Something is just not right and I can not figure it out, I need help

the below is a line from a Do Loop routine I have and I'm trying to
check how any characters there are after the decimal point. The CUnits
field is defined as DataType=Number, FieldSize=Double, Format=Fixed,
Decimals=2.
On the input form first record for CUnits I enter .01
on the second record for CUnits I enter .1

if I open the table I see that the entered .01 is stored as 0.01 and
the entered .1 is stored as 0.10

in Debug mode
for the first record
strLen = InStrRev(MyRec!CUnits, ".")

if I hold the cursor over CUnits it shows -> .01 and strLen=2

movenext, loop

if I hold the cursor over CUnits for the second record it shows -> 0.1
and strLen=2

which is what I don't understand,
first, why is the debug mode display different of the two records?
second, why is strLen 2 characters for this second record?
help, thanks
bobh.

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Man, don't worry about it. You can display the number in any format
using the Format() function and, if in a Control, the Format property:

If number = .01 (actually stored as 0.01):

Format(number, "#.##") = 0.01
Format(number, ".##") = .01

Format property: ".##" will display as above.

If the number = .1 (actually stored as 0.10):

Format(number, "#.##") = 0.10
Format(number, ".#") = .1

HTH,
--
MGFoster:::mgf00 <at> earthlink <decimal-point> net
Oakland, CA (USA)
** Respond only to this newsgroup. I DO NOT respond to emails **

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bobh wrote:

On Jan 23, 4:35*pm, MGFoster <m... (AT) privacy (DOT) com> wrote:

I wish I didn't have to worry about it but I do because this line is
not evaluating correctly or as how I think it should.

if I enter .1 as a value this line in my code will evaluate to a
"2" in the strLen and it should be "1" so the if statement does not
evaluate correctly
strLen = InStrRev(MyRec!CUnits, ".")
if strLen = 1 then
tmp1 = InStrRev(MyRec!CUnits, ".")

I need to know how many characters after the decimal point because if
I were to add a whole number to the decimal part I need it to work out
correctly.
example: a whole number say - 60 + InStrRev(MyRec!CUnits, ".") would
equal 61 and I need it to equal 70 since .1 equals 10 so, in the
above case if the InStrRev would calculate correctly then I could

strLen = InStrRev(MyRec!CUnits, ".")
if strLen = 1 then
tmp1 = InStrRev(MyRec!CUnits, ".") & 0
then tmp1 would have 10 and adding that to a value of 60 would equal
70
bobh.

bobh wrote:

Could you use Cstr()
Num = 123.4567
? Num
123.4567
Num = 12.34
? Instr(cstr(num),".")
3
Num = 1
? Instr(cstr(num),".")
0


Expanding this
Num = 123.4567
? num
123.4567
? len(Cstr(num)) -Instr(cstr(num),".")
4
Num = 12.345
? len(Cstr(num)) -Instr(cstr(num),".")
3


What if the number is an integer
Num = 1
? len(Cstr(num)) -Instr(cstr(num),".")
1
? len(Cstr(num)) - IIF(Instr(cstr(num),".") > 0 , Instr(cstr(num),"."),
Len(cstr(num)))
0

Would this help?

On Jan 26, 1:10*pm, Salad <o... (AT) vinegar (DOT) com> wrote:
Hi,
As you can imagine I've been working on this for a few days know and I
think I have what I need below. In all my debug mode testing it seems
to be calculating the right values for each user entered value, as in
if CUnits is .1 or .10 or 1 or 3.7 or 4.01 etc....

Else
'calculate the whole hours number
tmp1 = Left(MyRec!CUnits, InStr(1, MyRec!CUnits, ".") - 1) *
60
'calculate the mins and then add the two numbers and update the
field
strLen = Len(Mid(MyRec!CUnits, InStr(1, MyRec!CUnits, ".") +
1))
If strLen = 1 Then
tmp2 = Mid(MyRec!CUnits, InStr(1, MyRec!CUnits, ".") + 1)
tmp2 = tmp2 & 0
ElseIf strLen = 2 Then
tmp2 = Right(MyRec!CUnits, InStrRev(MyRec!CUnits, "."))
End If
MyRec.Edit
MyRec!TotMins = tmp1 + tmp2
MyRec.Update
End If
MyRec.MoveNext
Loop

bobh.