PHP: how to iterate through an array, and resume where you left off?

Question

Here's my array:

$array = array(1,2,3,4,5,6,7,8,9,10);

I want to iterate through the array 5 times, do something else, then resume iteration where I left off.

foreach ($array as $value) {
    //do something until key 5
}

//do something else now

//resume...
foreach ($array as $value) {
    //key should start at 6
}

How can I do this? Is there a way to achieve this with a foreach loop?

Update: I realized it would be silly to repeat the same code twice. The reason I was asking this is because I'm using a foreach loop to display table rows. I wanted to display the first five and hide the rest. So this is what I ended up with:

<?php
$counter = 1;
foreach ($array as $object): ?>

    <?php if ($counter > 5): ?>
        <tr style="display: none;">
    <?php else: ?>
        <tr>
    <?php endif; ?>
            <td><?php echo $object->name; ?></td>
        </tr>
    <?php $counter++; ?>

<?php endforeach; ?>

Answer 1

Just curious, but wouldn't calling a function in the array to do what you need done achieve the same result?

Answer 2

something like this:

$counter = 0;
foreach ($array as $value) 
{
if($counter == 5)
{
   do something random;
   $counter++;
   continue;
}
    //do something until key 5
$counter++;
}

Answer 3

Is there a reason you need to do this with foreach, as opposed to just for?

Answer 4

It's a little kloodgy, but it should work:

foreach($array as $key => $value)
{
    $lastkey = $key;

    // do things

    if($key == 5) break;
}

// do other things

foreach($array as $key => $value)
{
    if($key <= $lastkey) continue;

    // do yet more things
}

Answer 5

Hi, you can use each() to resume iterating.

$a = array(1,2,3,4);

foreach ($a as $v) {
  if ($v == 2) break;
}

while (list($k, $v) = each($a)) {
  echo "$k = $v\n";
}

Answer 6

Use two arrays:

$first_five = array_slice($array, 0, 5);
$remainder = array_slice($array, 5);

Answer 7

You need to use PHP's internal array pointer.

Something like:

$arr = range(0, 9);
for($i = 0; $i < 5; $i++) {
  print current($arr);
  next($arr);
}

//the pointer should be half way though the array here

Answer 8

I think jspcal had the best answer so far. The only modification I made to his code was to use a do-while loop instead, so it would not skip the element where first loop breaks.

$arr = array(1,2,3,4);

// Prints 1 2
foreach($arr as $v)
{
    if ($v == 3)
    {
        break;
    }
    echo "$v ";
}

// Prints 3 4
do
{
    echo "$v ";
}
while( list($k, $v) = each($arr) );

Answer 9

(For the problem as stated I wouldn't recommend it but:) You can also use the NoRewindIterator.

$array = array(1,2,3,4,5,6,7,8,9,10);
$it = new NoRewindIterator(new ArrayIterator($array));

foreach($it as $x) {
  echo $x;
  if ($x > 4) {
    break;
  }
}
// $it->next();
echo 'taadaa';
foreach($it as $x) {
  echo $x;
}

prints 12345taadaa5678910.
(Note the duplication of the element 5. Uncomment the $it->next() line to avoid that).