Php 从两个数组创建数组

我在尝试创建某个数组时遇到问题。基本上，我有这样一个数组：

[0] => Array
    (
        [id] => 12341241
        [type] => "Blue"
    )

[1] => Array
    (
        [id] => 52454235
        [type] => "Blue"
    )

[2] => Array
    (
        [id] => 848437437
        [type] => "Blue"
    )

[3] => Array
    (
        [id] => 387372723
        [type] => "Blue"
    )

[4] => Array
    (
        [id] => 73732623
        [type] => "Blue"
    )

[0] => Array
    (
        [id] => 34141
        [type] => "Red"
    )

[1] => Array
    (
        [id] => 253532
        [type] => "Red"
    )

[2] => Array
    (
        [id] => 94274
        [type] => "Red"
    )

接下来，我有一个这样的数组：

[0] => Array
    (
        [id] => 12341241
        [type] => "Blue"
    )

[1] => Array
    (
        [id] => 52454235
        [type] => "Blue"
    )

[2] => Array
    (
        [id] => 848437437
        [type] => "Blue"
    )

[3] => Array
    (
        [id] => 387372723
        [type] => "Blue"
    )

[4] => Array
    (
        [id] => 73732623
        [type] => "Blue"
    )

[0] => Array
    (
        [id] => 34141
        [type] => "Red"
    )

[1] => Array
    (
        [id] => 253532
        [type] => "Red"
    )

[2] => Array
    (
        [id] => 94274
        [type] => "Red"
    )

我想构造一个数组，它是上面两个的组合，使用以下规则：在3个蓝色之后，必须有一个红色：

Blue1
Blue2
Blue3
Red1
Blue4
Blue5
Blue6
Red2
Blue7
Blue8
Blue9
Red3

请注意，它们的颜色可以是红色多于蓝色，但也可以是蓝色多于红色。如果红色用完了，应该从第一个开始

示例：假设只有两个红色：

Blue1
Blue2
Blue3
Red1
Blue4
Blue5
Blue6
Red2
Blue7
Blue8
Blue9
Red1
...

如果蓝色的用完了，红色的应该追加，直到它们也用完为止。 示例：假设有5个蓝色和5个红色：

Blue1
Blue2
Blue3
Red1
Blue4
Blue5
Red2
Red3
Red4
Red5

注意：数组来自mysql抓取。也许在构建新阵列时获取它们更好

不管怎么说，当我遇到循环时，我想不出来

非常感谢您的帮助

$blue=数组(
$blue = array (
[0] => Array
(
    [id] => 12341241
    [type] => "Blue"
)

[1] => Array
(
    [id] => 52454235
    [type] => "Blue"
)

[2] => Array
(
    [id] => 848437437
    [type] => "Blue"
)

[3] => Array
(
    [id] => 387372723
    [type] => "Blue"
)

[4] => Array
(
    [id] => 73732623
    [type] => "Blue"
));

$red = array(
[0] => Array
    (
        [id] => 34141
        [type] => "Red"
    )

[1] => Array
    (
        [id] => 253532
        [type] => "Red"
    )

[2] => Array
    (
        [id] => 94274
        [type] => "Red"
    )
);
$mixed = array();
$index = $b = $r = 0;
if (count($blue) > count($red)){
   $counter = 0;
   for ($i = 0; $i<count($blue) ; $i++){
      $mixed [$index] = $blue[$b];
      $b++;
      $index++;
      if ($r == count($red))
         $r=0;
      if ($counter == 3){
          $counter =0;
          $mixed[$index] = $red[$r];
          $r++;
      }
    $counter++;
   }
}
else {
     $counter = 0;
   for ($i = 0; $i<count($red) ; $i++){
      $mixed [$index] = $blue[$b];
      $b++;
      $index++;
      if ($counter == 3 or $b == count($blue)){
          $counter =0;
          $mixed[$index] = $red[$r];
          $r++;
      }
    $counter++;
   }
}

[0]=>阵列
(
[id]=>12341241
[类型]=>“蓝色”
)
[1] =>阵列
(
[id]=>52454235
[类型]=>“蓝色”
)
[2] =>阵列
(
[id]=>848437437
[类型]=>“蓝色”
)
[3] =>阵列
(
[id]=>387372723
[类型]=>“蓝色”
)
[4] =>阵列
(
[id]=>73732623
[类型]=>“蓝色”
));
$red=数组(
[0]=>阵列
(
[id]=>34141
[类型]=>“红色”
)
[1] =>阵列
(
[id]=>253532
[类型]=>“红色”
)
[2] =>阵列
(
[id]=>94274
[类型]=>“红色”
)
);
$mixed=array（）；
$index=$b=$r=0；
如果（计数（$蓝色）>计数（$红色））{
$counter=0；
对于（$i=0；$i来说，这比你（和其他人）想象的要容易得多：

$r = 0;
foreach($blues as $c => $v) {
    $out []= $v;
    if(($c + 1) % 3 == 0)
        $out []= $reds[$r++ % count($reds)];
}

$out = array_merge($out, array_slice($reds, $r));

模数用于循环，最后一行将剩余的红色（如果有）附加到结果中

对于meSo来说，这似乎是一个非常简单的算法。你已经有了一个蓝色数组和一个红色数组，问题是它们只是交错在一起？为什么标记MySQL？如果两个数组都来自MySQL数据库，你可能可以用SQL创建最后一个数组，但是你需要向我们展示数据库structure@Reeno我有一种感觉，试着去做MySQL中的s会在数据库上产生不必要的开销case@binoculars：您希望遍历数组迭代器样式，例如使用
each
。这允许您只需使用
reset（$reds）即可响应即将用完的red
。这就是我咨询Stackoverflow的原因……当我的解决方案占用的行数超过它应该占用的行数时，总会有更智能的解决方案。非常感谢！