Question

我在尝试创建某个阵列时遇到了麻烦。基本上，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"
    )

...

接下来，我有一个这样的数组：

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

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

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

我想构建一个数组，这是上面两个的组合，使用这个规则：在3个蓝调之后，必须有一个红色：

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

请注意，他们的红色比蓝色更多，但蓝色也比红色更多。如果Red's耗尽，它应该从第一个开始。

示例：假设只有两个Red：

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

...

如果蓝色用完了，红色应该追加，直到它们用完为止。例如：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-fetches。也许在构建新阵列时获取它们会更好？

无论如何，while-loops到了我身边，我无法理解......

非常感谢任何帮助！

Answer 1

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));

模数关心循环，最后一行将剩余的红色（如果有的话）附加到结果中。

https://ideone.com/cxANRW

Answer 2

$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++;
   }
}

Answer 3

$blue_array = array(array('id'=>'121','type'=>'blue'),array('id'=>'122','type'=>'blue'),array('id'=>'123','type'=>'blue'),array('id'=>'124','type'=>'blue'),array('id'=>'125','type'=>'blue'),array('id'=>'125','type'=>'blue'));   
$red_array = array(array('id'=>'123','type'=>'red'),array('id'=>'123','type'=>'red'),array('id'=>'123','type'=>'red'),array('id'=>'123','type'=>'red'));
$result_array = array();
$blue_count = 1;
$red_count = 1;
// Go through $blue_array first
foreach($blue_array as $blue) {
    // Shove an item from $red_array for every 3 $blue_array items
    if($blue_count % 3 == 0) {
        $result_array[] = 'Blue' . $blue_count;
        if(count($red_array) > 0) {
            $temp = array_shift($red_array);
            $result_array[] = 'Red' . $red_count;
            $red_count++;
        }
    } else {
        $result_array[] = 'Blue' . $blue_count;
    }
    $blue_count++;
}
// Add remaining $red_array items to $result_array
foreach($red_array as $red) {
    $result_array[] = 'Red' . $red_count;
    $red_count++;
}
print_r($result_array);

从两个数组创建数组

3 个答案: