How to generate random numbers in timestamp format

Question

In my test I have to fill the application number for that I already have guid method to generate random numbers because for each time while running the test it should generate different numbers.

var DocumentNo = Guid.NewGuid().ToString();
editCasePage.FillDocumentNumber(DocumentNo);

but I need document number in time stamp format.How can i do that? Thanks in advance!

Your question is unclear. Please clarify what you want. Are you saying you want the document number to be an ISO8601 timestamp? Would `DateTime.UtcNow.ToString("o")` not suffice? — ProgrammingLlama, Jan 10 '18 at 10:49
@Mutu Arron a random number will give you, ex. 10, so this number will represent what, your question is unclear — styx, Jan 10 '18 at 10:51
Please provide an example of your input and what your expected output. — MakePeaceGreatAgain, Jan 10 '18 at 10:51
1. `DocumentNo` is already a string. No point of calling it's `ToString()` method. 2. What is a *time stamp format*? You can simply generate a random number (using in instance of the `Random` class) and add it to `DateTime.Now` and then use `ToString("yyyy-MM-dd HH:mm:ss")` — Zohar Peled, Jan 10 '18 at 10:51
If you already have a Guid, why not just use it instead of messing with some other format? — DavidG, Jan 10 '18 at 10:52
@john Exactly what i am looking for i need an alternative option for guid. — Mutu Arron, Jan 10 '18 at 10:57
I can´t imagine how a guid like `"F9168C5E-CEB2-4faa-B6BF-329BF39FA1E4"` should be transofmred to a timestamp or a date at all. — MakePeaceGreatAgain, Jan 10 '18 at 10:59
Possible duplicate of [Random date in C#](https://stackoverflow.com/questions/194863/random-date-in-c-sharp) — JeffC, Jan 11 '18 at 21:43

Oliver · Accepted Answer · 2018-01-10T11:19:24.823

1

If you need a random date then create one. One simple solution would be:

using System;

public class Program
{
    public static void Main()
    {
        for(int i = 0; i <= 100; i++)
        {
            var guid = Guid.NewGuid();
            var bytes = guid.ToByteArray();
            var rawValue = BitConverter.ToInt64(bytes, 0);
            var inRangeValue = Math.Abs(rawValue) % DateTime.MaxValue.Ticks;
            var date = new DateTime(inRangeValue);

            Console.WriteLine(date);
        }
    }
}

edited Jan 10 '18 at 11:19

answered Jan 10 '18 at 10:58

Oliver

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It's worth noting that this method does not guarantee unique results in any way. using a GUID [practically guarantees uniqueness.](https://stackoverflow.com/questions/1155008/how-unique-is-uuid) (though not completely guarantees it). – Zohar Peled Jan 10 '18 at 11:03
A guid has 128 bit and a datetime has 64 bit. So you always have duplicates when you try to convert guids to datetimes. – Oliver Jan 10 '18 at 11:07
While your comment is correct, I didn't say *convert guids to datetimes.* I only pointed out the fact that `Guid.NewGuid()` would return a unique value, while `random.Next` does not guarantee it in any way. – Zohar Peled Jan 10 '18 at 11:10
That's right. Just to get more uniqueness, I changed my answer to convert a guid (partially) into a DateTime. ;-) – Oliver Jan 10 '18 at 11:19

How to generate random numbers in timestamp format

1 Answers1