如果两个数组中都存在一项，如何合并两个数组？答案

【问题标题】：How to merge two array if one item exist in both array?如果两个数组中都存在一项，如何合并两个数组？
【发布时间】：2021-12-19 00:34:16
【问题描述】：

我想用邮政编码值扩展我的城市数组。如果 city_postcode 数组包含 city 数组名称记录，则将邮政编码值推送到 city 数组中。这就是我想以某种方式实现的目标。

城市数组：

Array
(
    [0] => Array
        (
            [id] => 1
            [city] => Budapest
            [population] => 1700000
        )
    [1] => Array
        (
            [id] => 2
            [city] => Szeged
            [population] => 160000
        )
)

city_postcode 数组：

Array
(
    [0] => Array
        (
            [name] => Budapest
            [post_code] => 12345
        )
    [1] => Array
        (
            [name] => Szeged
            [post_code] => 33356
        )    
)

我想要的结果：

Array
(
    [0] => Array
        (
            [id] => 1
            [city] => Budapest
            [population] => 1700000
            [post_code] => 12345
        )
    [1] => Array
        (
            [id] => 2
            [city] => Szeged
            [population] => 160000
            [post_code] => 33356
        )
)

【问题讨论】：

先写一段代码，这里没人会帮你做的。

标签： php html sql arrays

【解决方案1】：

如果您可以依赖城市和邮政编码的长度和排序相同，您可以这样做：

<?php

$cities = array(
  array("id"=>1,"city"=>"Budapest","Population"=>"1700000"),
  array("id"=>2,"city"=>"Szeged","Population"=>"160000")
);
$cityPostCode = array(
  array("name"=>"Budapest","post_code"=>12345),
  array("name"=>"Szeged","post_code"=>33356)
);

for($i = 0; $i < count($cities); $i++){
  $cities[$i]['post_code'] = $cityPostCode[$i]['post_code'];
}

print_r($cities);

否则，您可以做一些更动态的事情，例如：

function _parsePostCode($targetName,$targetArr){
  foreach($targetArr as $el){
    if($el['name'] == $targetName){
      return $el['post_code'];
    }
  }
  return false;
}

for($i = 0; $i < count($cities); $i++){
  $cities[$i]['post_code'] = _parsePostCode($cities[$i]['city'],$cityPostCode);
}
print_r($cities);

【讨论】：

【解决方案2】：

作为替代方案，您可以在 foreach 循环中使用“参考”PHP，如下所示

$city = array(
 0 => array(
'id' => 1,
'city' => "Budapest",
'population' => 1700000
),
1 => array(
'id' => 2,
'city' => "Szeged",
'population' => 160000
)
);

$city_postcode = array(
0 =>array(
'name' => 'Budapest',
'post_code' => 12345
),
1 => array(
'name' => 'Szeged',
'post_code' => 33356
)
);

foreach ($city as $ckey => &$cval) {
 $cval['post_code'] = $city_postcode[$ckey]['post_code'];
}
unset($cval);

var_dump($city);

【讨论】：

【解决方案3】：

你可以使用这个简单的方法：

<?php

$city = array(
    0 => array(
        'id' => 1,
        'city' => "Budapest",
        'population' => 1700000
    ),
    1 => array(
        'id' => 2,
        'city' => "Szeged",
        'population' => 160000
    )
);

$city_postcode = array(
    0 => array(
        'name' => 'Budapest',
        'post_code' => 12345
    ),
    1 => array(
        'name' => 'Szeged',
        'post_code' => 33356
    )
);

function apply_post_code(&$item, $key, $postcode)
{
    $item['post_code'] = $postcode[ $key ][ 'post_code' ];
}

array_walk($city, 'apply_post_code', $city_postcode);

var_dump($city);

注意 $city 变量是通过引用传递的

【讨论】：