以正确的方式创建JSON对象

Question

我试图从PHP数组中创建一个JSON对象.该数组如下所示:

$post_data = array('item_type_id' => $item_type,
    'string_key' => $string_key,
    'string_value' => $string_value,
    'string_extra' => $string_extra,
    'is_public' => $public,
    'is_public_for_contacts' => $public_contacts);

编码JSON的代码如下所示:

$post_data = json_encode($post_data);

JSON文件最终应该看起来像这样:

{
    "item": {
        "is_public_for_contacts": false,
        "string_extra": "100000583627394",
        "string_value": "value",
        "string_key": "key",
        "is_public": true,
        "item_type_id": 4,
        "numeric_extra": 0
    }
} 

如何将创建的JSON代码封装在"item"中:{JSON CODE HERE}.

Answer 1

通常,你会做这样的事情:

$post_data = json_encode(array('item' => $post_data));

但是,由于您似乎希望输出为" {}",因此最好json_encode()通过传递JSON_FORCE_OBJECT常量来确保强制编码为对象.

$post_data = json_encode(array('item' => $post_data), JSON_FORCE_OBJECT);

" {}"括号指定一个对象," []"根据JSON规范用于数组.

Answer 2

虽然这里发布的其他答案有效,但我发现以下方法更自然:

$obj = (object) [
    'aString' => 'some string',
    'anArray' => [ 1, 2, 3 ]
];

echo json_encode($obj);

Answer 3

你只需要php数组中的另一个图层:

$post_data = array(
  'item' => array(
    'item_type_id' => $item_type,
    'string_key' => $string_key,
    'string_value' => $string_value,
    'string_extra' => $string_extra,
    'is_public' => $public,
   'is_public_for_contacts' => $public_contacts
  )
);

echo json_encode($post_data);

Answer 4

您可以对通用对象进行 json 编码。

$post_data = new stdClass();
$post_data->item = new stdClass();
$post_data->item->item_type_id = $item_type;
$post_data->item->string_key = $string_key;
$post_data->item->string_value = $string_value;
$post_data->item->string_extra = $string_extra;
$post_data->item->is_public = $public;
$post_data->item->is_public_for_contacts = $public_contacts;
echo json_encode($post_data);