包含逗号分隔的字符串

您必须先拆分字符串，然后使用这些部分来构造带有两个WHERE ... LIKE子句的SQL命令. br/>

You would have to split the string, and use the parts to construct an SQL command with two WHERE ... LIKE clauses

SELECT ... WHERE SerialNo LIKE '%18%' OR SerialNo LIKE '%8AB%'

然后您将不得不执行该SQL命令.

真是一团糟:SQL不能很好地处理字符串，因此这不是一种不错"的方式.这会执行类似的操作:在SQL IN中使用逗号分隔的值参数字符串条款 [ ^ ]，您可以对其进行修改，但是...

诚实地?我会用我的表示语言而不是SQL来做到这一点.

You will then have to EXEC that SQL command.

It''s a mess: SQL is not good at string handling, so this isn''t a "nice" way to do it. This does something similar: Using comma separated value parameter strings in SQL IN clauses[^] and you could modify that, but ...

Honestly? I''d do it in my presentation language, rather than SQL.

在SQL Server服务器端实现此目标的另一种方法是使用名为 ^ ].
有关更多详细信息，请参见:
使用公用表表达式 [与common_table_expression(Transact-SQL)| Microsoft文档 [ OriginalGriff [

Another way to achieve that on SQL server server side is to use recursive queries named Common Table Expressions[^].
For further details, please see:
Using Common Table Expressions[^]
WITH common_table_expression (Transact-SQL) | Microsoft Docs[^]

I agree with OriginalGriff[^] (solution #1) that you should do that on presentation language (see example in c#).

Examples:

DECLARE @data TABLE (SerialNo VARCHAR(30));

INSERT INTO @data (SerialNo)
VALUES('8018A7400BC3'),
('8018A79B6433'),
('001DD55C6FAB'),
('HDE09B356634'),
('4650C7453C90'),
('3275D8AB544D'),
('4345D8AB544C');


DECLARE @find VARCHAR(30) = '18,8AB';

;WITH CTE AS 
(
	SELECT LEFT(@find, CHARINDEX(',', @find)-1) AS SerialPart, RIGHT(@find, LEN(@find) - CHARINDEX(',', @find)) AS Remainder
	WHERE CHARINDEX(',', @find)>0
	UNION ALL
	SELECT LEFT(Remainder, CHARINDEX(',', Remainder)-1) AS SerialPart, RIGHT(Remainder, LEN(Remainder) - CHARINDEX(',', Remainder)) AS Remainder
	FROM CTE
	WHERE CHARINDEX(',', Remainder)>0
	UNION ALL
	SELECT Remainder AS SerialPart, NULL AS Remainder
	FROM CTE
	WHERE CHARINDEX(',', Remainder)=0 
)
SELECT Orig.*
FROM @data AS Orig INNER JOIN CTE AS Parts ON Orig.SerialNo Like '%' + Parts.SerialPart + '%';

//create sample data
DataTable dt = new DataTable();
dt.Columns.Add(new DataColumn("SerialNo", typeof(string)));
dt.Rows.Add(new object[]{"8018A7400BC3"});
dt.Rows.Add(new object[]{"8018A79B6433"});
dt.Rows.Add(new object[]{"001DD55C6FAB"});
dt.Rows.Add(new object[]{"HDE09B356634"});
dt.Rows.Add(new object[]{"4650C7453C90"});
dt.Rows.Add(new object[]{"3275D8AB544D"});
dt.Rows.Add(new object[]{"4345D8AB544C"});

string find = @"18,8AB";

var result = dt.AsEnumerable()
	.Where(x => find.Split(new string[]{","}, StringSplitOptions.RemoveEmptyEntries).Any(y=> x.Field<string>("SerialNo").Contains(y)))
	.ToList();
	
foreach(DataRow dr in result)
{
	Console.WriteLine("{0}", dr.Field<string>("SerialNo"));
}

祝你好运！

Good luck!