Incorrect json output from mysql

Question

I made a request from MySQL and my code below is not working:

$email =$_POST['email'];  
$parola =$_POST['parola'];

$rez = mysqli_query($con, "SELECT * FROM utilizatori WHERE email='$email' AND parola='$parola'");

if($rez){
    $succes = 1;
} else {
   $succes = 0;
}

$data = array("succes" => $succes);
echo json_encode($data);

In postman it shows me this:

Which means your query failed, please echo `mysqli_error($con)` to see the error message. — catcon, Sep 16 '19 at 10:24
do `$rez = mysqli_query($con, "SELECT * FROM utilizatori WHERE email='$email' AND parola='$parola'") or die(mysqli_error($con));` and check error — Alive to die - Anant, Sep 16 '19 at 10:26
Also your SQL is open to SQL injection, please use prepare statement for better security — catcon, Sep 16 '19 at 10:26
See: https://stackoverflow.com/questions/15318368/mysqli-or-die-does-it-have-to-die — Dharman, Sep 16 '19 at 10:30
This output is correct because your code return, $data = array("succes" => $succes); json_encode($data); — Casper, Sep 16 '19 at 10:34
Use mysqli_fetch_assoc function to fetch result as an associative array, then combine with $data array. $row = mysqli_fetch_assoc($rez); $data = array("succes" => $succes, "data" => $row); — Casper, Sep 16 '19 at 10:42

score 0 · Answer 1 · edited Sep 16 '19 at 11:42

You missed the data variable or your query doesn't not return any output. Please try to use the MySQL query in Phpmyadmin or CLI. If it is OK then use the code like :

$email =$_POST['email'];  
$parola =$_POST['parola'];

$rez = mysqli_query($con, "SELECT * FROM utilizatori WHERE email='$email' AND parola='$parola'");

while($result = mysqli_fetch_array($rez))
{
 $data[] = $result;

}
if(mysqli_num_rows($rez) != 0) {
    $data = array("success" => "Success", "data" => $data );
} else {
    $data = array("success" => "Failed", "data" => array() );
}

echo json_encode($data);

Incorrect json output from mysql

1 Answers1