PHP:多维数组

Question

我有一个像这样的数组:

Array ( [today] => Array ( [0] => Array ( [hour] => 08/03/11 00:00)
                           [1] => Array ( [hour] => 08/03/11 11:00)
                           [n] => Array ( [hour] => 0xxxxxxxxx)
                         )
        [yesterday] => Array ( [0] => Array ( [hour] => 08/02/11 00:00)
                               [1] => Array ( [hour] => 08/02/11 11:00)
                               [n] => Array ( [hour] => 0xxxxxxxxx)
                             )
)

以此类推，今天有很多小时，昨天也有很多小时。

现在，我有点不知道如何在每个foreach中得到今天和昨天相同的小时。例如，我有:

foreach ($Array as $key => $data) { 

//display today's hour //display yesterday's hour value

如何逐行获取它们的值?

Answer 1

必须嵌套Forech

$array = array(
  'today'=>array(
    0 => Array ( 'hour' => '08/03/11 00:00'),
    1 => Array ( 'hour' => '08/03/11 11:00'),
    'n' => Array ( 'hour' => '0xxxxxxxxx')
  ),
  'yesterday'=>array(
    0 => Array ( 'hour' => '08/03/11 00:00'),
    1 => Array ( 'hour' => '08/03/11 11:00'),
    'n' => Array ( 'hour' => '0xxxxxxxxx')
  )
);

SO与上面的数组

foreach($array as $key => $arr){
  foreach($arr as $a_key => $a_arr){
    foreach($a_arr $b_key => $b_str){
      var_dump($b_str);
    }
  }
}

这将产生6个单行的小时值

无论如何，我认为你在这里设置了一个愚蠢的数组它只是为了好玩而浪费内存，

它应该保存相同的数据，没有第二个深度到你的数组

$array = array(
  'today'=>array(
    0 => '08/03/11 00:00',
    1 => '08/03/11 11:00',
    'n' => '0xxxxxxxxx'
  ),
  'yesterday'=>array(
    0 => '08/03/11 00:00',
    1 => '08/03/11 11:00',
    'n' => '0xxxxxxxxx'
  )
);

Answer 2

好奇怪的数组。试试这个:

foreach ($Array as $key => $data) {
    foreach($data as $v){
        echo $key."'s hour: ".$v[hour];
    }
}

Answer 3

如果count(array[today]) == count(array[yesterday])，那么这可能会有所帮助:

for ($i=0,$cnt=count(array['today']); $i<$cnt; $i++)
  echo $Array['today'][$i]['hour'] . ' : ' $Array['yesterday'][$i]['hour']
)

Answer 4

foreach($array['today'] as $k=>$v){
    //use $v
    //use $array['yesterday'][$key]
    //ex.
    print 'today'.$v['hour'].'; yesterday '.$array['yesterday'][$k]['hour'].EOLN;
}