json_decode轉換空數組字符串

Question

我JSON元素

{"element":{}, 
"gallery":{ 
    "images":[ 
    {"id":"1","description":{},"image_path":"1.jpg"}, 
    {"id":"2","description":"Test","image_path":"2.jpg"} 
    ]}, 
"additional_value":"Test"} 

PHP函數json_decode（$ JSON，TRUE）返回

Array 
(
    [element] => Array() 
    [gallery] => Array(
      [images] => Array(
        [0] => Array(
          [id] => 1 
          [description] => Array() 
          [image_path] => 1.jpg) 
        [1] => Array(
          [id] => 2 
          [description] => Test 
          [image_path] => 2.jpg) 
        ) 
      ) 
    [additional_value] => Test 
) 

我怎麼能代替/轉換空數組字符串？例如：

[0] => Array([id] => 1 
       [description] => "" 
       [image_path] => 1.jpg) 

謝謝！

Answer 1

我認爲這樣做的最好方法似乎是（但它不是！）解碼json，之後使用array_walk_recursive將空數組轉換爲空字符串。

該函數遞歸地遍歷數組的所有項。它們中的每一個都通過引用傳遞給指定的回調函數。

但是，事實證明，該函數不會調用本身爲數組的項目的回調，而只會針對這些數組中的項目調用回調。這種行爲使得使用array_walk_recursive無法找到空數組。

因此，我已經編寫了一個替換函數，它應該完全相同，除了在深入遞歸之前，它總是會調用本身爲數組的項的回調函數。

該函數和調用代碼可以在下面找到。

<?php 
// The replacement function for array_walk_recursive() 
function my_array_walk_recursive(&$array, $callback, $userdata = null) { 
    if (!is_array($array)) return false; 

    foreach ($array as $key => &$value) { 
    // Difference: PHP's array_walk_recursive will only call the callback 
    // for items that are not arrays themselves. Here, the callback is always called. 
    call_user_func_array($callback, array(&$value, $key, $userdata)); 
    if (is_array($value)) { 
     my_array_walk_recursive($value, $callback, $userdata); 
    } 
    } 
    return true; 
} 

// The calling code. 
$json = 
    '{"element":{}, 
    "gallery":{ 
     "images":[ 
     {"id":"1","description":{},"image_path":"1.jpg"}, 
     {"id":"2","description":"Test","image_path":"2.jpg"} 
     ]}, 
    "additional_value":"Test"}'; 

$yourArray = json_decode($json, TRUE); 

my_array_walk_recursive(
    $yourArray, 
    function(&$item, $key){ 
    if (is_array($item) && count($item) === 0) { 
     $item = "x"; 
    } 
    }); 

var_dump($yourArray); 

Answer 2

正確的工作下一個代碼/招

$json_upd = str_replace('{}', '""', $json); 
$result = json_decode($json_upd,TRUE); 

如果有人有更好的解決方案，請加註釋 - 我會很感激。