SQL not returning values that are in a table - datatype issue?

This statement returns all rows:
Select  column12 from mytable  

Column12  datatype is "real"

From this, I pick a random value from in the  rows returned:
for example,    -1.399726E+08  

copy an paste it into an this :
Select  column12 from mytable where column12 =-1.399726E+08

This returns a null,  no row found and returned

Is the query value  I'm seeing in the intial query,  ( -1.399726E+08 ) not exactly what's in the table?

What's going in here?
Thanks
awalkintheparkAsked:
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dsackerContract ERP Admin/ConsultantCommented:
You should try the following:

1. Either, widen the column, so that you can see the entire number.
2. Or, convert it to a character, so you can see the entire number (if #1 does not work).

Select CONVERT(varchar(40), column12) AS column12 from mytable

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Deepak ChauhanSQL Server DBACommented:
This should be work like this as well

Select  column12 from mytable where column12 ='-1.399726E+08'
Scott PletcherSenior DBACommented:
If you need to do equal comparisons against the value, you need to store is as decimal, not as real/float.  For example:

use tempdb

create table t1 (column12 real)
insert into t1 values(-1.399726E+08)
Select  column12 from t1 where column12 =-1.399726E+08 --no match found

create table t2 (column12 decimal(28, 6))
insert into t2 values(-1.399726E+08)
Select  column12 from t2 where column12 =-1.399726E+08 --match found

drop table t1
drop table t2

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awalkintheparkAuthor Commented:
thanks - works.
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