Append JSON code to a JSON created from MySQL and PHP

Question

I am trying to create dynamically a JSON from a MySQL, which seems fairly easy with this piece of code I found in this thread Create nested json object using php mysql. However, I want to add before and after the $json_response some piece of JSON code.

The main code

    $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);

Produces this 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"
                }
            ]
        }
    ]

I have somehow to add the following code before the already created JSON

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

And this code in the end

      }
    ]
  }
}

Blue · Accepted Answer · 2018-08-08T07:12:04.203

3

Pretty simple:

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

echo json_encode($final_json, JSON_PRETTY_PRINT);

Personally, I'd rename $json_response to $messages for clarity purposes.

edited Aug 08 '18 at 07:12

answered Aug 08 '18 at 06:34

Blue

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Cool, and how would be with adding the code in the end? Or it add it automatically – lStoilov Aug 08 '18 at 06:39
Yeah I have noticed after I wrote my comment :) – lStoilov Aug 08 '18 at 06:41
It throws an error --- syntax error, unexpected ':', expecting ']' --- that's on the line "version": "v2", – lStoilov Aug 08 '18 at 06:45
@lStoilov Sorry, fixed (Array syntax in PHP is `=>` not `:`). – Blue Aug 08 '18 at 06:46
@FrankerZ, thanks... I just noticed that the { "type": "cards", "elements": is missing :) Can you help adding it too. – lStoilov Aug 08 '18 at 06:50
I'm confused as to where you want the actual messages to go then @lStoilov. `"messages"` is an array, so where exactly do you want that object to go? Furthermore, use this as a jump start: If this isn't perfect, try modifying/changing the array yourself. It's fairly straightforward. – Blue Aug 08 '18 at 06:53
Thanks, I have just altered this line "messages" => [ to "messages" => [[ to serve my purpose. It works like a charm. – lStoilov Aug 08 '18 at 07:08

Chintan7027 · Answer 2 · 2018-08-08T06:55:14.827

1

While declaring array $json_response = array(); you can actually prepare it with your default values require like

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

echo json_encode($stdObj, JSON_PRETTY_PRINT);

edited Aug 08 '18 at 06:55

answered Aug 08 '18 at 06:50

Chintan7027

7,115
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Thanks, there is an error somewhere in the third line - it says: syntax error, unexpected ';', expecting ']' – lStoilov Aug 08 '18 at 06:54
Updated the answer ] was missing – Chintan7027 Aug 08 '18 at 06:55
1

Why convert everything from an array, back to an object? – Blue Aug 08 '18 at 06:57
We can achieve the result @FrankerZ by the methoed you have specified, but this is also one possible way. – Chintan7027 Aug 08 '18 at 07:11

Append JSON code to a JSON created from MySQL and PHP

2 Answers2