PHP脚本不会将变量返回给AJAX

Question

The code is for simple login validation.

The PHP script doesnt seem to run when returning values to JavaScript but runs fine when there are not variables to return. So is there anything wrong or do I need to add anything else to return the values from PHP.

<?php
    header('Content-type: application/json; charset=utf-8');
    include("config.php");
    $formd=array();
    //Fetching Values from URL
    $username2=$_POST['username1'];
    $password2=$_POST['password1'];
    $query = mysqli_query($db,"SELECT username FROM login WHERE username = '$username2'");
    $result=mysqli_fetch_assoc($query);
    $sql=mysqli_query($db,"SELECT password FROM login WHERE username = '$username2'");
    $resul=mysqli_fetch_assoc($sql);
    $row = mysqli_fetch_array($query,MYSQLI_ASSOC);
    $count = mysqli_num_rows($query);
    $pass=$resul['password'];

    if((password_verify($password2,$pass))and($count==1)) { 
        echo "ds";
    } else {
        echo "no";
        $formd['no']="Invalid password or username"
    }
    mysqli_close($db); // Connection Closed
    echo json_encode($formd);
    exit();
?>

JavaScript

<script>
    $(document).ready(function(){
        $("#submit").click(function(){
           var username = $("#username").val();
          var password = $("#password").val();
          // Returns successful data submission message when the entered information is stored in database.
          var dataString = 'username1='+ username + '&password1='+ password;
          if(username==''||password=='') {
             alert("Please Fill All Fields");
          } else {
             // AJAX Code To Submit Form.
            $.ajax({
               type: "POST",
               url: "ajaxsubmit.php",
               dataType: "json",
               data: dataString,
               success: function(data){
                  alert(data.no);
               }
            });
          }
        return false;
      });
  });
</script>

Answer 1

your Php code has 2 echo statements

if((password_verify($password2,$pass))and($count==1))
{   
   echo "ds";  // first 
}
else
{
   echo "no";  // first
   $formd['no']="Invalid password or username"
}
  mysqli_close($db); // Connection Closed
  echo json_encode($formd);   // second

This way, your php script returns malformed JSON data and hence $.ajax() cannot handle it. Also, as others pointed out, please use the developer console to verify your script returns the expected data.