blocks|key|27137|text|不知道你想要什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|27138|SELECT+MAX(SALARY)+FROM+EMPLOYEE+LEFT+JOIN+SALARY+ON+EMPLOYEE.ID=SALARY.ID+GROUP+BY+EMPLOYEE.ID|code-block|syntax|javascript|27139|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|$]]

Not sure what you wanted.

<pre><code>SELECT MAX(SALARY) FROM EMPLOYEE LEFT JOIN SALARY ON EMPLOYEE.ID=SALARY.ID GROUP BY EMPLOYEE.ID
</code></pre>

blocks|key|41356|text|基于salaries表(from_date+&+to_date)的定义，它是一个缓慢变化维数。您的数据可能如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|41357|Emp_no++salary+++from_date+++++to_date
++22+++++14000++2007-01-01++2008-03-31+--+or+2008-04-01
++22+++++16000++2008-04-01++2010-12-31+--+or+2011-01-01
++22+++++18000++2011-01-01++9999-12-31+--+or+NULL|code-block|syntax|javascript|41358|在这种情况下，您不需要MAX薪资，而是当前/最新的薪资。|41359|在SCD中，to_date通常设置为9999-12-31、3999-12-31或NULL之类的最大日期。若要获得当前薪资，请使用以下条件：|41360|WHERE+to_date+IS+NULL+
or
WHERE+to_date+=+DATE+'9999-12-31'+--+or+whatever+your+max+date+is
or+
WHERE+CURRENT_DATE+BETWEEN+from_date+AND+to_date|41361|在任何时间点获得工资：|41362|WHERE+whatever_date_you_want+BETWEEN+from_date+AND+to_date|41363|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Slowly_Changing_Dimension^0|2|8|C|9|O|7|14|6|0|0|0|B|3|0|6|7|I|A|T|A|14|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|G|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1P|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

Based on the definition of the <code>salaries</code> table (<code>from_date</code> &amp; <code>to_date</code>) it's a <a href="https://en.wikipedia.org/wiki/Slowly_Changing_Dimension" rel="nofollow">Slowly Changing Dimension</a>. Your data might look like this:

<pre><code>Emp_no salary from_date to_date
 22 14000 2007-01-01 2008-03-31 -- or 2008-04-01
 22 16000 2008-04-01 2010-12-31 -- or 2011-01-01
 22 18000 2011-01-01 9999-12-31 -- or NULL
</code></pre>

In that case you don't want the <code>MAX</code> salary but the current/latest salary. 

In a SCD <code>to_date</code> is usually set to either a maximum date like <code>9999-12-31</code> or <code>3999-12-31</code> or <code>NULL</code>. To get the current salary you use following conditions: 

<pre><code>WHERE to_date IS NULL 
or
WHERE to_date = DATE '9999-12-31' -- or whatever your max date is
or 
WHERE CURRENT_DATE BETWEEN from_date AND to_date
</code></pre>

To get the salary for any point in time:

<pre><code>WHERE whatever_date_you_want BETWEEN from_date AND to_date
</code></pre>

I have 2 different tables employees and salaries the salaries table has multiple duplicate id's on it my question is how can i combined the employee and salaries table and removed its duplicate but i want the max salary to be displayed for that employee.

Employees table
<a href="https://i.stack.imgur.com/9KOwD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9KOwD.png" alt="enter image description here"></a>

Salaries table
<a href="https://i.stack.imgur.com/xmP0o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xmP0o.png" alt="enter image description here"></a>

combined 2 different structure tables and remove duplicates

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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我有两个不同的表，雇员和薪水，工资表有多个重复的id在上面，我的问题是我如何组合员工和工资表，并删除它的副本，但我希望显示该雇员的最大工资。雇员表​薪金表​

问组合两个不同结构的表并删除重复项
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