一个特殊的双“for”循环

$bar contains 4 (in this example, but potentially much more than 4) values I want to copy into $foo and $blah in this way:

$foo[0] = $bar[0];
$foo[1] = $bar[2];
$blah[0] = $bar[1];
$blah[1] = $bar[3];

So, the even rows of $bar will be copied into $foo[i], and the odd rows will be copied into $blah[i].

I tried with:

for($i=0; $i<2; $i++)
{
    for($j=0; $j<4; $j++)
    {  
        if($j % 2 == 0)
            {  
            $foo[$i] = $bar[$j]; // EVEN
            }
            else
            {
            $blah[$i] = $bar[$j];  // ODD
        }
    }
}

However if I show the output of $foo and $blah, $foo[0] is equal to $foo[1], and $blah[0] is equal to $blah[1], while they should contain different values.

Example: Source:

$bar[0]: 27.8
$bar[1]: Napoli-Posillipo
$bar[2]: 29
$bar[3]: Stadio San Paolo di Napoli, Napoli

(wrong) result of the loop:

$foo[0] = 29 (it should be 27.8)  
$foo[1] = 29
$blah[0] = Stadio San Paolo di Napoli, Napoli (it should be Napoli-Posillipo)
$blah[1] = Stadio San Paolo di Napoli, Napoli

Any hints? Thanks

you have made it to complex, something simple like this should work:

foreach ($bar as $k=>$v){
    if($k % 2 == 0){  
         $foo[] = $v; // EVEN
    }else{
         $blah[] = $v;  // ODD
    }
}

what about sth like

foreach($bar as $key => $value) {
   if($key % 2 == 0) {
      //even
      $foo[] = $value;
   }
   else {
      // odd
      $blah[] = $value;
   }
}

You said it yourself:

"So, the even rows of $bar will be copied into $foo[i], and the odd rows will be copied into $blah[i]".

Cycle all $bar, and copy odd rows into $blah, even into $foo:

for ($i = 0; $i < 4; $i++)
  if ($i % 2)
     $blah[$i/2] = $bar[$i];
  else
     $foo[$i/2] = $foo[$i];

In the general case:

  switch($i % N) // "N" for Napoli :-D
  {
     case 0: $VAR1[$i/N] = $bar[$i]; break;
     case 1: $VAR2[$i/N] = $bar[$i]; break;
     ...
  }