JQuery 调用ajax方法 返回状态 canceled错误

问题：提交表单，ajax返回错误 状态为 canceled

原因：button按钮类型为type=submit ，script中又自定用botton按钮点击提交ajax，造成冲突。

解决方法：button按钮类型改为 type=button

出错代码

 <form>
 <!--表单内容代码省略-->   
    <div class="form-group">
        <div class="col-sm-offset-2 col-sm-10">
            <button type="submit" class="btn btn-primary" id="register">注册</button>
            <!--上面行代码错误，应该改为下面行代码-->
            <!--<button type="button" class="btn btn-primary" id="register1">注册</button>-->
        </div>
    </div>
</form>

javascript代码

<script type="text/javascript">
        $(function () {
            $('#register').on('click',function () {
//                alert($('#login').serialize());
                $.ajax({
                    type:'post',
                    url:"{:url('insert')}",
                    data:$('#login').serialize(),
                    dataType:'json',
                    success:function (data) {
                        switch (data.status)
                        {
                            case 1:
                                alert(data.message);
                                window.location.href="{:url('index/index/index')}";
                                break;
                            case 0:
                            case -1:
                                alert(data.message);
                                window.location.href="{:url('index/user/register')}";

                        }
//                        alert(data.status);
//                        if(data.status==1){
//                            alert(data.message);
//                            window.location.href="{:url('index/index/index')}";
//                        }else {
//                            alert(data.message);
//                            window.location.href="{:url('index/user/register')}";
//                        }
                    },
                })
            })

        })

    </script>