php代码没有给出正确的结果[关闭]

Question

the below code is not being executed

 (edited)
   <?php error_reporting(E_ALL); ini_set('display_errors', 1);


     print "Hello world!"; 

$con = mysqli_connect($host,$uname,$pwd,$db) or die(mysqli_error());


     $sql1="SELECT * FROM USERS WHERE username= 'aya'");


   $result = mysqli_query($sql1);
 if ($result && mysqli_num_rows($result) > 0) {
     print "Hea!"; 
 }



// json response array
$response = array("error" => FALSE);


    $email = $_POST['username'];
    $password = $_POST['password'];


    $sql="SELECT * FROM USERS WHERE username= 'aya'";
    if(mysqli_query($con,$sql))
    {
     echo json_encode($response);
    }



?>

the error is here in this part because when I remove it , the link give results

  $sql="SELECT * FROM USERS WHERE username= 'aya'") or die(mysqli_error())";


     $result = mysqli_query($sql);
     if ($result && mysqli_num_rows($result) > 0) {
     print "Hea!"; 
      }

Answer 1

The code in question has an obvious syntax and logical errors.
Correct your code as shown below:

 ...
 // stop script execution with error message if db connection fails
 $con = mysqli_connect($host, $uname, $pwd, $db) or die(mysqli_error());

 $sql1 = "SELECT * FROM USERS WHERE username= 'aya'";

 $result = mysqli_query($sql1);
 if ($result && mysqli_num_rows($result) > 0) {
     print "Hea!"; 
 }
...