Get SQL Results - JSONarray cannot be converted to JSONobject

Question

I have modified my code over a couple of hours and I just cannot seem to parse value from android and use it in php, I alway get JSONarray cannot be converted to JSONobject.

<?php

include 'config.php';

$con=mysql_connect("$servername", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$dbname")or die("cannot select DB");

if( isset($_POST["json"]) ) {

     $value = $_POST["json"]; 
     $decoded_value = json_decode($value);

    $sql = "SELECT * from table WHERE Code = '".$decoded_value->{'msg'}."'";
    $result = mysql_query($sql);
    $json = array();

    if(mysql_num_rows($result)){
        while($row=mysql_fetch_assoc($result)){
        $json['myarray'][]=$row;
        }
    }

mysql_close($con);
echo json_encode($json); 

}
?>

Is that a mysql_* function I see..

Liam Sorsby
– Liam Sorsby

2015-11-27 19:07:05 +00:00
Commented Nov 27, 2015 at 19:07 — Liam Sorsby
– Liam Sorsby, Commented Nov 27, 2015 at 19:07

Logan Bentley · Accepted Answer · 2015-11-28 11:28:48Z

1

json_encode($json, JSON_FORCE_OBJECT)

The json_encode function has multiple parameters that you can pass in order to modify the encoding method (check it out here). In your instance, there is the option to force the encoding to be in an object form, rather than the default of either array or object.

edited Nov 28, 2015 at 11:28

answered Nov 27, 2015 at 19:03

Logan Bentley

2741 silver badge7 bronze badges

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

1 Comment

Tobias Liefke Over a year ago

While this code may answer the question, it would be better to explain how it solves the problem and why to use it. Code-only answers are not useful in the long run.

Collectives™ on Stack Overflow

Get SQL Results - JSONarray cannot be converted to JSONobject

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related