如果下一个条目是整数，则在数组中应用Php函数(连接一个有效的FEN字符串)

Question

试图构建一个有效的分弦。

给定这个8*8数组示例，表示棋盘，("1“是空方块)：

$checkerboard = [["r","n","b","q","k","b","n","r"],["p","p","p","p","p","p","p","p"],["1","1","1","1","1","1","1","1"],["1","1","1","1","1","1","1","1"],["1","1","1","1","P","1","1","1"],["1","1","1","1","1","1","1","1"],["P","P","P","P","1","P","P","P"],["R","N","B","Q","K","B","N","R"]]

在原地，这是一种立场：

​

​

我正在寻找的有效结果是：

rnbqkbnr/pppppppp/8/8/4P3/8/PPPP1PPP/RNBQKBNR

到现在为止，我的产出是：

rnbqkbnr/pppppppp/11111111/11111111/1111P111/11111111/PPPP1PPP/RNBQKBNR

显然，数组中的任何整数条目都应该与下一个整数相加，只有当下一个条目是整数时，如果是的话，那么下一个条目应该被丢弃，直到结束。

通过迭代数组，我不知道如何简单地应用像步行()或地图()这样的东西。也许一个简单的字符串操作就足够了？

  $FEN = "";
  for ($i = 0;$i < 8;$i++){
    for ($j = 0;$j < 8;$j++){
      if ($checkerboard[$i][$j] === "1"){
        if ($checkerboard[$i][$j + 1] === "1"){
           /* How to iterate till the end */
           $FEN .= (int)$checkerboard[$i][$j] + (int)$checkerboard[$i][$j+1];
        }
      } else {
        $FEN .= $checkerboard[$i][$j];
      }      
    }
    $FEN .= "/";
  }

有什么见解吗？

联机示例： https://3v4l.org/tuqqo

Answer 1

$checkerboard = [["r","n","b","q","k","b","n","r"],["p","p","p","p","p","p","p","p"],["1","1","1","1","1","1","1","1"],["1","1","1","1","1","1","1","1"],["1","1","1","1","P","1","1","1"],["1","1","1","1","1","1","1","1"],["P","P","P","P","1","P","P","P"],["R","N","B","Q","K","B","N","R"]];

$parts =  array();
foreach ($checkerboard as $innerArray) {
    $num = null;
    $str = '';
    foreach($innerArray as $innerval){
        if(is_numeric($innerval)){
            $num += (int) $innerval;
        }
        else{
            if(!is_null($num)){
                $str .=$num;
                $num = null;
            }
            $str .=$innerval;
        }
    }
    if(!is_null($num)){
        $str .=$num;
    }
    array_push($parts,$str);
}
$result = implode('/',$parts);

上面的代码将生成所需的输出并将其存储在$result上。