如何在不重复的情况下显示数据库中的随机数。 [英] How to show the random number from database without repetition .

问题描述

在我的数据库中，我存储了500条记录，其主键是ID（数字）

现在我想随机显示用户的记录而不重复。根据我的要求每个用户我将显示一个id随机生成而不重复..

In my database i had stored 500 records its primary key is ID(number)
Now i want to show the record for user randomly without repetition.As per my requirement each user i will show one id randomly generaed without repetition..

推荐答案

创建一个ID数组，然后将数组洗牌。

查看我的解决方案： 纸牌游戏（在主要游戏中制作随机4组数字）组） [ ^ ]用于改组。

以随机顺序显示记录。

一旦出现，不要重复使用记录。

再次为下一系列记录洗牌。





首先得到一个记录对象的数组（或只是ID）。称之为 orderedData 。

调用 var shuffledData = Shuffle（orderedData）; （见下文） ）

按顺序使用 shuffledData [ 0到结束] 的值，它将成为记录对象（或如上所述的ID，没有重复的随机顺序。



Shuffle（）的代码转换为通用：

Make an array of the IDs and then shuffle the array.
Look at my solution in: playing cards game (make Random 4 groups of Numbers in Main Group)[^] for shuffling.
Present the records in the shuffled order.
Don't re-use a record, once presented.
Shuffle again for the next sequence of records.


First get an array of the record objects (or just the IDs). Call it orderedData.
Call var shuffledData = Shuffle(orderedData); (see below)
Use the values of shuffledData[0 thru end] in sequence and it will be the record objects (or IDs, as above) in a random order with no repeats.

Code for Shuffle() converted to generic:

private Random rand = new Random();
public static T[] Shuffle<T>(T[] data)
{
  T[] shuffled = new T[data.Length];
  shuffled[0] = data[0];
  for (int i = 1; i < data.Length; i++)
  {
    int j = rand.Next(i + 1);
    if (j != i)
      shuffled[i] = shuffled[j];
    shuffled[j] = data[i];
  }
  return shuffled;
}

：我在Shuffle方法之外重命名了这个变量，所以不会混淆名称。

: I renamed the variable outside of the Shuffle method so there's no confusion about names.

当然，它非常简单。

Sure, and its quite easy.

SELECT TOP 1 * FROM table ORDER BY NEWID()

希望我能得到信任，但答案来自 Stack Overflow [ ^ ]



Hogan

Wish I could take credit, but the answer comes from Stack Overflow[^]

Hogan

您好，



参考



在循环中生成随机数而不重复。 [ ^ ]

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