blocks|key|120198|text|将值放入表中，并使用join而不是in子句：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|120199|SELECT+*+FROM+emailaddresses+as+ea
INNER+JOIN+addresses+as+a
ON+ea.address+like+'%25'+%2B+a.address+%2B+'%25'|code-block|syntax|javascript|120200|entityMap^0|A|4|H|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

put the values into a table and use a <code>join</code> rather than an <code>in</code> clause:

<pre><code>SELECT * FROM emailaddresses as ea
INNER JOIN addresses as a
ON ea.address like '%' + a.address + '%'
</code></pre>

blocks|key|1737888|text|您可以使用LOWER+function|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1737889|SELECT+*+FROM+emailaddresses+where+LOWER(addr)+in+('john@google.com',+'jim@google.com')|code-block|syntax|javascript|1737890|它会将所有addr转换为小写，这样您就可以将结果与您想要的结果进行比较。|1737891|entityMap|0|LINK|mutability|MUTABLE|url|http://www.sqlinfo.net/sqlserver/sql_server_function_upper_lower.php^0|5|E|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

You can use the <a href="http://www.sqlinfo.net/sqlserver/sql_server_function_upper_lower.php" rel="nofollow">LOWER function</a>

<pre><code>SELECT * FROM emailaddresses where LOWER(addr) in ('john@google.com', 'jim@google.com')
</code></pre>

Which will convert all addr to lowercase, in which you can then compare the results to what you want.

blocks|key|2225141|text|请注意，LIKE将区分大小写还是不区分大小写，这取决于对表达式有效的排序规则，但在本例中，您没有指定通配符，因此使用LIKE的意义不大。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2225142|默认的SQL+Server安装不区分大小写。|2225143|如果您想要不区分大小写的比较，因为您有一个区分大小写的数据库，you+can+cast。我相信这是合适的语法(尽管我从来没有在表达式的一侧使用过它作为IN列表)。|2225144|SELECT+*
FROM+emailaddresses
WHERE+addr+COLLATE+SQL_Latin1_General_CP1_CI_AS+
IN+(
++++'john@google.com'+COLLATE+SQL_Latin1_General_CP1_CI_AS
++++,'jim@google.com'+COLLATE+SQL_Latin1_General_CP1_CI_AS
)|code-block|syntax|javascript|2225145|像addr+LIKE+'%25@google.com"这样的东西才是LIKE的真实案例|2225146|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/ms184391.aspx^0|4|4|1M|4|0|0|V|C|0|0|0|1|O|W|4|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@]|D|@$9|15|A|16|1|17]]|E|$]]|$1|J|3|K|5|L|7|18|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

Note that <code>LIKE</code> will work either case-sensitively or case-insensitively depending upon which collation is in effect for the expression, but in your case, you have specified no wildcards so there is little point looking to use <code>LIKE</code>.

The default SQL Server installation is case-insensitive.

If you want a case-insensitive compare because you've got a case-sensitive database, <a href="http://msdn.microsoft.com/en-us/library/ms184391.aspx" rel="nofollow">you can cast</a>. I believe this is the appropriate syntax (I've never used it for an IN list on one side of an expression, though).

<pre><code>SELECT *
FROM emailaddresses
WHERE addr COLLATE SQL_Latin1_General_CP1_CI_AS 
IN (
 'john@google.com' COLLATE SQL_Latin1_General_CP1_CI_AS
 ,'jim@google.com' COLLATE SQL_Latin1_General_CP1_CI_AS
)
</code></pre>

A real case for <code>LIKE</code> would be something for something like <code>addr LIKE '%@google.com"</code>

blocks|key|1738034|text|使用包含要搜索的电子邮件列表的表的交叉联接来尝试此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1738035|declare+@email+table(
email_check+nvarchar(500)+not+null)

insert+into+@email(email_check)
values('jack@google.com')
insert+into+@email(email_check)
values('john.@google.com')



select+hit,+ID,+EMAIL_ADDRESS+from+(
++++select+CHARINDEX(email_check,+lower(EMAIL_ADDRESS))+hit,+ID,+EMAIL_ADDRESS
++++from+Table_With_Email_Addresses
++++,@email
)+t
where+hit+>+0|code-block|syntax|javascript|1738036|不需要"like“，因为它会解析一个字符串来找到匹配项。干杯!|1738037|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|$]]

Try this using a cross join to a table containing a list of email's you want to search for:

<pre><code>declare @email table(
email_check nvarchar(500) not null)

insert into @email(email_check)
values('jack@google.com')
insert into @email(email_check)
values('john.@google.com')



select hit, ID, EMAIL_ADDRESS from (
 select CHARINDEX(email_check, lower(EMAIL_ADDRESS)) hit, ID, EMAIL_ADDRESS
 from Table_With_Email_Addresses
 ,@email
) t
where hit &gt; 0
</code></pre>

No need for a "like" since it will parse a string to find a match.
Cheers!

blocks|key|3810626|text|select+*+from+HotelAmenities_+
where+Pamenities+in+(
++++++++++++select+distinct(pamenities)+
++++++++++++from+HotelAmenities_+
++++++++++++where+pamenities+like+'%25Swimming%25'
++++++++)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3810627|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 * from HotelAmenities_ 
where Pamenities in (
 select distinct(pamenities) 
 from HotelAmenities_ 
 where pamenities like '%Swimming%'
 )
</code></pre>

blocks|key|3810643|text|我们可以在SQL中一起使用“like+-+in”方法，但风格略有不同，如下图所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3810644|SELECT+*+
FROM+emailaddresses+
WHERE+addr+LIKE+'john@google.com'+OR+addr+LIKE+'jim@google.com'|code-block|syntax|javascript|3810645|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|$]]

We can use the 'LIKE-In' approach together in SQL but in somewhat different style, like the one below:

<pre><code>SELECT * 
FROM emailaddresses 
WHERE addr LIKE 'john@google.com' OR addr LIKE 'jim@google.com'
</code></pre>

I want to do a soft string match in a table, like this:

<pre><code>SELECT * FROM emailaddresses where addr in ('john@google.com', 'jim@google.com')
</code></pre>

But if there is an addr value in the table 'JOHN@google.com', I want that returned.

Sort of like this:

<pre><code>SELECT * FROM emailaddresses where addr LIKE in ('john@google.com', 'jim@google.com')
</code></pre>

How do I do that?

Using 'LIKE' with an 'IN' clause full of strings

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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我想在表中做一个软字符串匹配，如下所示：SELECT * FROM emailaddresses where addr in ('john@google.com', 'jim@google.com')但是，如果表'JOHN@google.com‘中有addr值，我希望返回该值。有点像这样：SELECT * FROM emailaddresses where addr LIKE in ('john@

问将“LIKE”与充满字符串的“IN”子句一起使用
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