blocks|key|2102942|text|对于不在两个表中的用户，可以使用左连接而不是内连接，然后测试是否为null。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2102943|示例：|2102944|SELECT+a.member_id+
FROM+records+a
++LEFT+JOIN+records+b+ON++a.member_id+=+b.member_id
++++++++++++++++++++++AND+b.registered_year+=+'2013'+
WHERE+a.registered_year+=+'2012'+
++AND+b.registered_year+IS+NULL;|code-block|syntax|javascript|2102945|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

For users which are not in both tables, you can use a left join rather than an inner join and then test for null.

Example:

<pre><code>SELECT a.member_id 
FROM records a
 LEFT JOIN records b ON a.member_id = b.member_id
 AND b.registered_year = '2013' 
WHERE a.registered_year = '2012' 
 AND b.registered_year IS NULL;
</code></pre>

blocks|key|4419680|text|SELECT+member_id+FROM+records+a+
WHERE+a.registered_year='2012'+and++member_id+not+in+
(
++SELECT+member_id+FROM+records
++WHERE+registered_year='2013'
)+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4419681|内部select+SELECT+member_id+FROM+records+WHERE+registered_year='2013'获取2013年注册的所有用户|unstyled|offset|length|style|CODE|4419682|外层获取所有不在内部的用户，选择2012年注册的用户|4419683|entityMap^0|0|9|1M|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>SELECT member_id FROM records a 
WHERE a.registered_year='2012' and member_id not in 
(
 SELECT member_id FROM records
 WHERE registered_year='2013'
) 
</code></pre>

the inner select <code>SELECT member_id FROM records WHERE registered_year='2013'</code> gets all users who where registered in 2013

the outer one gets all users that where not in the inner select who where registered in 2012

blocks|key|1648345|text|SELECT+member_id
++FROM+records+a
+WHERE+registered_year+=+2012
+++AND+NOT+EXISTS+(SELECT+null
+++++++++++++++++++++FROM+records+b
++++++++++++++++++++WHERE+a.member_id+=+b.member_id
++++++++++++++++++++++AND+b.registered_year+=+2013)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1648346|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 member_id
 FROM records a
 WHERE registered_year = 2012
 AND NOT EXISTS (SELECT null
 FROM records b
 WHERE a.member_id = b.member_id
 AND b.registered_year = 2013)
</code></pre>

blocks|key|1653750|text|这通常使用NOT+EXISTS+(或等效的LEFT+JOIN+/+IS+NULL和NOT+IN构造)来解决。在这种特殊情况下，有一种使用GROUP+BY的方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1653751|SELECT+member_id+
FROM+records
WHERE+registered_year+IN+(2012,+2013)
GROUP+BY+member_id
HAVING+MAX(registered_year)+=+2012+;|code-block|syntax|javascript|1653752|我怀疑这是否有效，但您可以与其他版本一起测试。|1653753|entityMap^0|5|A|L|J|15|6|1W|8|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

This is usually solved with <code>NOT EXISTS</code> (or the equivalent <code>LEFT JOIN / IS NULL</code> and <code>NOT IN</code> constructions.) In this particular case, there is a way with <code>GROUP BY</code>:

<pre><code>SELECT member_id 
FROM records
WHERE registered_year IN (2012, 2013)
GROUP BY member_id
HAVING MAX(registered_year) = 2012 ;
</code></pre>

I doubt this will be efficient but you can test along with the other versions.

I'm trying to find 2 different result via JOINS.
Table look like;

<pre><code>SELECT id,member_id,registered_year FROM records;
</code></pre>

I can listing which members registered in 2012 and also in 2013 with;

<pre><code>SELECT member_id FROM records a
INNER JOIN records b ON a.member_id=b.member_id
WHERE a.registered_year='2013' AND b.registered_year='2012';
</code></pre>

But I can't list revers of It. How can I list which members were registered in 2012 but not in 2013?

Thnx in advance.

MySQL - JOINS (excepts)

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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我试图通过JOINS找到两个不同的结果。表格看起来像；SELECT id,member_id,registered_year FROM records;我可以列出哪些会员在2012年注册，也在2013年注册；SELECT member_id FROM records aINNER JOIN records b ON a...

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