Question

在查看了大量问题之后，我仍然找到了一个适合我情况的答案。

我试图将数组＃2中的2个字段连接到数组＃1

数组＃1

Array
(
    [0] => Array
        (
            [id] => 1
            [position] => top_banner_1
            [name] => Top Banner 1
            [order] => 1
        )

    [1] => Array
        (
            [id] => 2
            [position] => left_banner_1
            [name] => Left Banner 1
            [order] => 2
        )
)

数组＃2

Array
(
    [status] => 0
    [countries] => 
    [module_status] => 1
    [top_banner_1_status] => 1
    [top_banner_1_display] => 0
    [left_banner_1_status] => 1
    [left_banner_1_display] => 0
    [left_banner_2_status] => 1
    [left_banner_2_display] => 0
    [left_banner_3_status] => 1
    [left_banner_3_display] => 0
    [left_banner_4_status] => 
    [left_banner_4_display] => 0
    [left_banner_5_status] => 
    [left_banner_5_display] => 0
    [center_banner_1_status] => 
    [center_banner_1_display] => 0
    [center_banner_2_status] => 
    [center_banner_2_display] => 0
    [right_banner_1_status] => 
    [right_banner_1_display] => 0
    [right_banner_2_status] => 
    [right_banner_2_display] => 0
    [right_banner_3_status] => 
    [right_banner_3_display] => 0
    [right_banner_4_status] => 
    [right_banner_4_display] => 0
    [right_banner_5_status] => 
    [right_banner_5_display] => 0
    [bottom_banner_1_status] => 
    [bottom_banner_1_display] => 0 
)

我想要实现的目标是：

Array
(
    [0] => Array
        (
            [id] => 1
            [position] => top_banner_1
            [name] => Top Banner 1
            [order] => 1
            [top_banner_1_status] => 1
            [top_banner_1_display] => 0
        )
)

这两个数组都来自数据库。阵列＃1中有13个区域，所以到目前为止我所做的一切都是使用foreach循环，因为阵列＃2数据是从理想情况下无法编辑的函数中获取的。

我已经尝试了很多array_*功能，但我的速度不是很快。

Answer 1

假设有以下推理：

Array
(
    [0] => Array
        (
            [id] => 1
            [position] => top_banner_1
            [name] => Top Banner 1
            [order] => 1
            [top_banner_1_status] => 1 // Added because of key is [position]_status.
            [top_banner_1_display] => 0 // Added because of key is [position]_display.
        )
)

我愿意：

<?php

$array1 = // Array #1 from question.
$array2 = // Array #2 from question.

foreach ($array1 as $key => $item) {
    $position = $item['position'];
    $keySuffixes = array('_status', '_display');
    foreach ($keySuffixes as $suff) {
        if (array_key_exists($position . $suff, $array2)) {
            $array1[$key][$position . $suff] = $array2[$position . $suff];
        }
    }
}

＆GT;

不太优雅，我知道：（

Answer 2

诀窍是根据ID构造一个密钥以从$ arr2获取数据。

 foreach($arr1 as $key=>$val){
    $id = $val['id'];
    $tb_status_key = "top_banner_{$id}_status";
    $tb_display_key = "top_banner_{$id}_display";

    $arr1[$key][$tb_status_key] = $arr2[$tb_status_key];
    $arr1[$key][$tb_display_key] = $arr2[$tb_display_key];
    }

Answer 3

Array＃2的结构对于此任务并不十分适用。如果你有可能，你应该把它改成这样的东西：

Array
(
    [top_banner_1] => Array (
      [status] => 1
      [display] => 0
    )
    [left_banner_1] => Array (
      [status] => 1
      [display] => 0
    )
)
//and so on

但是如果 不可能，这应该有效：

foreach($array1 as &$info) { $status = $info['position'] . '_status'; $display = $info['position'] . '_display'; $info[$status] = $array2[$status]; $info[$display] = $array2[$display]; }

否则，这似乎是一种更优雅的方式：

foreach($array1 as &$info) { $info['status'] = $array2[$info['position']]['status']; $info['display'] = $array2[$info['position']]['display']; }

将数据从1个数组添加到另一个数组

3 个答案: