【jquery采坑】Ajax配合form的submit提交(微擎表单提交,ajax验证,submit提交)

1、采坑:实现form的submit提交,在提交之前,进行ajax的不同校验,然后onsubmit=return check(),进行提交

1/1 目的:可以实现以 from的submit提交,然后还能进行数据的ajax动态验证。

2、html:

<form class="form-horizontal" method="post" role="form" onsubmit='return check()'>

    <button type="submit" name="submit" id="submit_color" value="yes" >提交订单</button><br/>

    <input type="hidden" name="token" value="{$_W['token']}" />
    <input type="hidden" name="source_from" value="午夜专区订购" />
    <input type="hidden" name="can_buy" value="1" />
    </div>
</form>

 

3、js:

function check(){
        var self=false;
        var delivery_date = $('input[name=delivery_date]').val();
        var delivery_way=$('#dispatch option:selected').val();
;
        var area_name = $("#area_span").html();
//配送方式需要区域限单判断 -- addBy xzz 2018/09/30
        if(delivery_way=='4'||delivery_way=='8'){
            $.ajax({
                type: "POST",
                dataType: "JSON",
                async: false,        //这里必须是同步,否则程序逻辑错误
                url: "{php echo $this->createMobileUrl('checkself2',['op'=>'send_ajax_order_num_check'])}",
                data: {'delivery_date':delivery_date,'area_name':area_name},
                success: function(msg){
                    console.log(msg);
                    if(msg.error==0){
                        alert('抱歉!'+delivery_date+'日 配送量已经饱和,请选择自提');
                        self=true;
                    }else{
                        return true;
                    }
                }
            });
        }else{
            //自提不限
            return true;
        }
        //配送库存是否满足限购,true不允许再下单,false允许下单
        if(self){
            return false;
        }
}

 

posted @ 2018-09-30 15:45  PHP急先锋  阅读(2282)  评论(0编辑  收藏  举报