Question

我正在尝试从MySQL动态创建JSON，对于在线程Create nested json object using php mysql中找到的这段代码，这似乎相当容易。但是，我想在$ json_response之前和之后添加一些JSON代码。

主要代码

    $result = mysql_query("SELECT * FROM Places "); 
$json_response = array(); //Create an array
while ($row = mysql_fetch_array($result))
{
    $row_array = array();
    $row_array['title'] = $row['title'];        
    $row_array['image_url'] = $row['image_url'];
    $row_array['subtitle'] = $row['subtitle'];      
    $row_array['buttons'] = array();
    $id = $row['id'];


    $option_qry = mysql_query("SELECT * FROM Places where id=$id");
    while ($opt_fet = mysql_fetch_array($option_qry))
    {
        $row_array['buttons'][] = array(
            'type' => $opt_fet['type'],
            'caption' => $opt_fet['caption'],
            'url' => $opt_fet['url'],
        );

    }
    array_push($json_response, $row_array); //push the values in the array
}

echo json_encode($json_response,  JSON_PRETTY_PRINT);

产生此JSON

    [
        {
            "title": "Name of the place",
            "image_url": "image.jpg",
            "subtitle":Additional info",
            "buttons": [
                {
                    "type": "'url'",
                    "caption": "More Info",
                    "url": "https://some link "
                }
            ]
        },
        {
            "title": "Name of the place 2",
            "image_url": "image2.jpg",
            "subtitle":Additional info2",
            "buttons": [
                {
                    "type": "'url'",
                    "caption": "More Info",
                    "url": "https://some link 2"
                }
            ]
        }
    ]

我有某种方法可以在已创建的JSON之前添加以下代码

{
  "version": "v2",
  "content": {
    "messages": [
      {
        "type": "cards",
        "elements":

最后这段代码

      }
    ]
  }
}

Answer 1

非常简单：

$final_json = [
    "version" => "v2",
    "content" => [
        "messages" => [[
            "type" => "cards",
            "elements" => $json_response
        ]]
    ]
];

echo json_encode($final_json, JSON_PRETTY_PRINT);

为清晰起见，我个人将$json_response重命名为$messages。

Answer 2

在声明数组$json_response = array();的同时，您实际上可以使用默认值require来准备它

$stdObj=new \stdClass();
$stdObj->version="V2";
$stdObj->content=(object)["messages"=>(object)["type"=>'cards','elements'=>$row_array['buttons']]];

echo json_encode($stdObj, JSON_PRETTY_PRINT);

将JSON代码附加到通过MySQL和PHP创建的JSON

2 个答案: