随机获取问题 [英] Fetch Questions Randomly
问题描述
先生我做了一个在线测验项目。
问题编号是主键。
先生/女士
在测验中有一套30个问题。
我安排了15人的测验。
先生/女士
当我组织测验时,所有参与者都以相同的顺序或相同的序列号提出问题。 />
我想为所有参与者提取不同的问题(随机问题)。
我该怎么做。
请帮我解决这个问题。
随机选择问题ID ...使用randon no代表1-30 ...
ie ...
随机rand = new Random();
int ra = rand.Next( 1 , 30 );
并使用此ra ....这是一种简单的方法...... / blockquote>
HI,
结帐以下2个解决方案:
1:公开 类 Form1
< span class =code-keyword>私有 Sub Button1_Click( ByVal sender 作为系统。对象, ByVal e As System.EventArgs)句柄 Button1.Click
' 在VB.NET中创建一个新的Random类
Dim RandomClass As New Random()
Dim RandomNumber As 整数
生成1到2,147,483,647之间的随机数
RandomNumber = RandomClass。下一步()
MessageBox.Show(RandomNumber, www.interloper.nl,MessageBoxButtons.OK,MessageBoxIcon.Information)
Dim RandomNumber2 正如 整数
' 生成100到200之间的随机数
RandomNumber2 = RandomClass。下一步( 100 , 200 )
MessageBox.Show(RandomNumber2, www.interloper.nl,MessageBoxButtons.OK,MessageBoxIcon.Information)
结束 Sub
结束 班级
2.公开函数GetRandomNum
将myRandom调暗为新的随机
dim RandomNumber as Integer = myRandom.Next(''first value,''sencond value)''作为整数
返回RandomNumber
结束功能
谢谢
Hello Ankur,
Random是.net中的预定义类,因此您可以使用Random类生成随机数。
Ex:
Dim r As New Random
For i As Integer = 0 至 30
Dim r As New Random
r.Next()
下一页
使用这30个数字来获取问题,因为 Random 类每次生成不同的值执行。
希望这会对你有所帮助。
与Prafulla欢呼
Sir i make an online quiz project.
The question number is a primary key.
Sir/Madam
In the Quiz there is a set of 30 Questions.
And i schedule the quiz for 15 persons.
Sir/Madam
when i organize the quiz then all the participants have the question in the same order or in the same serial number.
I want to fetch the questions differently(Random Question) for all the participants.
How can i do it.
Please help me regarding this.解决方案choose question id randomly ... use randon no for 1-30...
i.e....
Random rand = new Random(); int ra = rand.Next(1, 30);
and use use this ra .... this is a easy way ....
HI,
checkout the following 2 solutions:
1:Public Class Form1 Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click 'Create a new Random class in VB.NET Dim RandomClass As New Random() Dim RandomNumber As Integer 'Generate a random number between 1 and 2,147,483,647 RandomNumber = RandomClass.Next() MessageBox.Show(RandomNumber, "www.interloper.nl", MessageBoxButtons.OK, MessageBoxIcon.Information) Dim RandomNumber2 As Integer 'Generate a random number between 100 and 200 RandomNumber2 = RandomClass.Next(100, 200) MessageBox.Show(RandomNumber2, "www.interloper.nl", MessageBoxButtons.OK, MessageBoxIcon.Information) End Sub End Class
2.Public Function GetRandomNum Dim myRandom as New Random dim RandomNumber as Integer = myRandom.Next(''first value, ''sencond value) ''both as integers Return RandomNumber End Function
Thanks
Hello Ankur,
Random is a predefined class in .net so you can use Random class to generate random numbers.
Ex:
Dim r As New Random For i As Integer = 0 To 30 Dim r As New Random r.Next() Next
use this 30 numbers to fetch the questions because Random class generate different values each time it will execute.
Hope this helps you.
Cheers with Prafulla
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