Question

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

$ 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
    )
)

所以我想将$ dates2合并到$ dates1中，忽略任何重复项，以使$ date3：

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

这样的事情应该有效

$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：我重构了代码。

如何合并两个日期数组，忽略重复日期，然后按日期排序

2 个答案: