如何检查varchar是否包含来自另一个varchar的所有字符,其中字符序列无关紧要?
例如:我在表@a = 'ABC'
中有varchar 'Col'
和列'Table'
,其中的行是'Col' = 'CBAD'
。我想选择此行,因为它包含@a
变量中的所有字符。请帮助。
我尝试过类似的东西:
DECLARE @a varchar(5) = 'ABCD'
DECLARE @b varchar(5) = 'DCA'
DECLARE @i int = 0
DECLARE @pat varchar(30) = ''
while @i <> len(@b) BEGIN
SET @i = @i + 1
SET @pat = @pat + '[' + @a + ']'
END
SELECT @pat
IF @b LIKE @pat SELECT 1
ELSE SELECT 0
但我不能把它放到WHERE
条件
答案 0 :(得分:9)
您首先需要将要检查的变量拆分为行,并删除重复项。对于只有几个字符,您可以简单地使用表值构造函数:
DECLARE @b varchar(5) = 'DCA';
SELECT DISTINCT Letter = SUBSTRING(@b, n.Number, 1)
FROM (VALUES(1),(2),(3),(4),(5),(6),(7),(8),(9),(10)) AS n (Number)
WHERE n.Number <= LEN(@b)
给出了:
Letter
----------
D
C
A
现在您可以将其与列进行比较,并将其仅限于列包含所有字母的列(在HAVING
子句中完成)
DECLARE @b varchar(5) = 'DCA';
WITH Letters AS
( SELECT DISTINCT Letter = SUBSTRING(@b, n.Number, 1)
FROM (VALUES(1),(2),(3),(4),(5),(6),(7),(8),(9),(10)) AS n (Number)
WHERE n.Number <= LEN(@b)
)
SELECT *
FROM (VALUES ('AA'), ('ABCD'), ('ABCDEFG'), ('CAB'), ('NA')) AS t (Col)
WHERE EXISTS
( SELECT 1
FROM Letters AS l
WHERE t.Col LIKE '%' + l.Letter + '%'
HAVING COUNT(DISTINCT l.Letter) = (SELECT COUNT(*) FROM Letters)
);
如果您的变量可能超过10个字符,那么您可能需要采用稍微不同的字符串拆分方法。我仍然会使用数字来执行此操作,而是使用Itzik Ben-Gan's stacked CTE method:
WITH N1 AS (SELECT N FROM (VALUES(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS n (N)),
N2 (N) AS (SELECT 1 FROM N1 AS N1 CROSS JOIN N1 AS N2),
N3 (N) AS (SELECT 1 FROM N2 AS N1 CROSS JOIN N2 AS N2)
SELECT ROW_NUMBER() OVER(ORDER BY N)
FROM N3;
这将为您提供1到10,000的一组数字,您可以根据需要添加更多CTE和交叉连接以扩展该过程。因此,如果使用更长的字符串,您可能会:
DECLARE @b varchar(5) = 'DCAFGHIJKLMNEOPNFEDACCRADFAE';
WITH N1 AS (SELECT N FROM (VALUES(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS n (N)),
N2 (N) AS (SELECT 1 FROM N1 AS N1 CROSS JOIN N1 AS N2),
N3 (N) AS (SELECT 1 FROM N2 AS N1 CROSS JOIN N2 AS N2),
Numbers (Number) AS (SELECT TOP (LEN(@b)) ROW_NUMBER() OVER(ORDER BY N) FROM N3),
Letters AS (SELECT DISTINCT Letter = SUBSTRING(@b, n.Number, 1) FROM Numbers AS n)
SELECT *
FROM (VALUES ('ABCDDCAFGHIJKLMNEOPNFEDACCRADFAEEFG'), ('CAB'), ('NA')) AS t (Col)
WHERE EXISTS
( SELECT 1
FROM Letters AS l
WHERE t.Col LIKE '%' + l.Letter + '%'
HAVING COUNT(DISTINCT l.Letter) = (SELECT COUNT(*) FROM Letters)
);
答案 1 :(得分:5)
您可以尝试这样:
SELECT * FROM yourTable where colname like '%[A]%'
AND colname like '%[B]%'
AND colname like '%[C]%'
或者您可以尝试使用PATINDEX
SELECT * FROM yourTable WHERE PATINDEX('%[ABC]%',colname) > 1
答案 2 :(得分:2)
还有一个版本:
DECLARE @a varchar(5) = 'ABCD'
DECLARE @b varchar(5) = 'DCA'
;WITH cte AS(
SELECT ROW_NUMBER() OVER( ORDER BY (SELECT NULL)) rn
FROM (VALUES(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) a(n)
CROSS JOIN (VALUES(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) b(n)
),
astring AS(SELECT DISTINCT SUBSTRING(@a, rn, 1) AS l FROM cte WHERE rn <= LEN(@a)),
bstring AS(SELECT DISTINCT SUBSTRING(@b, rn, 1) AS l FROM cte WHERE rn <= LEN(@b))
SELECT CASE WHEN EXISTS(SELECT * FROM bstring WHERE l NOT IN(SELECT * FROM astring))
THEN 0 ELSE 1
END AS result
答案 3 :(得分:1)
SELECT * FROM yourTable WHERE PATINDEX('%A%',colname) >= 1 and PATINDEX('%B%',colname) >= 1 AND PATINDEX('%C%',colname) >= 1