SQL Repeating Character String

Question

Is there a way to evaluate a field if it has repeating values in a cell. For example if someone holds down a number key and it returns some variation of \'00000\' or \'222222222

Answer 1

As you are only interested in complete repeatings, such as '3333' and not in, say, '43333', this is rather simple: Find strings longer than one character, where you end up with an empty string when you remove all characters that equal the first one:

select *
from mytable 
where len(value) > 1 and len(replace(value, left(value,1), '')) = 0

Answer 2

The following finds patterns i.e. 333... or 123... or 987...

Think of it like Rummy 500... Runs and groups of 3's or more.

Declare @Table table (col int)
Insert into @Table values
(4141243),(4290577),(98765432),(78635389),(4141243),(22222),(4290046),(55555555),(4141243),(6789),(77777),(45678),(4294461),(55555),(4141243),(5555)

Declare @Num table (Num int);Insert Into @Num values (0),(1),(2),(3),(4),(5),(6),(7),(8),(9)

Select Distinct A.*
  From @Table A
  Join (
        Select Patt=replicate(Num,3) from @Num
        Union All
        Select Patt=right('000'+cast((Num*100+Num*10+Num)+12 as varchar(5)),3) from @Num where Num<8
        Union All
        Select Patt=reverse(right('000'+cast((Num*100+Num*10+Num)+12 as varchar(5)),3)) from @Num where Num<8
       ) B on CharIndex(Patt,cast(col as varchar(25)))>0

Returns

col
5555
6789
22222
45678
55555
77777
55555555
98765432

Now, If you don't care to identify "runs" (123...)", just remove the following:

    Union All
    Select Patt=right('000'+cast((Num*100+Num*10+Num)+12 as varchar(5)),3) from @Num where Num<8
    Union All
    Select Patt=reverse(right('000'+cast((Num*100+Num*10+Num)+12 as varchar(5)),3)) from @Num where Num<8