PHP比较今天的日期与（多维）数组中的日期？

Question

我有一个日期数组，我想知道是否可以搜索一个数组并删除今天之前的所有项目，并保留其余的？

这是我的数组

Array
(
    [0] => Array
        (
            [date] => 2013-07-14
        )

    [1] => Array
        (
            [date] => 2013-08-31
        )

    [2] => Array
        (
            [date] => 2013-09-15
        )

    [3] => Array
        (
            [date] => 2013-10-12
        )

    [4] => Array
        (
            [date] => 2013-10-16
        )

    [5] => Array
        (
            [date] => 2013-10-19
        )

    [6] => Array
        (
            [date] => 2013-10-23
        )

    [7] => Array
        (
            [date] => 2013-10-26
        )

    [8] => Array
        (
            [date] => 2013-10-30
        )

    [9] => Array
        (
            [date] => 2013-09-07
        )

    [10] => Array
        (
            [date] => 2013-08-14
        )

    [11] => Array
        (
            [date] => 2013-08-24
        )

    [12] => Array
        (
            [date] => 2013-09-11
        )

    [13] => Array
        (
            [date] => 2013-09-28
        )

    [14] => Array
        (
            [date] => 2013-10-05
        )

)

Answer 1

像这样：

$buffer = array();
foreach ($dates as $element) {
    if (strtotime($element['date']) >= time()) {
        $buffer[] = $element;
    }
}

Answer 2

循环循环并将日期与实际所需的数据进行比较并构建新的过滤后的数组呢？

$newarray = array();
foraeach($array as $data)
{
     if($data['date'] >= $yourdate)
     {
         $newarray] = $data;
     }
}

Answer 3

array_walk函数使您可以定义在数组中的每个项目上运行的回调。