How to decode a json object sent through jquery ajax method in php file?

Question

This is the jquery ajax part in which i have sent json data to the php file.

$(function () {
 $('form').submit(function () {
   var fields = $(this).serializeArray();
   var ans_json = JSON.stringify(fields);
   console.log(ans_json);

   $.ajax({
     url: "results.php",
     type: "POST",
     data: ans_json,
     dataType: "json",
     success: function (result) {
       console.log(result);
     }
   });
   return false;
 });
});

Now i want to use this json data sent to the php page.How can i do it? I have done it this way but it returns null.

<?php
   echo json_decode('ans_json');
?>

I have a set of 10 questions which need to be answered. 3 questions were answered so got the below result.This is what i got in my console.

 [{"name":"answer_9","value":"a"},{"name":"answer_10","value":"a"}] quizzes.php:14

null

Alexander · Accepted Answer · 2013-01-13 11:51:55Z

2

You don't need to decode any JSON string at server-side if you encode properly your parameters.

You can use .serialize() to do the form serialization for you, and it's ready to send.

$(function () {
  $('form').submit(function () {
    var serialized = $(this).serialize();

    $.ajax({
      url: "results.php",
      type: "POST",
      data: serialized,
      ...
    });

    return false;
  });
});

Your parameters will be available in your $_POST as in any normal POST request. For example,

$ninth_answer = $_POST["answer_9"];

answered Jan 13, 2013 at 11:51

Alexander

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2 Comments

TeamA1 Over a year ago

Thanx it worked now, but what should be the dataType now?Is it json or somrthing else?

Alexander Over a year ago

dataType is for the server's response, not for the request. It depends of what you are returning

Community · Accepted Answer · 2023-11-17 19:38:25Z

1

You need to decode the POST variable. Currently you're decoding just a string which even isn't valid JSON.

<?php
$json_arr = json_decode($_POST['my_json'], true);
var_dump($json_arr);
echo "First name in json is:". $json_arr[0]['name'];
?>

and edit your javascript to reflect following: This posts my_json parameter with your json as an value. This makes it easy for PHP to recieve it using $_POST.

$.ajax({
 url: "results.php",
 type: "POST",
 data: {"my_json": ans_json},
 dataType: "json",
 success: function (result) {
   console.log(result);
 }
});

I suggest to read a little about those things:

http://api.jquery.com/jQuery.ajax/

https://www.php.net/manual/en/function.json-decode.php

edited Nov 17, 2023 at 19:38

CommunityBot

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answered Jan 13, 2013 at 11:36

Henrik Peinar

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