blocks|key|1836327|text|如果多名员工的薪资相同，则查询将不起作用.|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1836328|尝试：|1836329|SELECT+emp_id,
+++++++salary,
+++++++dense_rank()+over(order+by+salary+desc)+AS+rk+
FROM+test_qaium|code-block|syntax|javascript|1836330|partition+by子句就像group+by，这里没有什么需要分组的。|offset|length|style|CODE|1836331|entityMap^0|0|0|0|0|C|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

If you have multiple employees on the same salary, your query will not work...

Try:

<pre><code>SELECT emp_id,
 salary,
 dense_rank() over(order by salary desc) AS rk 
FROM test_qaium
</code></pre>

The <code>partition by</code> clause works like a group by, and there's nothing here that needs grouping.

blocks|key|2505378|text|Oracle安装|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2505379|CREATE+TABLE+test_qaium+(+emp_id,+salary,+department_id+)+AS
++SELECT++1,+10000,+1+FROM+DUAL+UNION+ALL
++SELECT++2,+20000,+1+FROM+DUAL+UNION+ALL
++SELECT++3,+30000,+1+FROM+DUAL+UNION+ALL
++SELECT++4,+40000,+1+FROM+DUAL+UNION+ALL+--+One+highest,+one+2nd+highest
++SELECT++5,+10000,+2+FROM+DUAL+UNION+ALL
++SELECT++6,+20000,+2+FROM+DUAL+UNION+ALL
++SELECT++7,+30000,+2+FROM+DUAL+UNION+ALL
++SELECT++8,+30000,+2+FROM+DUAL+UNION+ALL+--+Two+highest,+one+2nd+highest
++SELECT++9,+10000,+3+FROM+DUAL+UNION+ALL
++SELECT+10,+10000,+3+FROM+DUAL+UNION+ALL+--+Two+highest,+no+2nd+highest
++SELECT+11,+10000,+4+FROM+DUAL+UNION+ALL+--+One+highest,+no+2nd+highest
++SELECT+12,+20000,+5+FROM+DUAL+UNION+ALL
++SELECT+13,+20000,+5+FROM+DUAL+UNION+ALL
++SELECT+14,+30000,+5+FROM+DUAL;++++++++++--+One+highest,+Two+2nd+highest|code-block|syntax|javascript|2505380|查询：|2505381|这将为每个部门获得第二高工资的所有行：|2505382|SELECT+*+
FROM+++(
++SELECT+t.*,
+++++++++DENSE_RANK()+OVER+(PARTITION+BY+department_id
++++++++++++++++++++++++++++ORDER+BY+salary+DESC)+AS+rnk+
++FROM+++test_qaium+t
)+
WHERE++rnk=2;|2505383|输出|2505384|EMP_ID+SALARY+DEPARTMENT_ID+RNK
------+------+-------------+---
+++++3++30000+++++++++++++1+++2
+++++6++20000+++++++++++++2+++2
++++12++20000+++++++++++++5+++2
++++13++20000+++++++++++++5+++2|2505385|entityMap^0|0|8|0|0|0|2|0|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|13|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|V|$]]

Oracle Setup:

<pre><code>CREATE TABLE test_qaium ( emp_id, salary, department_id ) AS
 SELECT 1, 10000, 1 FROM DUAL UNION ALL
 SELECT 2, 20000, 1 FROM DUAL UNION ALL
 SELECT 3, 30000, 1 FROM DUAL UNION ALL
 SELECT 4, 40000, 1 FROM DUAL UNION ALL -- One highest, one 2nd highest
 SELECT 5, 10000, 2 FROM DUAL UNION ALL
 SELECT 6, 20000, 2 FROM DUAL UNION ALL
 SELECT 7, 30000, 2 FROM DUAL UNION ALL
 SELECT 8, 30000, 2 FROM DUAL UNION ALL -- Two highest, one 2nd highest
 SELECT 9, 10000, 3 FROM DUAL UNION ALL
 SELECT 10, 10000, 3 FROM DUAL UNION ALL -- Two highest, no 2nd highest
 SELECT 11, 10000, 4 FROM DUAL UNION ALL -- One highest, no 2nd highest
 SELECT 12, 20000, 5 FROM DUAL UNION ALL
 SELECT 13, 20000, 5 FROM DUAL UNION ALL
 SELECT 14, 30000, 5 FROM DUAL; -- One highest, Two 2nd highest
</code></pre>

Query: 

This will get all the rows with the 2nd highest salary for each department:

<pre><code>SELECT * 
FROM (
 SELECT t.*,
 DENSE_RANK() OVER (PARTITION BY department_id
 ORDER BY salary DESC) AS rnk 
 FROM test_qaium t
) 
WHERE rnk=2;
</code></pre>

Output:

<pre><code>EMP_ID SALARY DEPARTMENT_ID RNK
------ ------ ------------- ---
 3 30000 1 2
 6 20000 2 2
 12 20000 5 2
 13 20000 5 2
</code></pre>

I already get it by doing this query:

<pre><code> SELECT * 
 FROM
 (
 SELECT emp_id,salary,row_number() over(order by salary desc) AS rk 
 FROM test_qaium
 ) 
 where rk=2;
</code></pre>

But one of my friend ask me to find second <code>MAX</code> salary from employees table must using "<code>over(partition by )</code>" in oracle sql. Anybody please help me.
And clear me the concept of "<code>Partition by</code>" in oracle sql.

How I can get Second max salary using "over(partition by)" in oracle SQL?

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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我已经通过执行以下查询获得了它：   SELECT *     FROM      (      SELECT emp_id,salary,row_number() over(order by salary  desc) AS rk       FROM test_qaium      )     where rk=2...

问如何使用oracle SQL中的"over(partition )“获得第二个最高工资？
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问如何使用oracle SQL中的"over(partition )“获得第二个最高工资？EN