generating numbers in sequence

Question

Hi I want to generate numbers automatically in sequence that is from 1 to 100 in c# windows application can any one help me how to generate numbers. Thanks, saikumar

Suraj Kumar · Answer

You can try this int [] noOfSeries = Enumerable.Range(1, 100).ToArray(); Hope this will help you.

cheguri saikumar · Answer

namespace DocSort_CPA.Forms { public partial class TicketScreen : Form { public TicketScreen() { InitializeComponent(); } public int ticketNo = 0; public string orderedTicketNum = string.Empty; private void btnSubmit_Click_1(object sender, EventArgs e) { { string From = string.Empty; string FromPWD = string.Empty; string ToAddress = string.Empty; string BCCAddress = string.Empty; string Name = string.Empty; string Phone = string.Empty; string Email = string.Empty; string Comments = string.Empty; string ticket = string.Empty; string MessageBody = string.Empty; try { ConfirmLicenseManager objConfirmLicenseManager = new ConfirmLicenseManager(); DocSortResult getAllConfigValues = objConfirmLicenseManager.GetAllConfigValues(); if (getAllConfigValues.resultDS != null && getAllConfigValues.resultDS.Tables[0].Rows.Count > 0) { foreach (DataRow dr in getAllConfigValues.resultDS.Tables[0].Rows) { switch (Convert.ToString(dr["Config_Name"].ToString()).ToUpper()) { case "FROMEMAIL": From = (this.Decrypt(dr["Config_Value"].ToString())); break; case "FROMPASSWORD": FromPWD = (this.Decrypt(dr["Config_Value"].ToString())); break; case "TOADDRESS": ToAddress = dr["Config_Value"].ToString(); break; case "BCCADDRESS": BCCAddress = dr["Config_Value"].ToString(); break; } } } { txtticket.Text = ticketNo.ToString(); ticketNo = Convert.ToInt32(txtticket.Text)+1; MessageBox.Show("Your ticket has been created." ,ticketNo.ToString()); txtticket.Text = ticketNo.ToString(); orderedTicketNum = orderedTicketNum + ","; }

Jaish Mathews · Answer

I love algorithms and any custom logics better to have in a separate class like below. Whenever you need, you call as Sequence.Next. This starts from 1 and up to 100, then again start from 1. You can change the logic in below class itself. static class Sequence { private static int _value = 0; private static readonly object m_lock = new object (); public static int Next { get { lock (m_lock) { if (_value == 100) _value = 0; return ++_value; } } } }

Sreekanth Reddy · Answer

Try This please : namespace yourapp : Form { public int ticketNo =0; public string orderedTicketNum = string.Empty; private void button1_Click(object sender, System.EventArgs e) { txtticket.Text = ticketNo.ToString(); } private void button1_Click(object sender, System.EventArgs e) { ticketNo = Convert.ToInt32(txtticket.Text)+ 1; MessageBox.Show("Your ticket has been created." ticketNo.ToString()); txtticket.Text = ticketNo.ToString(); orderedTicketNum = orderedTicketNum+","; } }

Sreekanth Reddy · Answer

This helps you to get the ordered ticket Numbers separated by comma into a separate string "orderedTicketNum": namespace yourapp : Form { public int ticketNo =1; public string orderedTicketNum = "1"; private void button1_Click(object sender, System.EventArgs e) { } private void button1_Click(object sender, System.EventArgs e) { MessageBox.Show("Your ticket has been created." ticketNo.ToString()); txtticket.Text = ticketNo.ToString(); ticketNo = ticketNo + 1; orderedTicketNum = orderedTicketNum+","; } } Mark it as answer, If it helps else come with query. Thank you.

cheguri saikumar · Answer

Thanks it is working but the thing is Iam unbale to generate numbers in order Thanks,

Sreekanth Reddy · Answer

If you want to generate numbers in random dont use "Random" class. Because "Random" class is used to a random value witin specified range. Directly do with For loop which I had provided in my previous answers. namespace yourapp : Form { public int ticketNo =1; private void button1_Click(object sender, System.EventArgs e) { } private void button1_Click(object sender, System.EventArgs e) { MessageBox.Show("Your ticket has been created." ticketNo.ToString()); txtticket.Text = ticketNo.ToString(); ticketNo = ticketNo + 1; } }

Sanwar Ranwa · Answer

Hello cheguri Please check these Links https://www.youtube.com/watch?v=gog3_xnf86M https://www.youtube.com/watch?v=QnbIfuQV8S4 https://www.codeproject.com/Questions/550351/HowplustoplusGenerateplusAuto-Incrementplusplusnum https://stackoverflow.com/questions/26931528/random-number-generator-with-no-duplicates

cheguri saikumar · Answer

I want to generate numbers automatically and put those numbers in textbox control. For example i have generated random numbers but i am unable to generate numbers in ascending order here is my random number generator code Random r = new Random(); int i; i = r.Next(1, 100); MessageBox.Show("Your ticket has been created"., ,i.ToString()); txtticket.Text = i.ToString(); But i want to generate as 1,2,3....98,99,100.

Sreekanth Reddy · Answer

This works for your requirement : private void button1_Click(object sender, System.EventArgs e) { // On load of form displaying 0 in textbox txtnum.Text = "0"; } private void button1_Click(object sender, System.EventArgs e) { //Get the value from the text box int num = Convert.Toint32(txtnum.Text); txtnum.Text = num.ToString(); } Mark it as answer, If it helps. Else come up with next query.

cheguri saikumar · Answer

Can you please send me the full code i am new to .net c#. Thanks,

Sreekanth Reddy · Answer

Then the logic is like I stated in my previous answer : for(int i =1; i { // "i" contains number. It loops upto 100 }

cheguri saikumar · Answer

I want to print in text box control for example when i click ok button it has to show 1 and again when i click on ok button it has to show 2 and so on.... Thanks,

cheguri saikumar · Answer

I dont want to print just i want to generate numbers

Sreekanth Reddy · Answer

Where you needs to print in windows application(on which control) ? Any how the logic is : for(int i=0; i { }