blocks|key|2865052|text|去除子选择。IN()子句完全没问题。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|2865053|在AJ要求写下更正后的查询后：|2865054|SELECT+`r`.`id`
FROM+`tips`+AS+`r`
LEFT+JOIN+`tips_showed`+AS+`a`
ON+(`a`.`user`='8'+AND+`r`.`id`=`a`.`tip`)
WHERE+`r`.`id`+IN+('3','4','5','2','6','7','8','9','10','18',
+++++'11','12','13','14','15','16','17','20')
AND+`a`.`tip`+IS+NULL
LIMIT+1|code-block|syntax|javascript|2865055|上面的代码没有经过测试，可能包含一些严重的错误(临时编写)-如果您发现任何错误，请在评论中给我反馈。|2865056|加号|2865057|不要忘记添加适当的索引。|2865058|entityMap^0|0|5|6|4|0|0|0|0|0|0|C|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|D]]|E|@]|F|$]]|$1|G|3|H|5|6|7|10|8|@]|E|@]|F|$]]|$1|I|3|J|5|K|7|11|8|@]|E|@]|F|$L|M]]|$1|N|3|O|5|6|7|12|8|@]|E|@]|F|$]]|$1|P|3|Q|5|6|7|13|8|@]|E|@]|F|$]]|$1|R|3|S|5|6|7|14|8|@$9|15|A|16|B|C]]|E|@]|F|$]]|$1|T|3|-4|5|6|7|17|8|@]|E|@]|F|$]]]|U|$]]

Get rid of sub-selects. The <code>IN()</code> clause is perfectly ok.

Following AJ's request for writing down the corrected query:

<pre><code>SELECT `r`.`id`
FROM `tips` AS `r`
LEFT JOIN `tips_showed` AS `a`
ON (`a`.`user`='8' AND `r`.`id`=`a`.`tip`)
WHERE `r`.`id` IN ('3','4','5','2','6','7','8','9','10','18',
 '11','12','13','14','15','16','17','20')
AND `a`.`tip` IS NULL
LIMIT 1
</code></pre>

The above code is not tested and may contain some serious mistakes (was written ad hoc) - if you find any, please give me feedback in the comments.

Plus

Do not forget to add proper indexes.

blocks|key|4654604|text|我相信您可以将查询简化为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4654605|SELECT+t.id
++++FROM+`tips`+t
++++++++LEFT+JOIN+`tips_showed`+a
++++++++++++ON+t.id+=+a.tip
++++++++++++++++AND+a.user+=+'8'
++++WHERE+a.tip+IS+NULL
++++++++AND+t.id+IN+('3','4','5','2','6','7','8','9','10','18',
+++++++++++++++++++++'11','12','13','14','15','16','17','20')
++++LIMIT+1|code-block|syntax|javascript|4654606|使用NOT+EXISTS而不是LEFT+JOIN可能还会为您带来一些额外的性能。尝试每种方法并进行比较。|offset|length|style|CODE|4654607|+SELECT+t.id
+++++FROM+`tips`+t
+++++WHERE+t.id+IN+('3','4','5','2','6','7','8','9','10','18',
++++++++++++++++++++'11','12','13','14','15','16','17','20')
++++++++AND+NOT+EXISTS(SELECT+NULL+
+++++++++++++++++++++++++++FROM+`tips_showed`+a+
+++++++++++++++++++++++++++WHERE+a.tip+=+t.id
+++++++++++++++++++++++++++++++AND+a.user+=+'8')
++++LIMIT+1|4654608|entityMap^0|0|0|2|A|F|9|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

I believe you can simplify the query to:

<pre><code>SELECT t.id
 FROM `tips` t
 LEFT JOIN `tips_showed` a
 ON t.id = a.tip
 AND a.user = '8'
 WHERE a.tip IS NULL
 AND t.id IN ('3','4','5','2','6','7','8','9','10','18',
 '11','12','13','14','15','16','17','20')
 LIMIT 1
</code></pre>

Using <code>NOT EXISTS</code> instead of the <code>LEFT JOIN</code> may also buy you some additional performace. Try each and compare.

<pre><code> SELECT t.id
 FROM `tips` t
 WHERE t.id IN ('3','4','5','2','6','7','8','9','10','18',
 '11','12','13','14','15','16','17','20')
 AND NOT EXISTS(SELECT NULL 
 FROM `tips_showed` a 
 WHERE a.tip = t.id
 AND a.user = '8')
 LIMIT 1
</code></pre>

blocks|key|4654667|text|我注意到IN是值的连续序列|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4654668|2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+and+an+extra+number+20.|code-block|syntax|javascript|4654669|您可以将此处的IN替换为|4654670|WHERE+(r.id+=+20)+OR+(r.id+BETWEEN+2+AND+18)|4654671|更快，因为你只需要做3次最坏的比较。|4654672|如果有随机值，则为|4654673|使用临时内存表。|offset|length|style|BOLD|4654674|使用散列主键。|4654675|并在主散列键上执行内部联接。|4654676|计时并告诉我它有多快:-)|4654677|entityMap^0|0|0|0|0|0|0|0|8|0|0|7|0|0|E|0|0|D|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@$Q|19|R|1A|S|T]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1B|8|@$Q|1C|R|1D|S|T]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1E|8|@$Q|1F|R|1G|S|T]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1H|8|@$Q|1I|R|1J|S|T]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|11|$]]

I noticed that the IN is a sequential series of the values 

<pre><code>2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 and an extra number 20.
</code></pre>

You can just replace the IN here with 

<pre><code>WHERE (r.id = 20) OR (r.id BETWEEN 2 AND 18)
</code></pre>

Much faster, because you only have to do 3 comparisons worst case. 

If you have random values 

Use a temp memory table. 
With a hash primary key. 
And do an inner join on the primary hash key. 

Time this and tell me how much faster it went :-)

I know that I have to avoid to use "IN" clause, but I don't know how to do in this case. The scenario is this: from PHP I get a list of values and then I construct a query like this:

<pre><code>SELECT id FROM
 ( SELECT * FROM `tips` r 
 LEFT JOIN (SELECT tip FROM `tips_showed` WHERE user='8') AS a ON r.id=a.tip
 WHERE a.tip IS NULL ) AS t
WHERE t.id IN
 ('3','4','5','2','6','7','8','9','10','18',
 '11','12','13','14','15','16','17','20') LIMIT 1
</code></pre>

This makes my site very slow (because it is executed each time a user visits it). How can make it faster?

MySQL: How to simplify a "IN" query

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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我知道我必须避免使用" in“子句，但我不知道在这种情况下该怎么做。场景是这样的:我从PHP获得一个值列表，然后构造一个查询，如下所示：SELECT id FROM    ( SELECT * FROM `tips` r       LEFT JOIN (SELECT tip FROM `tips_showed` WH...

问MySQL:如何简化"IN“查询
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