$_POST variable as value of array

Question

I have problem, that i want to get variable $_POST to add to value of an array. Then i will encode it to json. but the $_POST variable didn't give any value. EDIT i did edit the codes: Here my codes:

    <?php
    $temp = isset($_POST['temp']) ? $_POST['temp'] : '';
    $arr["temp_display"] = $temp;
    echo json_encode($arr);

and i hope the result will be like this

  {"temp_display": value_of_$temp}

I try to use same case but I add the value of $_POST to MySQL database and its Works. Here the code:

   <?php
    include("connect.php");
    $link=Connection();
    $temp=$_POST["temp"];
    $query = "INSERT INTO `templog` (`temperature`) 
    VALUES ('".$temp."')"; 
    mysql_query($query,$link);
    mysql_close($link);
    ?>

EDIT Here my jquery code:

     $(document).ready(function() {

        setInterval( update, 200);

          function update(){

               var cache = $('.deneme').children();

               $.getJSON('json.php', function(data) {
                $('.deneme').text(data.temp).append(cache);

             })

        }

    });

Thank you for Ur help...

whats the jquery code you used to retrieve date. post that too — J.K, Mar 31 '16 at 05:58
$arr["temp_display"] = isset($_POST['temp']) ? $_POST['temp'] : ''; — Rahul Vyas, Mar 31 '16 at 06:02
@RomanPerekhrest yeah undefined, may that the problem? Thanks — markonah, Mar 31 '16 at 06:18
**Danger**: You are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) and should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). You are **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that a modern API would make it easier to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from. — Quentin, Mar 31 '16 at 08:44
Your JavaScript is making a GET request and not passing any data at all, so I don't understand why you expect that put to data into `$_POST`. — Quentin, Mar 31 '16 at 08:45

Imran · Answer 1 · 2016-03-31T06:46:18.733

0

In the first code block this line $temp=$_POST["temp"]; is unnecessary as you have done the same thing in the previous line. You can try by displaying the json data using var_dump method like this:

echo "<pre>";
var_dump(json_encode('$arr');
echo "</pre>";

edited Mar 31 '16 at 06:46

answered Mar 31 '16 at 05:57

Imran

4,582
2
18
37

`string(19) "{"temp_display":""}"` The result like this, no value. Thanks – markonah Mar 31 '16 at 06:28

score 0 · Answer 2 · edited Mar 31 '16 at 08:41

0

<?php
  $temp = isset($_POST['temp']) ? $_POST['temp'] : '';
  $arr["temp_display"] = $temp;
  echo json_encode($arr);
?>

edited Mar 31 '16 at 08:41

Natalie Hedström

2,607
3
25
36

answered Mar 31 '16 at 05:58

J.K

1,382
1
11
27

$_POST variable as value of array

2 Answers2