JavaScript解决一个带验证的Form两个Submit事件（一个页面保持不动【AJAX实现】，一个页面提交并跳转）的场景

<form class="form-horizontal" action="/biz/patent/edit" method="post" enctype="multipart/form-data" data-toggle="validator" role="form" novalidate="true" id="patentEditForm"> 
                    <button class="btn btn-primary" type="button" id="but_sub">保存</button>
                    <button class="btn btn-primary" type="button">保存返回列表</button>
                    <button class="btn btn-warning" type="button" onclick="window.location=‘/biz/project/list‘;">返回</button>
</form> 

$(document).ready(function(){
    var buttonId;

    $(‘#patentEditForm‘).validator().on(‘submit‘, function (e) {
        if (e.isDefaultPrevented()) {
            // handle the invalid form...
        } else {
            // everything looks good!
            if(buttonId == "but_sub"){
                var formData = new FormData($("#patentEditForm")[0]);
                $.ajax({
                    type:"POST",
                    url:"/biz/",
                    data: formData,
                    dataType:"JSON",
                    nsync:false,
                    cache: false,
                    contentType: false, 
                    processData: false,
                    success:function(date){
                        switch(date.status){
                        case 1:
                            $success("保存成功！");
                            break;
                        case 1009:
                            $error("保存失败！");
                            break;
                        default :
                            $error("没有提交信息");
                        }
                    },
                    error:function(){
                        $error("没有提交信息");
                    }
                });

                return false;
            }
        }
    });

    $(".btn-primary").click(function(){
        buttonId = $(this).attr("id");
        $(‘#patentEditForm‘).trigger(‘submit‘);
    });