want to generate random no. through coding in some specific pattern.

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hi, I am working in C#.net and taking ms-access as backend for windows application. I want to auto generate one number in a particular pattern(eg. AB/2014/2/2).
Here AB: will be some fixed string
2014: present year
2:month
5:auto generate no.(will increment after submission of form)

whenever some one will fill the form and will click on submit button then this no. has to be generated. last digit in this no. is sequence no. which will be incremented after checking previous no.
this no. will act as a primary key in database.

can somebody help me out with this? I am not getting how to code this.....

Posted 4-Feb-14 23:26pm

Diya Bh.

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Karthik_Mahalingam · Answer 1 · 2014-02-04T23:40:00

Try like this,
do some little homework..

C#

public string GenerateNext(string text)
   {
       string check = string.Format("{0}/{1}/{2}/",text,DateTime.Now.Year,DateTime.Now.Month);
       // query the DB to filter the primary key by passing the 'check' as Start with 'Like' Search
       // and return only the primary key values in a list of string or datatable
       List<string> lstValues = new List<string>();
      // lstValues =   GetDBValues(check); // you need to write logic for this..
       if (lstValues.Count == 0)
           return check + "0";
       else
       {
          int nextID =  lstValues.Select(k=>  k.Replace(check,"")).Select(k0=> Convert.ToInt32(k0)).Max() + 1;
           return check + nextID.ToString();
       }


   }

OriginalGriff · Answer 2 · 2014-02-04T23:45:00

Use an identify value in the database, and a computed column to generate the whole item.
Assuming you have an Identity field called Ident, a fixed text field call Prefix, and a DateTime field called InsertDate:

SQL

[ID]  AS (((((([Prefix]+'/')+CONVERT([varchar],datepart(year,[InsertDate]),0))+'/')+CONVERT([varchar],datepart(month,[InsertDate]),0))+'/')+CONVERT([varchar],[Ident],0))

Pikoh · Answer 3 · 2014-02-04T23:50:00

Solution 4

Something like this? (i'm using a textbox just to try)

C#

string number = "";
string[] split = this.textBox1.Text.Split('/'); //I suppose here you'll get the last generated number from you database
number += split[0] + "/";

if (split[1] == DateTime.Now.Year.ToString("0000") && split[2] == DateTime.Now.Month.ToString("0"))
{
      //if actual year/month is the same as last generated number,generate next secuencial number
      number += split[1] + "/" + split[2] + "/" + (int.Parse(split[3]) + 1).ToString();
}
else
{      
      //month has changed,start with 0
      number += DateTime.Now.Year.ToString("0000") + "/" + DateTime.Now.Month.ToString("0") + "/0";
}
this.textBox1.Text = number;

Posted 4-Feb-14 23:50pm

Pikoh

Comments

Diya Bh. 5-Feb-14 6:00am

this one is not working...

Pikoh 5-Feb-14 6:03am

It's working..but i forgot to initialize the textbox with "AB/2014/2/1" for example,or even "AB/0000/0/0". If it's empty it doesn't work

Diya Bh. 5-Feb-14 6:18am

yeah now its working.
but you had taken textbox for doing this.
I don't have 2 use textbox. This no. has to be generated only in code and not to be shown to the user.
Is there any other way?

Pikoh 5-Feb-14 6:40am

well..just use a string variable (string pattern="AB/0000/0/0") and replace all occurences of this.textbox1.text with pattern. Textbox was just to try it,but the logic is valid.

Diya Bh. 5-Feb-14 6:51am

thanks....:)

NitishBharadwaj · Answer 4 · 2014-02-07T22:33:00

Solution 5

You can use Random class. That is a predefined class, make its object and use it.

Posted 7-Feb-14 22:33pm

NitishBharadwaj

**Dnyaneshwar**@Pune · Answer 5 · 2014-02-04T23:34:00

you may refer this function...
private string GenerateRandomNumber()
{
Random rnd = new Random();
List<char> availableNumbers = new List<char>{'0','1','2','3','4','5','6','7','8','9'};
Dictionary<char,> counts = new Dictionary<char,int>();
for (int i = 0; i < 10; i++)
{
counts.Add(availableNumbers[i],0);
}
char[] generatedCharacters = new char[10];
for (int i = 0; i < generatedCharacters.Length; i++)
{
char digit = availableNumbers[rnd.Next(availableNumbers.Count)];
generatedCharacters[i] = digit;
counts[digit]++;
if (counts[digit] == 3)
availableNumbers.Remove(digit);
}
return new string(generatedCharacters);
}

want to generate random no. through coding in some specific pattern.

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