CODEWITHSUNDEEP
sqlsql-servervarcharcontain
Przemysław Niedziela
Przemysław Niedziela

Reputation: 588

check chars in varchar

How can I check that varchar contains all chars from another varchar, where sequence of characters is irrelevant?

For example: I have varchar @a = 'ABC' and column 'Col' in table 'Table' where is row with 'Col' = 'CBAD'. I want to select this row, because it contains all characters from @a variable. Please for your help.

I tried something like that:

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

But I can not put this to WHERE condition

Upvotes: 10

Views: 2752

Answers (4)

Sanjay Bhardwaj
Sanjay Bhardwaj

Reputation: 412

SELECT * FROM yourTable WHERE PATINDEX('%A%',colname) >= 1 and PATINDEX('%B%',colname) >= 1 AND PATINDEX('%C%',colname) >= 1

Upvotes: 1

GarethD
GarethD

Reputation: 69759

Your first need to split your variable that you are checking for into rows, and remove duplicates. For only a few characters you could simply use a table valued constructor:

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)

Which gives:

Letter
----------
D
C
A

Now you can compare this to your column, and limit it only to columns where the column contains all the letters (done in the HAVING clause)

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)
        );

If your variable can be longer than 10 characters, then you may need to adopt a slightly different string splitting method. I would still use numbers to do this, but would instead use 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;

This will give you a set of numbers from 1 to 10,000, and you can simply add more CTE's and cross joins as necessary to extend the process. So with a longer string you might have:

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)
        );

Upvotes: 9

Giorgi Nakeuri
Giorgi Nakeuri

Reputation: 35780

One more version:

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

Upvotes: 2

Rahul Tripathi
Rahul Tripathi

Reputation: 172398

You can try like this:

SELECT * FROM yourTable where colname like '%[A]%' 
                        AND colname like '%[B]%'
                        AND colname like '%[C]%'

or you can try using PATINDEX

SELECT * FROM yourTable WHERE PATINDEX('%[ABC]%',colname) > 1

Upvotes: 5

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