How to add a data to each JSON object PHP

Question

I'm getting posts details from 'postsTable' with php and encoding it in JSON Like this way

$result_json = mysqli_fetch_all ($result, MYSQLI_ASSOC);
echo json_encode($result_json);

each post has a unique ID

Then I have another table called 'postsLikes' I want to see how many Likes the post have using mysqli_num_rows()

But my question is how can I add the data it returns to each object in Encoded JSON ?

$query_checkup = "SELECT * FROM postsTable WHERE Post_AgeFrom < $age AND $age < Post_AgeTo AND Post_Reviewed = 1";


$result=mysqli_query($con, $query_checkup);
$result_json = mysqli_fetch_all ($result, MYSQLI_ASSOC);

echo "{\"result\":";
echo json_encode($result_json);
echo "}";

You should be getting this in the same query, with a JOIN or something like that. Have you considered this possibility? — Ed de Almeida, Mar 21 '16 at 14:30

score 4 · Accepted Answer · answered Mar 21 '16 at 14:36

4

You should append extra data to your array before encoding to json.

foreach ($result_json as $key => $result) {
    $result_json[$key]['likes'] = getLikes();
}
echo json_encode($result_json);

And you need to implement getLikes function as you wish or can do the operation inside foreach loop.

There a note that you need to pay attention: you need to query for each product to get likes. It is better to join tables and format your array as your need in a loop.

answered Mar 21 '16 at 14:36

Ali Farhoudi

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Could you please just paste your code in [pastebin](http://pastebin.com) and show that how are you trying to get likes or what you expect to be the result? – Ali Farhoudi Mar 21 '16 at 15:14
You suppose this is my getlikes function : function getlikes(){ return 2; } – M4HdYaR Mar 21 '16 at 15:23
If its necessary tell me and I'll upload my code for ya. – M4HdYaR Mar 21 '16 at 15:24
Yes, please. Because I have tested the code. It's working. Have you tested this foreach loop? Consider that i am assigning to ```$result_json[$key]['likes']```, and **not** ```$result['likes']```. – Ali Farhoudi Mar 21 '16 at 18:26
This made it working : `foreach ($result_json as $key => &$result) { $result['likes'] = getlikes(); }` – M4HdYaR Mar 21 '16 at 22:18
You mean 'likes' index is available out of foreach loop? – Ali Farhoudi Mar 21 '16 at 22:26
Oh, you used it by value (&$result) instead of by reference. It's the same as setting the [$key] in main loop. – Ali Farhoudi Mar 21 '16 at 22:48
Whatever , Damet garm haji :)) – M4HdYaR Mar 22 '16 at 10:06

score 0 · Answer 2 · answered Mar 21 '16 at 14:41

Maybe you can create a new array and use the foreach construct to add the previous values and the new values.

$new_array = array();

foreach($result_json as $result){
   $new_array[] = array(
      'id' => $result['id'],
      'likes' => getlikes($result['id'])
   );
}

function getlikes($id){
   // your code to get likes number with mysqli_num_rows()
}

echo json_encode($new_array);

score 0 · Answer 3 · answered Mar 21 '16 at 14:50

You should join the 'postsLikes' table in your original query to pull in all the data you need with a single trip to the database.

Something like this: (Guessing on how your tables are setup)

$query = "
SELECT 
  P.* ,
  L.Likes

FROM postsTable P
LEFT JOIN postsLikes L ON L.Post_id = P.Post_Id

WHERE 
  P.Post_AgeFrom < :age AND 
  :age < P.Post_AgeTo AND 
  P.Post_Reviewed = 1 ";

$params = array( "age" => $age );

$pdo = new PDO('mysql:dbname=testdb;host=127.0.0.1', $user, password);
$stmt = $pdo->prepare( $query );
$stmt->execute( $params );
$results = $stmt->fetchAll( PDO::FETCH_ASSOC );

$response = array( "results" => $results );
echo json_encode( $response );

Other bits of advice:

1) Use bound parameters to prevent SQL injection.

2) Don't manually create any of the JSON in your response, create the response array first and let json_encode do the rest of the work

Isn't mysqli_real_escape_string() enough for preventing sql injection ? — M4HdYaR, Mar 21 '16 at 15:07
@M4HdYaR - No, not all the time - [http://stackoverflow.com/a/5741264/1725265](http://stackoverflow.com/a/5741264/1725265). Also, the mysql* functions have been deprecated - [http://stackoverflow.com/q/12859942/1725265](http://stackoverflow.com/q/12859942/1725265) — syndicate_software, Mar 21 '16 at 16:28

How to add a data to each JSON object PHP

3 Answers3