PHP组合二乘二

$players = array('1','2','3','4','5');

while(count($players) != 0){
    $currentPlayer = array_shift($players);
    foreach($players as $player){
        echo $currentPlayer.' - '.$player.'<br/>';
    }
}

编辑: 我的代码可以工作,即使playerNumber是非连续的。$players数组可以如下所示 $players = ('1','5','209','42'); 并且仍然打印出所需的输出。

简单方式:--

function combinations($arr, $n)
{
    $res = array();

    foreach ($arr[$n] as $item)
    {
        if ($n==count($arr)-1)
            $res[]=$item;
        else
        {
            $combs = combinations($arr,$n+1);

            foreach ($combs as $comb)
            {
                $res[] = "$item $comb";
            }
        }
    }
    return $res;
}

$words = array(array('A','B'),array('C','D'), array('E','F'));

$combos = combinations($words,1);  
echo '<pre>';
print_r($combos);
echo '</pre>';
?>

输出:--

Array
(
    [0] => C E
    [1] => C F
    [2] => D E
    [3] => D F
)

  $players = array('1', '2', '3', '4', '5');

          for($i=0;$i<sizeof($players)-1;$i++){
               for($j=$i;$j<sizeof($players);$j++){
                   if( $players[$i]!= $players[$j]){
                          echo  '<br/>';
                       echo $players[$i].' - '.$players[$j];
                   }
              }
          }

输出-

1 - 2
1 - 3
1 - 4
1 - 5
2 - 3
2 - 4
2 - 5
3 - 4
3 - 5
4 - 5

试试这个:

<?php
  $arr = array('1','2', '3', '4', '5');

  for($i;$i<=count($arr);$i++){
    for($y=$i+1;$y<count($arr); $y++){
       echo ($i+1).'-'.$arr[$y].PHP_EOL;
    }
  }
?>

    $players = array('1', '3', '5', '7', '9');

    for($i=0; $i<count($players); $i++) {
        for($j=$i+1; $j<count($players); $j++){
            echo ($players[$i]).' - '.$players[$j].PHP_EOL;
        }
    }

这适用于数组中的任何值

一个foreach循环和一个从foreach值+1开始的for循环。

$players =array ('1', '2', '3', '4', '5');

Foreach($players as $player){

    For($i=$player+1; $i<=count($players); $i++){
        Echo $player ."-" .$i."\n";
    }
}

https://3v4l.org/Bunia