在ajax通话后Json没有回来

Question

HTML PAGE

$("#submit_login").click(function() {
    var username = $('input[name=user_email]').val();
    var password = $('input[name=user_password]').val();


    $.ajax({
        type: "POST",
        dataType: "json",
        url: "newExam.php",
        data: {
            name: username,
            pwd: password
        },
        cache: false,
        beforeSend: function() {
            $("#submit_login").val('Connecting...');
        },
        success: function(dataa) {
            if (dataa) {
                alert(dataa);
                $("#submit_login").val('success')
            } else {
                $("#submit_login").val('Login')
                alert('nodata')
            }
        }
    });
});

PHP PAGE

 <?php  
$servername = "localhost";
$username = "root";
$password = "12345";
$dbname = "uvm_server";
// Create connection
 $conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
  }  
  $email_php = $_POST['name'];  
  $pwd_php=$_POST['pwd'];
  echo $email_php;
 echo $pwd_php;
    $sql = "select name from user where email='$email_php'";  
  $result = mysqli_query($conn,$sql);  
  while($row=mysqli_fetch_array($result))  
   {  
   echo "<p>".$row['name']."</p>";  
   $rows[]=$row;
   } 
   echo '{"members":'.json_encode($rows).'}';
   ?> 

当我查看资源时，我能够获得输出。但成功的功能并没有得到充分发挥。我需要SQL查询的输出作为JSON，可以在javascript中访问。这样我就可以验证用户名和密码了。 它总是显示正在连接...我想知道我的ajax调用是否正确

Answer 1

摆脱<p>输出，这不是JSON。

while($row=mysqli_fetch_array($result))  
{  
    $rows[]=$row;
} 
echo json_encode(array('members' => $rows));