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I am trying to merge a static data with json encode array data for output. Here is my php code:

$arr = array();

$rs = mysql_query("SELECT id, name, picture, mail, gender, birthday FROM users WHERE id='$logged_id' ");

while($obj = mysql_fetch_object($rs)) {
   $arr[] = $obj;
}

echo '{"users":'.json_encode($arr).'}';

Now I want to merge other data with it:

$user_ip = array("user_ip" => $user_ip_address);

I have tried array_merge($arr, $user_ip). But it didn't work. I think this is not correct json array format if I merge with existing data array. Please let me know what to do how to output other data as well as current data coming from mysql with json encode.

I am getting such output with my existing code, which is correct:

{"users":[{"id":"14","name":"Sonu Roy","picture":"image012.jpg","mail":"[email protected]","gender":"Male","birthday":"1983-01-11"}]}

But now I want to add other variable e.g $user_ip_address as user's data joining with current output data like this:

{"users":[{"id":"14","name":"Sonu Roy","picture":"image012.jpg","mail":"[email protected]","gender":"Male","birthday":"1983-01-11",user_ip:"127.0.0.1"}]}.

I want to get it in this way. How to do it? Please let me know. Thanks in advance.

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6  
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. – Neal Jan 11 at 15:33

2 Answers

up vote 0 down vote

try this:

echo json_encode(array('users' => $arr, 'user_ip' => $user_ip_address));

on a side note: you should use PHP PDO class to connect and query the database.

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up vote 0 down vote

mysql_fetch_object returns an object, not an array. So, what are you doing by $arr[] = $obj; is just adding an object into an array. So, the actual structure of the $arr is something like

$arr => [
    [0] => Object(....),
    [1] => Object(....),
    ....
]

In your particular case, I assume you are fetching single row by primary key, so there are only one object.

THe simpliest way to fix this is to add a field to an object. I haven't worked with PHP since 5.3, so can't be sure, but it's as simple as adding

$obj->user_ip = $user_ip_address;

inside the loop.

Btw, a couple of questions for you:

  1. Why do you use loop if it should result in a single row?
  2. Why you are embedding the variable directly into SQL query. It's quite vulnerable, read about sql-injections and how to prevent it (one day I was really tired telling people here on SO to use PDO and prepared statements, so just go read about it),
  3. Have you read the documentation about mysql_fetch_object and array_merge? Why not (because if you have read it you wouldn't be asking the question).
  4. Tried debugging? E.g. attaching a debugger and watching for variables contents? Inserting logging or (God forgive me) debug print with echo?
  5. "Didn't work" is a quite bad error description. This time you was lucky, because the error was obvious. Next time, please, be more specific.
share|improve this answer
{"users":[{"id":"14","name":"Sonu Roy","picture":"image012.jpg","mail":"[email protected]","gender":"Male","birthd‌​ay":"1983-01-11"}]}. I am getting such output with my existing code, which is correct. But now I want to add other variable e.g $user_ip_address as user's data joining with current output data like this: {"users":[{"id":"14","name":"Sonu Roy","picture":"image012.jpg","mail":"[email protected]","gender":"Male","birthd‌​ay":"1983-01-11",user_ip:"127.0.0.1"}]}. – Neeraj Kumar Jan 11 at 16:34
Please, read the answer carefully. It is an answer. – J0HN Jan 11 at 18:58

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