相关疑难解决方法(0)

我知道用一个foreach,然后一个while- >列表等程序很容易实现,(我已经完成了它),但我觉得我的代码有点脏,看起来不是最好的解决方案......我想使用本机PHP数组函数来执行以下操作:

我有两个看起来像这样的数组:

[0] (Array)#2
  [rank] "579"
  [id] "1"
[1] (Array)#4
  [rank] "251"
  [id] "2"

[0] (Array)#2
  [size] "S"
  [status] "A"
  [id] "1"
[1] (Array)#15
  [size] "L"
  [status] "A"
  [id] "2"

我需要的结果如下:

[0] (Array)#2
  [size] "S"
  [status] "A"
  [id] "1"
  [rank] "579"

[1] (Array)#2
  [size] "L"
  [status] "A"
  [id] "2"
  [rank] "251"

有没有办法能够合并两个数组与id值(或其他),而无需进入无限的foreachs？

我有两个具有id键的数组数组,我想根据该数组的键和键值将数据合并在一起.数据看起来像:

    $color = [
        ['id' => 1, 'color' => 'red'],
        ['id' => 2, 'color' => 'green'],
        ['id' => 3, 'color' => 'blue'],
    ];

    $size = [
        ['id' => 1, 'size' => 'SM'],
        ['id' => 2, 'size' => 'XL'],
        ['id' => 3, 'size' => 'MD'],
        ['id' => 4, 'size' => 'LG'],
    ];

    $combined = [
        ['id' => 1, 'color' => 'red', 'size' => 'SM'],
        ['id' => 2, 'color' => 'green', 'size' => 'XL'],
        ['id' => 3, 'color' => 'blue', …

我有两个这样的数组：

数组1

Array
(
    [0] => Array
        (
            [ID] => 101
            [Code] => 1075
            [Date] => 2012-03-03 17:13:12.433
        )

    [1] => Array
        (
            [ID] => 103
            [Code] => 175
            [Date] => 2012-09-05 20:30:02.217
        )

    [2] => Array
        (
            [ID] => 109
            [Code] => 178
            [Date] => 2012-07-05 20:30:02.217
        )

)

阵列2

Array
(
    [0] => Array
        (
            [Amount] => 1234
            [ID] => 101
        )

    [1] => Array
        (
            [Amount] => 1342
            [ID] => 103
        )

    [2] => Array
        (
            [Amount] …