php与Validform() AJAX验证提交如何判断给予提示信息?
我想实现在用户提交时用Validform插件验证,根据后台操作页面返回的信息,进行判断并提示用户 !没有用过Validform插件验证,不知道怎么用,烦请各位大侠,给指导一下,谢谢了!
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in.php
if($tagstype == "custinput"){ $name = addslashes($_REQUEST['txtname']); $pw = addslashes($_REQUEST['txtpw']); $sql = "insert $name,$pw form userinfo" $result = $db->insert($sql); if (!$result){ //返回提交信息和状态 print "<script> alert('数据插入失败!');history.back();</script>"; }else{ print "<script> alert('数据插入成功!');history.back();</script>"; //打印提示信息 }
user.php
<body><label id="spsn">用户名:</label><input id="txtname" name="txtname" type="text" size="40" datatype="*2-8" nullmsg="请填写用户名!" errormsg="用户名范围在2-8个字!" tip="请填写用户名!" /><label id="spsn">密 码:</label><input id="txtpw" name="txtpw" type="text" size="40" datatype="*6-16" nullmsg="请填写密码!" errormsg="密码范围在6-16个字!" /><input type="reset" id="btn_reset" name="btn_reset" value="重填" /><input type="submit" id="btn_sub" name="btn_sub" value="提交" /> </body><script type="text/javascript">$(function(){ /*$(".registerform").Validform(); //就这一行代码!; *【自定义提示效果】固定位置显示提示信息 **/ $.Tipmsg.r=null; var showmsg=function(msg){//假定你的信息提示方法为showmsg, 在方法里可以接收参数msg,当然也可以接收到o及cssctl; alert(msg); } $(".registerform").Validform({ tiptype:function(msg){ showmsg(msg); }, tipSweep:true, ajaxPost:true //根具后台返回值,判断是否成功,并提示用户! });})</script>
回复讨论(解决方案)
Validform 是 JQuery 的插件,与 php 毫无关系
无论客户端验证的如何严密,到了 php 都还是要进行验证的
盲目的相信客户端验证,必将遭到 CSRF攻击
Validform 是 JQuery 的插件,与 php 毫无关系
无论客户端验证的如何严密,到了 php 都还是要进行验证的
盲目的相信客户端验证,必将遭到 CSRF攻击
我刚接触,只是想在客户端这边,实现信息提示!让用户知道是否提交成功!知道错误地方!
试了好多的方案,都不怎么有用,有哪位大侠,能给个好的方案和实例!谢谢了!

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