Passing a Jquery array of objects into a php array

Question

I have an array in jQuery that I am trying to convert to a PHP array by using Post:

$.post("http://www.samepage.com", {final_slip: JSON.stringify(final_slip)});

When I pass this dynamically created array "final_slip" into it :

[final_bet_game { final_user_id="1",  final_game_id="1",  final_game_type="spread_2",final_price="10", final_odds="1.8"},  final_bet_game { final_user_id="2",  final_game_id="3",  final_game_type="spread_2",final_price="1", final_odds="2.8"},  final_bet_game { final_user_id="3",  final_game_id="14",  final_game_type="spread_32",final_price="140", final_odds="1.8"},  final_bet_game { final_user_id="4",  final_game_id="1",  final_game_type="spread_2",final_price="10", final_odds="2.8"}, ]

I get this php outputted :

$data =  $_POST['final_slip'];
print_r ( $data);

[{\"final_user_id\":\"1\",\"final_game_id\":\"1\",\"final_game_type\":\"spread_2\",\"final_price\":\"211\",\"final_odds\":\"1.8\"},{\"final_user_id\":\"1\",\"final_game_id\":\"2\",\"final_game_type\":\"spread_2\",\"final_price\":\"212\",\"final_odds\":\"1.8\"},{\"final_user_id\":\"1\",\"final_game_id\":\"parlay\",\"final_game_type\":\"\",\"final_price\":\"021\",\"final_odds\":\"\"}]

I tried to use the json_decod, but Im not getting any results. How can I get this to a usable php array? Would I be better off using ajax?

You're already using AJAX. $.post() is a jQuery shorthand AJAX method. — Jay Blanchard
– Jay Blanchard, Commented Dec 9, 2015 at 18:40
that's totally invalid json. what exactly is in final_slip before you stringify it? — Marc B
– Marc B, Commented Dec 9, 2015 at 18:41

Community · Accepted Answer · 2020-06-20 09:12:55Z

3

$.post("http://www.samepage.com", {myJsonString: JSON.stringify(myJsonString)});
But when I then try to access it in PHP I am not seeing any results.

And that tells you that you have another problem - PHP is not reporting errors. You want to turn error_reporting on.

In this case the likely cause is that your data is arriving into the $_POST array, and $myJsonString is not defined (automatic definition of parameters has been deprecated since PHP 5.3.0, and no longer available since PHP 5.4.0).

You should either do

if (array_key_exists('myJsonString', $_POST)) {
    $myJsonString = $_POST['myJsonString'];
} else {
    die("myJsonString not in _POST");
}

or try straight

$result = json_decode($_POST['myJsonString'], true);

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Dec 9, 2015 at 21:32

LSerni

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Comments

nick · Accepted Answer · 2015-12-09 23:33:50Z

0

The problem was the backslashes in the outout needed to be stripped out:

  $result = $_POST['json_string'];  
  $test =  urldecode(str_replace("\\","",$result));  
  $test2 = (json_decode($test,true));
  print_r($test2);

answered Dec 9, 2015 at 23:33

nick

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