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how to use random generate number with linq

Posted By:      Posted Date: September 15, 2010    Points: 0   Category :ASP.Net
 
how to use random generate number with linqthanks in Advance:)


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need help to generate unique random numbers

  

I want to generate a unique random number everytime a user submits a form. As the form is submited the data of the form should be placed in a database and a unique random number should be generated so that the user can later use this unique random number to reterieve his details from the database that refer to his unique random id. The unique random number should be atleast 9 digits long. Please help me on this.


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Random rand = new Random((int)DateTime.Now.Ticks);
        int numIterations = 0;
        numIterations = rand.Next(1, 50);
        Response.Write(numIterations.ToString());
        Response.Write(' ');
        numIterations = rand.Next(1, 50);
        Response.Write(numIterations.ToString());
        Response.Write(' ');
        numIterations = rand.Next(1, 100);
        Response.Write(numIterations.ToString());
        Response.Write(' ');
        numIterations = rand.Next(50, 100);
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Hello,

Not sure if I'm approaching this correctly but this is what I've got and what I'm trying to do

I have a table with 10 Labels in it (Labels Id A1 thru A10)

I'm trying to have each label assigned a random number using 0-9

Need to have each labels number value different from the other

Code Below:

Randomize()

A1.Text = Int(Rnd() * 10)

A2.Text = Int(Rnd() * 10)

A3.Text = Int(Rnd() * 10)

A4.Text = Int(Rnd() * 10)

A5.Text = Int(Rnd() * 10)

A6.Text = Int(Rnd() * 10)

A7.Text = Int(Rnd() * 10)

A8.Text = Int(Rnd() * 10)

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Thanks in Advance

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Is that available (Linq Query with Infinite number of conditions) ?

  

Hello all ,

 

is that available to make dynamic query with LINQ ? 

to be more clear 

i have a query like this 

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In the past I have generated a "Reference Number" for new forms submitted by setting the default value for the Reference Number-(integer Data type) to "max(@Reference_Number) +1". I now have a request to have the Reference Number begin with "REQACH" followed by the number that is generated as above. 

I assume that the data type needs to be set to string but that seems to disqualify the logic that I use above for generating the next "number".  I've tried to generate this value using several different iterations of the Concat function with no luck.

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