如何合并两个日期数组忽略重复日期，然后按日期排序

Question

例如，我有$ dates1和$ dates2如下所示，并且想要创建$ dates3，它将所有日期保留在$ dates1中，并添加任何唯一日期在$ dates2，忽略任何重复。数组包含其他值，但我只是显示日期，因为这是我想合并/排序。

$ dates1 =

Array 
(
[0] => Array 
    (
     [day] => 2012-01-01 
     [foo] => "bar" 
    ) 

[1] => Array 
    (
     [day] => 2012-01-02 
     [foo] => "bar" 
    ) 

[2] => Array 
    (
     [day] => 2012-01-03 
     [foo] => "bar" 
    ) 
) 

$ dates2 =

Array 
(
[0] => Array 
    (
     [day] => 2011-12-31 
    ) 

[1] => Array 
    (
     [day] => 2012-01-01 
    ) 

[2] => Array 
    (
     [day] => 2012-01-02 
    ) 
[3] => Array 
    (
     [day] => 2012-01-03 
    ) 
[4] => Array 
    (
     [day] => 2012-01-04 
    ) 
) 

所以我想合并$ dates2到$ dates1，忽略任何重复，使$ dates3：

Array 
(
[0] => Array 
    (
     [day] => 2011-12-31 
    ) 

[1] => Array 
    (
     [day] => 2012-01-01 
     [foo] => "bar" 
    ) 

[2] => Array 
    (
     [day] => 2012-01-02 
     [foo] => "bar" 
    ) 
[3] => Array 
    (
     [day] => 2012-01-03 
     [foo] => "bar" 
    ) 
[4] => Array 
    (
     [day] => 2012-01-04 
    ) 
) 

Answer 1

我不确定我是否正确理解你，你的意思是这样吗？

$old_array = array_merge($dates1, $dates2); 
    $new_array = array(); 
    foreach ($old_array as $item){ 
     if (!array_key_exists($item['day'], $new_array)){ 
      $new_array[$item['day']] = $item; 
     } 
    } 
    return $new_array; 

Answer 2

像这样的东西应该工作

$a1 = array(array('day' => '2012-01-01', 'foo' => 'bar1'), 
      array('day' => '2012-01-02', 'foo' => 'bar2'), 
      array('day' => '2012-01-03', 'foo' => 'bar3')); 

$a2 = array(array('day' => '2011-12-31'), 
      array('day' => '2012-01-01'), 
      array('day' => '2012-01-02'), 
      array('day' => '2012-01-03'), 
      array('day' => '2012-01-04')); 

$orderedArray = array(); 
foreach (array_merge($a1, $a2) as $v) 
{ 
    if (count($v) == 1 && isset($orderedArray[$v['day']])) 
     continue; 

    $orderedArray[$v['day']] = $v; 
} 
ksort($orderedArray); 

print_r($orderedArray); 

编辑：我重构代码。