php与Validform() AJAX验证提交如何判断给予提示信息？-PHP Tutorial-php.cn

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php与Validform() AJAX验证提交如何判断给予提示信息？

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Jun 23, 2016 pm 01:33 PM

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

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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