blocks|key|2094340|text|你不需要这个HAVING+COUNT(*)=2。它只给你两个匹配的电影(一个人有两个片头或者同时有condition1和condition2)。不仅如此-当您使用distinct时(在本例中)，您根本不需要group+by。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2094341|SELECT+DISTINCT+movie
FROM+credits
WHERE+person+=+'condition1'+OR+person+=+'condition2'|code-block|syntax|javascript|2094342|entityMap^0|6|H|1D|A|1O|A|29|8|2V|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$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]]|$1|K|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|L|$]]

You no need this <code>HAVING COUNT(*)=2</code>. It gives you only movies with two matches (one person with two credits or with both <code>condition1</code> and <code>condition2</code>). More than that - you no need <code>group by</code> at all when you using <code>distinct</code> (in this case).

<pre><code>SELECT DISTINCT movie
FROM credits
WHERE person = 'condition1' OR person = 'condition2'
</code></pre>

blocks|key|474897|text|select+distinct+cr1.movie
from+credits+as+cr1
join+credits+as+cr2
using+(movie)
where+cr1.person+=+{person+1+goes+here}
and+cr2.person+=+{person+2+goes+here}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|474898|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 distinct cr1.movie
from credits as cr1
join credits as cr2
using (movie)
where cr1.person = {person 1 goes here}
and cr2.person = {person 2 goes here}
</code></pre>

blocks|key|4168283|text|SELECT+DISTINCT+c1.movie
FROM+credits+AS+c1
JOIN+credits+AS+c2+on+(c1.movie+=+c2.movie)
WHERE+c1.person+=+'john'
AND+c2.person+=+'kate'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4168284|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 DISTINCT c1.movie
FROM credits AS c1
JOIN credits AS c2 on (c1.movie = c2.movie)
WHERE c1.person = 'john'
AND c2.person = 'kate'
</code></pre>

blocks|key|2404509|text|这应该是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2404510|SELECT+DISTINCT+movie
FROM+(
++++++SELECT+movie,+COUNT(person)+AS+contributors
++++++FROM+credits
++++++WHERE+person+IN('person1','person2')
++++++GROUP+BY+movie,+person
++++++HAVING+contributors>1
+++++)+t1|code-block|syntax|javascript|2404511|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|$]]

This should work:

<pre><code>SELECT DISTINCT movie
FROM (
 SELECT movie, COUNT(person) AS contributors
 FROM credits
 WHERE person IN('person1','person2')
 GROUP BY movie, person
 HAVING contributors&gt;1
 ) t1
</code></pre>

I am trying to figure out a way to find shared credits by two people in a movie database, for example:

table: 'credits' 
columns: 'id','movie','person' 

My other problem is a person might have multiple credits for the same movie, so how do I filter out duplicates? Any help would be appreciated. The following is what I have so far. Am I on the right track?

<pre><code>SELECT DISTINCT movie
FROM credits
WHERE person = 'condition1' OR person = 'condition2'
GROUP BY movie
HAVING COUNT(*)=2
</code></pre>

MySQL query on single table to find multiple rows

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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我正在尝试找出一种在电影数据库中找到两个人共享的片尾的方法，例如：表：“信用”列：'id'，‘电影’，‘人’我的另一个问题是，一个人可能对同一部电影有多个片头，那么我如何过滤掉重复的内容呢？任何帮助都将不胜感激。以下是我到目前为止所掌握的。我在正确的轨道上吗？SELECT DISTINCT movieFROM cred...

问对单个表进行MySQL查询以查找多行
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问对单个表进行MySQL查询以查找多行EN