Someone in the forums asked for finding out a string in which each character is same. Consider the following example

declare @t table(data varchar(20))
insert into @t
select '222222222222' as data union all
select '666666663466' union all
select 'aaaaaaaa'+CHAR(32) union all
select 'jjkjkhdg' union all
select '0----------' union all
select '9999999999999' union all
select ']##########' union all
select ']    ' union all
select ']]]]]]]]]]]]]'

As you see only three values have all same characters. You can use many methods to find out this but the simplest method is

select data from @t
where PATINDEX('%[^'+left(replace(data,']',char(0)),1)+']%',replace(data,']',char(0)))=0

The result is

data
--------------------
222222222222
9999999999999
##########

The logic is to see if there is a character which is different than the first character. If there is different character the patindex function will return a value greater than 0 otherwise all characters are same and it will return 0

Edit : As per the comment by Ramesh Saive, the code does not work well if the first character is ] (a closing sqaure bracket). So I have modified the code to handle that too