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Home Backend Development PHP Tutorial 急PHP+Ajax实现用户登录验证问题求解?

急PHP+Ajax实现用户登录验证问题求解?

Jun 23, 2016 pm 02:08 PM

表单里用户名和密码两个属性,单击按钮触发AJAX函数,这个AJAX函数提取表单中“用户名”和“密码”的值发送到后台程序验证,php写的后台程序,php程序分别对“用户名”和“密码”进行验证,用户名出错给出提示,传回给ajax,显示到

中,同理密码出错,显示到
中,现在用户名和密码能够正确传到PHP程序中,做出正确验证,但是xmlHttp.responseText,如何能实现上面的效果呀?急!!!


回复讨论(解决方案)

贴出你的代码!

你需要在回调函数中有条件的对不同的目标赋值,那就要回传有控制信息

// JavaScript Documentvar xmlHttp;function S_xmlhttprequest(){	if(window.ActivXObject){				xmlHttp=new ActiveXObject('Mcrosoft.XMLHTTP');		}else if(window.XMLHttpRequest){			xmlHttp=new XMLHttpRequest;		}	}function php100(){	S_xmlhttprequest();	user=document.myform.user.value;	pwd=document.myform.pwd.value;   var postStr = 'user='+ user+'&pwd='+pwd;	xmlHttp.open('POST','login3.php',true);	 //定义传输的文件HTTP头信息 	xmlHttp.onreadystatechange=byphp;	xmlHttp.setRequestHeader('Content-Type','application/x-www-form-urlencoded');	 //发送POST数据	 xmlHttp.send(postStr);	}function byphp(){	var byphp100=xmlHttp.responseText;	document.getElementById('userInfo').innerHTML=byphp100;	}// JavaScript Document// JavaScript Document
Copy after login
form name="myform" method="post" action=""> 用户名:<input type="text"  name="user" value=""  size=10;/>   <br> 密 码:<input type="password" name="pwd" value="" size=10;/>   <br> <input type="button"  name="button" value="登陆"  onclick="php100();"/> </form><div id="userInfo"></div><div id="passInfo"></div>
Copy after login
<?php $username=$_POST['user']; //获取用户编号 $password=$_POST['pwd']; //获取密码编号 if($username=='09430102'){	  echo "用户名正确";	 }else{		     echo "用户名不正确";		 } if($password=='my0917'){	 echo "密码正确";	 }else{		     echo "密码错误";		 }?>
Copy after login

“用户名”验证返回的信息显示到userinfo那个div中,密码显示到passinfo那个div中;代码应该怎么改?希望详细些,新手谢谢了!

你没有回传啊

var byphp100=xmlHttp.responseText;
document.getElementById('userInfo').innerHTML=byphp100;这不是回传处理了吗?

function byphp(){  var byphp100=xmlHttp.responseText;  if(/用户/.test(byphp100))    document.getElementById('userInfo').innerHTML=byphp100;  else    document.getElementById('passInfo').innerHTML=byphp100;}
Copy after login
Copy after login

index.html





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密码







ajax.js
var http_request;
function send_request(url,method) {
  http_request = false;
  if(window.XMLHttpRequest) {
  http_request = new XMLHttpRequest();
  if (http_request.overrideMimeType) {
   http_request.overrideMimeType('text/xml');
  }
  }
  else if (window.ActiveXObject) {
  try {
   http_request = new ActiveXObject("Msxml2.XMLHTTP");
  } catch (e) {
   try {
   http_request = new ActiveXObject("Microsoft.XMLHTTP");
   } catch (e) {}
  }
  }
  if (!http_request) {
  window.alert("不能创建XMLHttpRequest对象实例.");
  return false;
  }
  switch(method){
case 1:http_request.onreadystatechange = chk;break;
   }
  http_request.open("GET", url, true);
  http_request.send(null);
}

function check(){
send_request("check.php?action="+document.getElementById('pw').value,1);


function chk() {
if (http_request.readyState == 4) {
if (http_request.status == 200) {
document.getElementById("user").innerHTML="";
document.getElementById("user").value=http_request.responseText;
    } else {
     alert("您所请求的页面异常。");
    }
   }else {
document.getElementById("user").innerHTML="正在读取数据中……";
}
}


check.php


if(存在){
echo"正确";
}else{
echo "不正确";
}
?>

check.php中判断下$_POST['username']和$_POST['pw']

function byphp(){  var byphp100=xmlHttp.responseText;  if(/用户/.test(byphp100))    document.getElementById('userInfo').innerHTML=byphp100;  else    document.getElementById('passInfo').innerHTML=byphp100;}
Copy after login
Copy after login

代码运行,咋还是不能将“用户名”和“密码”验证返回信息显示到不同的div中?回传控制信息怎么写?谢谢!--新手!

传json吧

留名!   用jquery不是更好吗

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