PHP 4.4.1 JSON ENCODE

Question

I have a jQuery login form that uses $post function to send credentials to the php script that checks the credentials againts the users in the database.

I have been using PHP 5.2 and have been using echo json_encode($data); in my php scripts to send back error messages to the login form witout any issues. The problem is now I have a client that has PHP 4.4.1 installed and it doesn't recognise this command and throw the following error.

Fatal error: Call to undefined function: json_encode()

Is there an equivalent to json_encode() in version PHP 4.4.1?

Any help would be great.

Thanks in advance.

PHP 4.4.1 was released October 31, 2005. There is no reason why they haven't updated by now. I would suggest telling them to PHP rather than changing your code. — user142162
– user142162, Commented Dec 17, 2011 at 17:08

Richard · Accepted Answer · 2012-10-26 07:55:48Z

2

You should use this wrapper function: http://www.boutell.com/scripts/jsonwrapper.html. All you need to do with this solution is to include that json_encode.php and it will do a function_exists check. If it is doesn't exist, it adds the function.

edited Oct 26, 2012 at 7:55

answered Dec 17, 2011 at 17:09

Richard

4,4267 gold badges38 silver badges58 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

plaes · Accepted Answer · 2011-12-17 17:09:09Z

1

JSON support has been in PHP core since version 5.2.0. So you need to use PECL::json instead.

Also, consider updating to PHP5 as support for PHP4 has been discontinued since 2007-12-31 and it might have open security issues.

answered Dec 17, 2011 at 17:09

plaes

32.9k11 gold badges95 silver badges89 bronze badges

Comments

Explosion Pills · Accepted Answer · 2011-12-17 17:11:34Z

0

You can use Services_JSON (part of PEAR), which is php4 compatible. If you don't have access to that either, then I'm afraid you will just have to write your own function:

function my_json_encode($arrayus) {
   $newarr = '{';
   foreach ($arrayus as $key => $val) {
      $newarr .= '"' . $key . '":"' . $val . '",';
   }
   $newarr = substr($newarr, 0, strlen($newarr) - 1);
   $newarr .= '}';
   return $newarr;
}

The above is not compatible with json_encode(), but it may work for your purposes.

answered Dec 17, 2011 at 17:11

Explosion Pills

192k56 gold badges341 silver badges417 bronze badges

Collectives™ on Stack Overflow

PHP 4.4.1 JSON ENCODE

3 Answers 3

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related