How to use the json_decode/json_encode in PHP?

Question

I am getting a problem in PHP. I am trying to pass the username/password from Android and checking the value in MySQL through PHP. While using json_decode and json_encode , json_decode works but json_encode does not work. But when I remove the json_decode , json_encode works, but I want both of them to work in my program.

Here is my code:

$a = $_POST['userpwd_value']; //Accesing the value from Android.

$b = json_decode($a); //Decoding android value using JSON.

$username = $b->{'username'}; //Assigning username from android to a variable.
$password = $b->{'password'}; //Assigning password from android to a variable.

echo $username.$password;

$check = mysql_query("select username,password from user where id=1");
$row = mysql_fetch_assoc($check);

//if($row['username']==$username && $row['password']==$password)
    $output[]=$row;
//else
    //$output[]=array("value"=>"false");
print(json_encode($output));

Where is the problem?

Answer 1

json_encode fails if the variable content is not correctly UTF-8 sequenced. If your database uses another charset, the variables contain special characters, then you should get an error there. (raise the error_reporting level or check json_last_error to find out)

Another problem with your specific code is that you first output something else:

 echo $username.$password;

This will invalidate the JSON output as a whole. If you have leading garbage, your browser will not decode the returned variables correctly. Also don't forget to send the appropriate header with your result using header("Content-Type: application/json");

---

  $b=json_decode($_POST['userpwd_value']); 
  $username=$b->{'username'}; 
  $password=$b->{'password'};

  $check = mysql_query("select username,password from user where id=1");
  $row = mysql_fetch_assoc($check);
  $row = array_map("utf8_encode", $row);

  if ($row['username']==$username && $row['password']==$password) {
      $output[] = $row;
  } else {
      $output[] = array("value"=>"false");
  }
  header("Content-Type: application/json");
  print(json_encode($output));

Answer 2

I would think there may be another level in the JSON-encoded data ( $a ). Does your echo statement show the correct username and password? If not, place var_dump($b); after your $b = json_decode($a); statement.

Answer 3

Replace the below two lines to the following.

Your original lines:

  $username=$b->{'username'};
  $password=$b->{'password'};

New lines:

  $username=$b->username;
  $password=$b->password;

After the replacement, check if it's working or not.