Help with Generating Random ID

[Chris]

Hi,
I need to generate IDs for entries and I would Like to have it in the
format

MMDDYYYHHMMTTMS-RANDOM NUMBER

I am using the following to generate the random number

Threading.Thread.Sleep(1)
Dim rnd As New Random((CInt(DateTime.Now.Millisecond)))
GenerateTransactionID = rnd.Next().ToString()
Return GenerateTransactionID

How can I get the datetime in the format MMDDYYYHHMMTTMS?

Thanks

[Siva M]

Assuming MS is millisecond, try this:

DateTime dt = DateTime.Now;
String rnd = dt.ToString ("MMddyyyyHHmmtt") + dt.Millisecond.ToString());

Note that the millisecond part may have one, two or three digits based on
its value.

"Chris" <de*****@gmail.com> wrote in message
news:11**********************@g47g2000cwa.googlegr oups.com...
Hi,
I need to generate IDs for entries and I would Like to have it in the
format

MMDDYYYHHMMTTMS-RANDOM NUMBER

I am using the following to generate the random number

Threading.Thread.Sleep(1)
Dim rnd As New Random((CInt(DateTime.Now.Millisecond)))
GenerateTransactionID = rnd.Next().ToString()
Return GenerateTransactionID

How can I get the datetime in the format MMDDYYYHHMMTTMS?

Thanks

[rviray]

There some overloads on the the "ToString" function, one allows you to
format the DateTime string...so it would be

dt.ToString(string format)...

try that...
--
rviray
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