php中的随机洗牌？

发布于 2024-12-02 10:46:20 字数 449 浏览 0 评论 0原文

我正在创建一个彩票来将人们配对。所以我想要一种方法来随机排列数组中的字符串，其中没有项目最终出现在同一位置。（你无法与自己配对）

    public function shuffleSantas(){
    $query = $this->db->get('person');
    $givers = array();
    $recievers = array();
    foreach($query->result() as $row):          
        $givers[] = $row->name;
        //here i want a random order, but no name can be on the same place as in $givers!
        $recievers[] = '';
    endforeach;

原文

I'm creating a lottery to pair up people. So I want a way to shuffle the strings in an array where no item ends up on the same place. (You can't pair up with yourself)

    public function shuffleSantas(){
    $query = $this->db->get('person');
    $givers = array();
    $recievers = array();
    foreach($query->result() as $row):          
        $givers[] = $row->name;
        //here i want a random order, but no name can be on the same place as in $givers!
        $recievers[] = '';
    endforeach;

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陌伤ぢ 2024-12-09 10:46:20

将数组打乱一次，然后将第一个元素与第二个元素配对，第二个元素与第三个元素配对，等等，最后一个元素与第一个元素配对。

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空城旧梦 2024-12-09 10:46:20

$src = $query->result();
$givers = array();
$receivers = array();
foreach ($src as $idx=>$first_person){
    $count = 0; //infinite loop guard
    do{
        ++$count;
        $sec_idx = rand(0,count($src)-1);
        $second_person = $src[$sec_idx];
    } while ($second_person==$first_person && $count<5); 
    $givers[] = $first_person;
    $receivers[] = $second_person; 
}

在这种情况下，一个人将能够从一个人那里获得并给予另一个人。可以吗？而且这个算法也不是最优的，如果数组中只有一个人，肯定会陷入无限循环。

$src = $query->result();
$givers = array();
$receivers = array();
foreach ($src as $idx=>$first_person){
    $count = 0; //infinite loop guard
    do{
        ++$count;
        $sec_idx = rand(0,count($src)-1);
        $second_person = $src[$sec_idx];
    } while ($second_person==$first_person && $count<5); 
    $givers[] = $first_person;
    $receivers[] = $second_person; 
}

In this case one will be able to receive from one person and to give to other person. Is it OK? Also, this algorithm is not optimal and will definitely fall into infinity loop if there is only one person in an array.

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