帶有數字鍵的PHP array_merge

Question

如何才能讓array_merge（）覆蓋兩個數組中具有不同值但密鑰索引相同的鍵？

例如，合併：

[0] => 'whatever'

與

[0] => 'whatever', [1] => 'a', [2] => 'b'

應該產生

[0] => 'whatever', [1] => 'a', [2] => 'b'

基本上我想array_merge到bahave如果陣列具有字符串鍵它的工作方式......

Answer 1

使用+運算符。

比較array_merge到+操作：

<?php 

$a1 = array(0=>"whatever",); 
$a2 = array(0=>"whatever",1=>"a",2=>"b"); 

print_r(array_merge($a1,$a2)); 
print_r($a1+$a2); 
?> 

輸出：

Array 
(
    [0] => whatever 
    [1] => whatever 
    [2] => a 
    [3] => b 
) 
Array 
(
    [0] => whatever 
    [1] => a 
    [2] => b 
) 

的+操作仍然作品，如果你的關聯數組具有數字鍵外的順序：

<?php 

$a1 = array(0=>"whatever",); 
$a2 = array(1=>"a",0=>"whatever",2=>"b"); 

print_r(array_merge($a1,$a2)); 
print_r($a1+$a2); 
?> 

輸出：

Array 
(
    [0] => whatever 
    [1] => a 
    [2] => whatever 
    [3] => b 
) 
Array 
(
    [0] => whatever 
    [1] => a 
    [2] => b 
) 

注意array_merge在這種情況下創建一個新的關鍵。不可取...

Answer 2

很容易手工編寫：

function array_merge_custom($first, $second) { 
    $result = array(); 
    foreach($first as $key => $value) { 
     $result[$key] = $value; 
    } 
    foreach($second as $key => $value) { 
     $result[$key] = $value; 
    } 

    return $result; 
} 

更新：此行爲與聯合運營商（return $first + $second;），因爲在不同的這種情況下第二個數組在兩個元素具有相同的密鑰時獲勝。

但是，如果切換參數的位置，並且在發生衝突時將想要「贏」的數組放置爲第一個操作數，則可以獲得相同的行爲。所以上面的函數的行爲完全像return $second + $first;。

Answer 3

您可以使用array_merge()，然後使用array_unique()。

Answer 4

function array_merge_custom() 
{ 
    $array  = array(); 
    $arguments = func_num_args(); 
    foreach($arguments as $args) 
     foreach($args as $key => $value) 
      $array[$key] = $value; 
    return $array; 
} 

Answer 5

您應該使用$a2+$a1得到相同的結果與array_merge($a1,$a2);

$a1 = array(
    'k1' => 1, 
    'k2' => 2, 
    'k3' => 3, 
); 

$a2 = array(
    'k1' => 11, 
    'k2' => 22, 
    'k4' => 44, 
); 

代碼：

print_r(array_merge($a1,$a2)); 

輸出：

Array ( 
    [k1] => 11 
    [k2] => 22 
    [k3] => 3 
    [k4] => 44 
) 

代碼：

print_r($a1+$a2); 

輸出：

Array ( 
    [k1] => 1 
    [k2] => 2 
    [k3] => 3 
    [k4] => 44 
) 

代碼：

print_r($a2+$a1); 

輸出：

Array ( 
    [k1] => 11 
    [k2] => 22 
    [k4] => 44 
    [k3] => 3 
) 

Answer 6

the solution could be this: 
function array_merge_custom($array1, $array2) { 
    $mergeArray = []; 
    $array1Keys = array_keys($array1); 
    $array2Keys = array_keys($array2); 
    $keys = array_merge($array1Keys, $array2Keys); 

    foreach ($keys as $key) { 
     $mergeArray[$key] = array_merge_recursive(isset($array1[$key]) ? $array1[$key] : [], isset($array2[$key]) ? $array2[$key] : []); 
    } 

    return $mergeArray; 
} 

$array1 = [ 
    '66_' => [ 
     'k1' => 1, 
     'k2' => 1, 
    ], 
    '67_' => [ 
     'k1' => 1, 
     'k2' => 1, 
    ], 
    '68_' => [ 
     'k1' => 1, 
     'k2' => 1, 
    ], 
    68 => [ 
     'k1' => 1, 
     'k2' => 1, 
    ] 
]; 
$array2 = [ 
    '66_' => [ 
     'a1' => 1, 
     'a2' => 1, 
    ], 
    '68_' => [ 
     'b1' => 1, 
     'b2' => 1, 
    ], 
    68 => [ 
     'b1' => 1, 
     'b2' => 1, 
    ] 
]; 
echo '<pre>'; 
print_r(array_merge_custom($array1, $array2)); 

Answer 7

array_replace正是這樣做的。請參閱：http://php.net/manual/de/function.array-replace.php

Answer 8

$arrA = [10, 11, 12]; 
$arrB = [12, 13]; 

$arrCommon = array_keys(array_flip($arrA) + array_flip($arrB)); 

print_r($arrCommon); 
Array 
(
    [0] => 10 
    [1] => 11 
    [2] => 12 
    [3] => 13 
) 

比較，以WRONG使用 「+」

$arrCommon = $arrA + $arrB; 

print_r($arrCommon); 
Array 
(
    [0] => 10 
    [1] => 11 
    [2] => 12 
)