Finding ASCII values in a STRING variable

[netbumbler]

Hi all,
I'm pulling some data from the active user list table using the openschema function to enumerate the active user list. For example I'm trying to take each entry for 'login_name' and drop it to a text file. The problem is this: I'm trying to create a modified string that includes data from the openschema method with some other data - but when I write the line it adds line-feeds for each reference from the openschema function - THEN - when I do a len(.fields("login_name") it reports a length of 32 even though the visible result is only 'Admin'... there must be some additional characters appended to the end of the string - but my string manipulation skills in VBA are pretty thin. Can anyone help me decipher this? Code snip follows at the bottom.
Thanks,
Chris

set rst1 = cnn1.openschema(adschemaproviderspecific, , _
"{947bb102-5d43-11d1-bdbf-00c04fb92675}")
with rst1
do until .EOF
str3 = .Fields("login_name")
msgbox str3
rem note: will show 'admin'
msgbox len(Str3)
rem note: will show 32
str1 = date & ":" & .fields("login_name") & _
environ("USERNAME")
print #1, str1

.movenext
loop
end with


NOTE: results of the print command will be as follows:

03/04/2004 : Admin
chris

I'm trying to get the result to look like this:

03/04/2003 : Admin chris

 

[LostInCode]

Try just using:

strRoster = rst1.GetString

 

[netbumbler]

Thanks for the reply LostInCode... it pointed me to the right direction... I figured out the results of the openschema is fixed-width... Actually, I ended up using the rst.getrows method. Your idea pointed me to the array process - and it helped me learn a few things as well as realize the total length of the row was 70 characters... there are 4 elements in the array (32 char, 32 char, 4, and 2 lengths respectively)... Which made me realize I could probably use the mid function to pick out the good data. I then ran the rtrim on the first element and it showed the length at 6 (for the admin username instead of 32 as it was showing previously) - then I ran the mid with the len(str)-1 to clear out whatever annoying special character is at the end of the string ... is there a way to find out what that 6th character is? (not sure if it's a cr, crlf, or what...??)
Thanks for your suggestion!
Chris