId union all
 select 2 , 'Mark' , 123 union all
 select 3 , 'Tom' , 231 union all
 select 4 , 'Debra' , 321 union all
 select 5 , 'Debra' , 321 union all
 select 6 , 'Debra' , 321 union all
 select 7 , 'Tom' , 321 union all
 select 8 , 'Tom' , 322
)
</code></pre>
I think the test data needs another row for Tom, with the <code>max()/group by</code> solution Tom 322 is returned but according to the specification should not be as it's a single CardId
<pre><code> select distinct name, CardId from (
 select name, CardId, Count(*) over(partition by CardId) c
 from t
 )x
 where c&gt;1
</code></pre>
You need to count the number of distinct CardIds and filter single rows

blocks|key|4388014|text|使用exists|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4388015|select+distinct+t.*
from+t
where+exists+(select+1
++++++++++++++from+t+t2
++++++++++++++where+t2.cardid+=+t.cardid+and+t2.name+<>+t.name
+++++++++++++);|code-block|syntax|javascript|4388016|使用(cardid,+name)上的索引，这可能是处理查询的最快方法。|4388017|entityMap^0|2|6|0|0|2|E|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

Use <code>exists</code>:
<pre><code>select distinct t.*
from t
where exists (select 1
 from t t2
 where t2.cardid = t.cardid and t2.name &lt;&gt; t.name
 );
</code></pre>
With an index on <code>(cardid, name)</code>, this is probably the fastest way to approach the query.

Based on below table
<pre><code>id name cardId
--------
1 John 123
2 Mark 123
3 Tom 231
4 Debra 321
5 Debra 321
6 Debra 321
7 Tom 321
</code></pre>
How can I select records with same cardId but with different names?
Output should looks like this
<pre><code> name cardId
--------
 John 123
 Mark 123
 Debra 321
 Tom 321
</code></pre>
EDIT:
Just to be clear. One user can have multiple cardId, but one cardId cannot be shared between different users. How can I find this?

Find duplicates based on non-duplicates column

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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基于下表id  name  cardId--------1   John    1232   Mark    1233   Tom     2314   Debra   3215   Debra   3216   Debra   3217   Tom     321如何选择cardId相同但名称不同的记录？输出应如下所...

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