Display tip on home page by day number in asp.net

Scenario:.I have 30 tips in xml file, now I want to display it on home page by day number.

For example, today is the 12th then tip number 12 will be displayed开发者_如何学运维 on the home page and after that 13 and so on. After 30 tips it will start from tip 1.

I have tried following code

public class TipReader
{
    private const string TIP_XML_PATH = @"C:\temp\tips.xml";
    public static int GetTipNumber()
    {
        int tipNo = 1;
        if (DateTime.Now.Day >0)
        {
            if (DateTime.Now.Day == 31)
            {
                tipNo = 1;
            }
            else if (DateTime.Now.Day == 28)
            {
                tipNo = DateTime.Now.Day + 1;
            }
            else
            {
                tipNo = DateTime.Now.Day;
            }
        }
        return tipNo;
    }
    public IEnumerable<Tip> GetTipByID(string ID)
    {
        var result = from tip in XDocument.Load(TIP_XML_PATH).Descendants("Tip")
                     where (string)tip.Element("ID")==ID
                     select new Tip
                     {
                         ImageUrl = (string)tip.Element("ImageUrl"),
                         Title = (string)tip.Element("Title").Value,
                         Desciption = (string)tip.Element("Description"),
                     };
        return result;
    }
}

Now, I am facing issue on GetTipNumber. How would I handle date greater than 30 and less than 30. If you have better solution for this please suggest me. Thanks in advance

---

What is the harm in doing it in this fashion

public static int GetTipNumber()
{
     int tipNo = 1;
     if (DateTime.Now.Day > 0 && DateTime.Now.Day <= 30)
     {
         tipNo = DateTime.Now.Day;
     }
     else 
     {
         tipNo = 1; // 
     }                

     return tipNo;
}

---

int tipNo = DateTime.Now.Day > 30 ? 1 : DateTime.Now.Day;

---

Another option is to display a random tip each time you view the page:

public static int GetTipNumber()
{
   Random rand = new Random();
   return rand.Next(1, 30);
}