Database

-- return 90% of a string
declare @s varchar(6)
set @s = '123456'
select substring(@s, 1, cast(floor(0.9 * len(@s)) as int))

-- usage
select * 
from [table] 
where charindex(substring(@s, 1, cast(floor(0.9 * len(@s)) as int)), [field]) > 0

RyanEK wrote:

-- return 90% of a string declare @s varchar(6)
set @s = '123456' select substring(@s, 1, cast(floor(0.9 * len(@s)) as int))

-- usage
select *
from [table]
where charindex(substring(@s, 1, cast(floor(0.9 * len(@s)) as int)), [field]) > 0

<br />
-- place each character in a temp table<br />
declare @temp table (ch char)<br />
insert into @temp values ('a')<br />
insert into @temp values ('b')<br />
insert into @temp values ('c')<br />
insert into @temp values ('d')<br />
insert into @temp values ('e')<br />
insert into @temp values ('f')<br />
<br />
-- some test data<br />
declare @temp2 table (string varchar(10))<br />
insert into @temp2 values ('Sabaschede')<br />
insert into @temp2 values ('aaaaaaaaaa')<br />
insert into @temp2 values ('a1b2c3d4e5')<br />
<br />
-- use a cross join to determine how many times each character appears in the string field<br />
-- you just need to determine that 90% of 'abcdef' is 5 characters<br />
select string<br />
from (<br />
  select ch, string, case when charindex(ch,string) > 0 then 1 else 0 end chrcount<br />
  from @temp<br />
  cross join @temp2<br />
) as t<br />
group by string<br />
having sum(chrcount) >= 5<br />

bgates1970 wrote:

My depth is 5.

bgates1970 wrote:

I have gotten what I expected out of this forum

bgates1970 wrote:

I have gotten what I expected out of this forum... smart a$$ comments and nothing really substantive.

Luc Pattyn wrote:

Unless you forget about version numbers until you actually need them