数据已在数据库上更新，但Ajax失败

Question

我想更新数据库信息。当我使用这一行代码时，它会更新。不幸的是，这告诉我我的ajax无法成功执行，因为它没有继续执行成功功能。请帮忙。

HTML代码

<form class="EditEmployees" method="POST">
                                <span>Employee ID:</span><span id="employeeid"></span>
                                        <span>First Name</span>
                                        <input type="text" id="editfirstname" name="editfirstname" placeholder="" class="tb-style">

                                        <span>Last Name</span>
                                        <input type="text" id="editlastname" name="editlastname" placeholder="" class="tb-style">

                                            <span class="label" for="">Gender</span>
                                            <select class="tb-style" name="editgender" id="editgender">
                                                <option>Please Select</option>
                                                <option value = "male">Male</option>
                                                <option value="female">Female</option>
                                            </select>

                                                    <span class="label" for="dob">Date of Birth</span>
                                                    <input type="date" id="editdob" name="editdob" class="tb-style">

                                            <div class='input-group date' id='datetimepicker1' >
                                                <div class="input-group-addon">
                                                    <span class="label" for="dob">Date Hired</span>
                                                    <input type="date" id="editdatehired" name="editdatehired" class="tb-style">

                                <button type="" class="btn-EditSave w-100" name="">Save Changes</button>
                    </form>

jQuery代码。

$('.btn-EditSave').click(function(){
            // alert($('#editfirstname').val())
            $.ajax({
                   url:'ajax.php',
                   data:{id:id,
                        type:'update-records-via-id',
                        FirstName:$('#editfirstname').val(),
                        LastName:$('#editlastname').val(),  
                        dob:$('#editdob').val(),
                        datehired:$('#editdatehired').val(),
                        gender:$('#editgender').val()
                    },
                   dataType:'JSON',
                   type:'POST',
                   success: function(data){
                    alert("data saved")
                   }
                })
        })

PHP代码：更新数据库上的数据

function updaterecordviaid($id,$FirstName,$LastName,$gender,$dob,$datehired,$position){
    $conn = mysqli_connect("localhost","root","","test");
    $result = "update employees set FirstName = '$FirstName', LastName='$LastName', Gender = '$gender', DateOfBirth = '$dob', DateHired= '$datehired', PositionId='$position' where EmployeeId='$id'"; 
    if(mysqli_query($conn,$result)){
        echo json_encode(array('message'=>'Record Succesfully Updated!','FirstName'=>$FirstName,'LastName'=>$LastName));
        return;
    }
        echo json_encode(array('message'=>'Something went wrong'));
        return;
}