我怎样才能在 LIKE 运算符中引入多个条件？

Here is an alternative way:

select * from tbl where col like 'ABC%'
union
select * from tbl where col like 'XYZ%'
union
select * from tbl where col like 'PQR%';

Here is the test code to verify:

create table tbl (col varchar(255));
insert into tbl (col) values ('ABCDEFG'), ('HIJKLMNO'), ('PQRSTUVW'), ('XYZ');
select * from tbl where col like 'ABC%'
union
select * from tbl where col like 'XYZ%'
union
select * from tbl where col like 'PQR%';
+----------+
| col      |
+----------+
| ABCDEFG  |
| XYZ      |
| PQRSTUVW |
+----------+
3 rows in set (0.00 sec)

This is a good use of a temporary table.

CREATE TEMPORARY TABLE patterns (
pattern VARCHAR(20)
);


INSERT INTO patterns VALUES ('ABC%'), ('XYZ%'), ('PQR%');


SELECT t.* FROM tbl t JOIN patterns p ON (t.col LIKE p.pattern);

In the example patterns, there's no way col could match more than one pattern, so you can be sure you'll see each row of tbl at most once in the result. But if your patterns are such that col could match more than one, you should use the DISTINCT query modifier.

SELECT DISTINCT t.* FROM tbl t JOIN patterns p ON (t.col LIKE p.pattern);

This might help:

select * from tbl where col like '[ABC-XYZ-PQR]%'

I've used this in SQL Server 2005 and it worked.

select * from tbl where col like 'ABC%'
or col like 'XYZ%'
or col like 'PQR%';

This works in toad and powerbuilder. Don't know about the rest

Oracle 10g has functions that allow the use of POSIX-compliant regular expressions in SQL:

• REGEXP_LIKE
• REGEXP_REPLACE
• REGEXP_INSTR
• REGEXP_SUBSTR

See the Oracle Database SQL Reference for syntax details on this functions.

Take a look at Regular expressions in Perl with examples.

Code :

    select * from tbl where regexp_like(col, '^(ABC|XYZ|PQR)');

I also had the same requirement where I didn't have choice to pass like operator multiple times by either doing an OR or writing union query.

This worked for me in Oracle 11g:


REGEXP_LIKE (column, 'ABC.*|XYZ.*|PQR.*');

Even u can try this

Function

CREATE  FUNCTION [dbo].[fn_Split](@text varchar(8000), @delimiter varchar(20))
RETURNS @Strings TABLE
(
position int IDENTITY PRIMARY KEY,
value varchar(8000)
)
AS
BEGIN


DECLARE @index int
SET @index = -1


WHILE (LEN(@text) > 0)
BEGIN
SET @index = CHARINDEX(@delimiter , @text)
IF (@index = 0) AND (LEN(@text) > 0)
BEGIN
INSERT INTO @Strings VALUES (@text)
BREAK
END
IF (@index > 1)
BEGIN
INSERT INTO @Strings VALUES (LEFT(@text, @index - 1))
SET @text = RIGHT(@text, (LEN(@text) - @index))
END
ELSE
SET @text = RIGHT(@text, (LEN(@text) - @index))
END
RETURN
END

Query

select * from my_table inner join (select value from fn_split('ABC,MOP',','))
as split_table on my_table.column_name like '%'+split_table.value+'%';

If your parameter value is not fixed or your value can be null based on business you can try the following approach.

DECLARE @DrugClassstring VARCHAR(MAX);
SET @DrugClassstring = 'C3,C2'; -- You can pass null also


---------------------------------------------


IF @DrugClassstring IS NULL
SET @DrugClassstring = 'C3,C2,C4,C5,RX,OT'; -- If null you can set your all conditional case that will return for all
SELECT dn.drugclass_FK , dn.cdrugname
FROM drugname AS dn
INNER JOIN dbo.SplitString(@DrugClassstring, ',') class ON dn.drugclass_FK = class.[Name] -- SplitString is a a function

SplitString function

SET ANSI_NULLS ON;
GO
SET QUOTED_IDENTIFIER ON;
GO
ALTER FUNCTION [dbo].[SplitString](@stringToSplit VARCHAR(MAX),
@delimeter     CHAR(1)      = ',')
RETURNS @returnList TABLE([Name] [NVARCHAR](500))
AS
BEGIN


--It's use in report sql, before any change concern to everyone


DECLARE @name NVARCHAR(255);
DECLARE @pos INT;
WHILE CHARINDEX(@delimeter, @stringToSplit) > 0
BEGIN
SELECT @pos = CHARINDEX(@delimeter, @stringToSplit);
SELECT @name = SUBSTRING(@stringToSplit, 1, @pos-1);
INSERT INTO @returnList
SELECT @name;
SELECT @stringToSplit = SUBSTRING(@stringToSplit, @pos+1, LEN(@stringToSplit)-@pos);
END;
INSERT INTO @returnList
SELECT @stringToSplit;
RETURN;
END;

SELECT * From tbl WHERE col LIKE '[0-9,a-z]%';

simply use this condition of like in sql and you will get your desired answer

I had to add all to Asaph's answer to make it work.

select * from tbl where col like 'ABC%'
union all
select * from tbl where col like 'XYZ%'
union all
select * from tbl where col like 'PQR%';