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Home Backend Development PHP Tutorial php 使用ajax发送密码过去,接收端发现密码被裁剪了

php 使用ajax发送密码过去,接收端发现密码被裁剪了

Jun 23, 2016 pm 01:42 PM

没怎么用过PHP。特来此论坛问下。
现在遇到的问题是,用户输入用户名与密码,进行登录,如果密码简单点,是可以传过去的。但如果密码是ab@;&1这种,则发现服务端得到的密码值被裁剪了,变成ab@;了。但是代码中紧接着,下面的部分也是这么操作的,居然又可以,实在没找出两处的差异,求解惑。

client部分的PHP

function getFirstU8Accid(){	var sUserName = document.getElementById("user_name").value.toLowerCase();	var sPassword = document.getElementById("password").value;	sUserName = escape(sUserName);	if(sUserName=="")	{		return("");	}	var sUrl = '/Server/a.php?t='+escape(Date())+'&a='+sUserName+'&b='+sPassword;        var xml = CreateXMLHttp();	xml.open("GET",sUrl,false);	xml.send();	var sLoginUser = xml.responseText;        //处理.......}                                        //具体调用处如下代码所示                                        //此处调用getFirstU8Accid()失败,具体错误是服务端的PHP并没有得到完整的密码内容。					Accid = getFirstU8Accid();					try{var guid = obj.GetClientGUID();}catch(e){return " Error:failed to call GetClientGUID,error message:" + e.message;}					var sUrl = '/Server/b.php?t='+escape(Date())+'&a='+user+'&b='+password+'&c='+sDomain;					var xml = CreateXMLHttp();					xml.open("GET",sUrl,false);					xml.send(null);                                        //此处能正常返回,因为b.php中的代码能正常获取到密码内容					return xml.responseText;[/color]a.php   b.php中的代码大概如下,其实基本一样,只是b中参数多了,多了一个解析的。[code=php]<?php $langs=$_COOKIE['LANGUAGES'];$langs=empty($langs) ? 'cn' : $langs;$language=include"lang-$langs.php";require("CmxConsts.php");require("CmxCommon.php");try {	$sUserName = $_GET["a"];	$sUserName = str_replace("|YJ|","&",$sUserName);	$sUserName = str_replace("|YJJ|","#",$sUserName);        //下面获取的密码就被裁剪了。	$sPassword = $_GET["b"];       //...................?>
Copy after login


给我的感觉,就是第一次失败后,再做第二次,只是多带了参数,就成功了。虽然服务端a,b两个PHP文件,但其实没啥区别。


回复讨论(解决方案)

& 字符在 url 参数串中是作为字段分隔符出现的,如果你的值中有该字符,则需要做 url 编码处理(%26)

版主真快哈,可是第二次提交时,好像也没有做处理啊。

另外,URL编码处理,有没有通用的代码呢?我是中英繁三种语言的,会有影响不?PHP不熟。我也刚用没多久,请问有相关的资料或小例子没?

js 不是有 encodeURI、encodeURIComponent 方法吗

问题解决了。第一处处理时,对密码没有escape操作,而接收处则进行了unescape处理。而第二处是配对的。谢谢版本的提示。

var sUrl = '/Server/a.php?t='+escape(Date())+'&a='+sUserName+'&b='+sPassword;

改?

var sUrl = '/Server/a.php?t='+escape(Date())+'&a='+encodeURIComponent(sUserName)+'&b='+encodeURIComponent(sPassword);

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