洗牌的随机数在Javascript / PHP不重复 [英] Shuffles Random Numbers with no repetition in Javascript/PHP

问题描述

我已经通过这里的一些答案搜查，但似乎我需要或我只是不知道如何申请，虽然它的东西没有。

I've searched through some of the answers here but it doesn't seem the thing that I needed or I just don't know how to apply it though.

我还没有开始任何codeS和我只是在想如何做到这一点，我不知道该怎么做。我需要你的帮助球员。

I haven't started any codes and I'm only thinking on how to do it and I have no idea how to do it. I need your help guys.

让我们假设我有一个由这些值低于

Let's assume that I have an array which consists of these values below

[1,2,3,4,5,6,7,8,9]

和我需要它洗不重复的最后结果的每个数字的位置。所以它可能会像

And I need to shuffle it without repeating the position of each numbers of the last result. so it would probably like

[5,3,9,6,2,8,1,4,7]

如果我再次洗牌它，它会像

if I shuffle it again it would be like

[4,7,2,1,8,3,6,9,5]

等。

好吧，我不知道如果有任何相关性，但，宁可不使用兰特（）虽然。这个东西任何解决方案？

Well I don't know if there' any relevance to it but, would rather not to use rand() though. Any solution for this stuff?

推荐答案

试试这个，

$count = 15;
$values = range(1, $count);
shuffle($values);
$values = array_slice($values, 0, 15);

或

$numbers = array();
do {
   $possible = rand(1,15);
   if (!isset($numbers[$possible])) {
      $numbers[$possible] = true;
   }
} while (count($numbers) < 15);
print_r(array_keys($numbers));

这可能会帮助你。

may this help you.

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