我正在尝试从mysql动态创建一个json，使用我在这个线程中找到的这段代码使用php-mysql创建嵌套的json对象似乎相当容易。但是，我想在$json\u响应前后添加一些json代码。
主代码

$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);

生成此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"
                }
            ]
        }
    ]

我不得不在已经创建的json之前添加以下代码

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

最后这个代码

}
    ]
  }
}