blocks|key|5566558|text|select+*+from+table+group+by+Subject+having+count(dept)+>+1;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5566559|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>select * from table group by Subject having count(dept) &gt; 1;
</code></pre>

blocks|key|5566554|text|尝试执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|5566555|select+a.dept,a.Subject+from+tableName+a,b+where+a.dept=b.dept|code-block|syntax|javascript|5566556|entityMap^0|0|9|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|$]]

Try Following:

<pre><code>select a.dept,a.Subject from tableName a,b where a.dept=b.dept
</code></pre>

blocks|key|4098378|text|尝试执行以下查询：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4098379|SELECT+dept,+Subject+FROM+tableName+WHERE+Subject+IN+(SELECT+Subject+FROM+tableName+GROUP+BY+Subject+HAVING+COUNT(Subject)>1);|code-block|syntax|javascript|4098380|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try the following query:

<pre><code>SELECT dept, Subject FROM tableName WHERE Subject IN (SELECT Subject FROM tableName GROUP BY Subject HAVING COUNT(Subject)&gt;1);
</code></pre>

blocks|key|5566578|text|+select+*+from+tabelname+where+subject+in+(+select+subject+from+tablename+
+group+by+subject+having+count(subject)+>+1)+++++++++++++++++|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5566579|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> select * from tabelname where subject in ( select subject from tablename 
 group by subject having count(subject) &gt; 1) 
</code></pre>

blocks|key|1781080|text|你可以试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781081|SELECT+DEPT,+Subject+FROM+Table1+WHERE+Subject+IN+
+(SELECT+Subject+FROM+Table1+GROUP+BY+Subject+HAVING+COUNT(Subject)>1);|code-block|syntax|javascript|1781082|这里有一个真实的例子：|1781083|http://sqlfiddle.com/#!2/b5885/12|offset|length|1781084|输出如下：|1781085|DEPT++++SUBJECT
ECE+++++++++Maths
EEE+++++++++Maths
EEE+++++++++ET
MEC+++++++++Maths
MEC+++++++++ET|1781086|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|X|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|10|8|@]|9|@$K|11|L|12|1|13]]|A|$]]|$1|M|3|N|5|6|7|14|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|J]]]]

You can try this out:

<pre><code>SELECT DEPT, Subject FROM Table1 WHERE Subject IN 
 (SELECT Subject FROM Table1 GROUP BY Subject HAVING COUNT(Subject)&gt;1);
</code></pre>

Live example here:

<a href="http://sqlfiddle.com/#!2/b5885/12" rel="nofollow">http://sqlfiddle.com/#!2/b5885/12</a>

The output is as follows:

<pre><code>DEPT SUBJECT
ECE Maths
EEE Maths
EEE ET
MEC Maths
MEC ET
</code></pre>

blocks|key|1487558|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1487559|SELECT+*+FROM+XYZ+WHERE+SUBJECT+IN+
(SELECT+SUBJECT+FROM+XYZ+HAVING+COUNT(SUBJECT)>1+
GROUP++++BY+SUBJECT);|code-block|syntax|javascript|1487560|这是最后一张桌子|1487561|DEPT+++++++++++++YEAR+SUBJECT
----------+----------+----------
EEE+++++++++++++++++1+MATHS
EEE+++++++++++++++++1+ET
MEC+++++++++++++++++1+MATHS
MEC+++++++++++++++++1+ET|1487562|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

TRY THIS 

<pre><code>SELECT * FROM XYZ WHERE SUBJECT IN 
(SELECT SUBJECT FROM XYZ HAVING COUNT(SUBJECT)&gt;1 
GROUP BY SUBJECT);
</code></pre>

HERE IS THE END TABLE

<pre><code>DEPT YEAR SUBJECT
---------- ---------- ----------
EEE 1 MATHS
EEE 1 ET
MEC 1 MATHS
MEC 1 ET
</code></pre>

blocks|key|1487577|text|SELECT+
++++DEPT,
++++Subject
FROM+tablename
WHERE
+++++Subject+IN+(
++++++++++++++++SELECT+[Subject]+FROM+(
+++++++++++++++++++++++++++++++++++++++SELECT+[Subject],count([Subject])+AS+CountSubj+FROM+tablename+GROUP+BY+[Subject]
++++++++++++++++++++++++++++++++++++++)+AS+subtbl
++++++++++++++++WHERE+subtbl.CountSubj>1)
ORDER+BY+Subject|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1487578|尝试上面的sql查询。|unstyled|1487579|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>SELECT 
 DEPT,
 Subject
FROM tablename
WHERE
 Subject IN (
 SELECT [Subject] FROM (
 SELECT [Subject],count([Subject]) AS CountSubj FROM tablename GROUP BY [Subject]
 ) AS subtbl
 WHERE subtbl.CountSubj&gt;1)
ORDER BY Subject
</code></pre>

try above sql query.

blocks|key|1781150|text|使用带有EXISTS运算符的选项|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781151|SELECT+*
FROM+dbo.test110+t
WHERE+EXISTS+(
++++++++++++++SELECT+1
++++++++++++++FROM+dbo.test110+t2
++++++++++++++WHERE+t.Subject++=+t2.Subject
++++++++++++++HAVING+COUNT(t2.Subject)+>+1
++++++++++++++)|code-block|syntax|javascript|1781152|上的演示|1781153|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Use option with EXISTS operator

<pre><code>SELECT *
FROM dbo.test110 t
WHERE EXISTS (
 SELECT 1
 FROM dbo.test110 t2
 WHERE t.Subject = t2.Subject
 HAVING COUNT(t2.Subject) &gt; 1
 )
</code></pre>

Demo on <a href="http://sqlfiddle.com/#!3/567f7/1" rel="nofollow">SQLFiddle</a>

I have a table like 

<pre><code>DEPT Years Subject
ECE 1 Maths
ECE 1 PDC
EEE 1 Maths
EEE 1 ET
MEC 1 Maths
MEC 1 ET
</code></pre>

I want the query to find all DEPT (departments) in which subjects are repeated like Maths in ECE, EEE &amp; MEC and ET repeated in EEE &amp; MEC i.e., only repeated ones

Please advice,

SQL Query required

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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我有一张像这样的桌子DEPT   Years   SubjectECE      1       MathsECE      1       PDCEEE      1       MathsEEE      1       ETMEC      1       MathsMEC      1       ET我希望查询找到所有部门(部门)，其中科目重复，如数学在欧洲教育，EEE和MEC和EEE在

问需要SQL查询
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回答 8

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