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IEnumerable.skip() method
Marcel Kloubert edited this page Sep 25, 2015
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2 revisions
Skips a number of elements (s. Skip()).
public function skip(int $count) : IEnumerable;
Name | Type | Description |
---|---|---|
$count | int | The number of elements to skip. |
The number of elements to skip.
The current or new sequence.
use \System\Linq\Enumerable;
$seq = Enumerable::create([239, 5979, 22]);
foreach ($seq->skip(1) as $item) {
// [0] 5979
// [1] 22
}