php 以正确的方式创建 JSON 对象
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Create JSON-object the correct way
提问by Mark Denn
I am trying to create an JSON object out of a PHP array. The array looks like this:
我正在尝试从 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);
The code to encode the JSON look like this:
编码 JSON 的代码如下所示：
$post_data = json_encode($post_data);
The JSON file is supposed to look like this in the end:
JSON 文件最终应该是这样的：
{
"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
}
}
How can I encapsulate the created JSON code in the "item": { JSON CODE HERE }.
如何将创建的 JSON 代码封装在“项目”中：{ JSON CODE HERE }。
回答by Cristian
Usually, you would do something like this:
通常，你会做这样的事情：
$post_data = json_encode(array('item' => $post_data));
But, as it seems you want the output to be with "{}", you better make sure to force json_encode()to encode as object, by passing the JSON_FORCE_OBJECTconstant.
但是，由于您似乎希望输出带有“ {}”，因此您最好确保json_encode()通过传递JSON_FORCE_OBJECT常量来强制将其编码为对象。
$post_data = json_encode(array('item' => $post_data), JSON_FORCE_OBJECT);
"{}" brackets specify an object and "[]" are used for arrays according to JSON specification.
" {}" 括号指定一个对象，" []" 用于根据 JSON 规范的数组。
回答by theDmi
Although the other answers posted here work, I find the following approach more natural:
尽管此处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);
回答by zeusstl
You just need another layer in your php array:
你只需要你的 php 数组中的另一个层：
$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);
回答by Alauddin Afif Cassandra
$post_data = [
"item" => [
'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
]
];
$post_data = json_encode(post_data);
$post_data = json_decode(post_data);
return $post_data;
回答by Mitesh
You could json encode a generic object.
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);
