blocks|key|5184314|text|它不应该是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5184315|select+Address+FROM+Addresses+WHERE+Address+NOT+LIKE+'%25'+%2B+HouseNumber+%2B+'%25'|code-block|syntax|javascript|5184316|?？|5184317|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Shouldn't it be:

<pre><code>select Address FROM Addresses WHERE Address NOT LIKE '%' + HouseNumber + '%'
</code></pre>

??

blocks|key|5668274|text|也许你应该换一种方式试试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5668275|select+Address+FROM+Addresses+WHERE+Address+NOT+LIKE+'%25+'+%2B+HouseNumber+%2B+'+%25'|code-block|syntax|javascript|5668276|这样，它将尝试将“1”(房间号)与“2+the+way”进行匹配。而不是反过来。|5668277|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Maybe you should try it the other way around

<pre><code>select Address FROM Addresses WHERE Address NOT LIKE '% ' + HouseNumber + ' %'
</code></pre>

This way it will try to match '1' (housenumber) with '2 the way'. Rather than the other way around.

blocks|key|5184416|text|您必须交换Address和HouseNumber。此外，因为您的Address以HouseNumber开头，所以您可以使您的查询模式正确且可转换。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5184417|select+Address
FROM+++Addresses
WHERE++Address+NOT+LIKE+HouseNumber+%2B+'+%25'|code-block|syntax|javascript|5184418|entityMap^0|5|7|D|B|W|7|14|B|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]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

You must swap <code>Address</code> and <code>HouseNumber</code>. Also, because your <code>Address</code> starts with <code>HouseNumber</code> you can make your query mode correct and also sragable.

<pre><code>select Address
FROM Addresses
WHERE Address NOT LIKE HouseNumber + ' %'
</code></pre>

blocks|key|5184458|text|您需要从地址中提取门牌号，并将其与门牌号列(未测试)进行比较：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5184459|select+Address
from+++Addresses
where++cast(Housenumber+as+int)+!=+cast(left(Address,+instr(Address,+"+")-1)+as+int)|code-block|syntax|javascript|5184460|这假设门牌号始终位于地址的开头，并且门牌号和地址的其余部分之间有一个空格。|5184461|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You need to extract the house number from the address, and check it against the house number column (not tested):

<pre><code>select Address
from Addresses
where cast(Housenumber as int) != cast(left(Address, instr(Address, " ")-1) as int)
</code></pre>

This assumes the house number is always at the beginning of the address, and there is a space between the house number and the rest of the address.

I have some data in a table called Addresses:

<pre><code>HouseNumber Address
32 32 The Street
1 2 The Way
4 4 The Street
5 5 The Street
</code></pre>

I am trying to find the House Numbers that are incorrect. 
I thought the query below would return '2 The Way' but it does not.

<pre><code>select Address
FROM Addresses
WHERE HouseNumber NOT LIKE '%' + Address + '%'
</code></pre>

Is there any easy way to do this like in the query above, or do I have to write a stored procedure?

Variable in LIKE clause

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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我在一个名为Addresses的表中有一些数据：HouseNumber Address32          32 The Street1           2 The Way4           4 The Street5           5 The Street我在找不正确的门牌号码。我原以为下面的查询会返...

问LIKE子句中的变量
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