php登录系统身份验证

通过html页面提供登陆窗口,根据用户的输入,用php执行sql语句来查找当前用户帐号的密码,找不到,显示用户名不存在,找到了根据用户输入的密码对比,不一致显示错误,一致就跳转到登陆后的页面。
求上面程序的代码!!!

login.html

<form id="form1" name="form1" method="post" action="checklogin.php">
<table width="400">
<tr>
<th width="27%" scope="row">用户名</th>
<td width="73%"><input name="user_name" type="text" id="user_name" /></td>
</tr>
<tr>
<th scope="row">密码</th>
<td><input name="user_pwd" type="password" id="user_pwd" /></td>
</tr>
<tr>
<th scope="row">请填写答案 </th>
<td><input name="yzm" type="text" id="yzm" size="5" />
123</td>
</tr>
<tr>
<td colspan="2" scope="row"><div align="center">
<input type="submit" name="Submit" value=" 登录 " />
</div></td>
</tr>
</table>
</form>

checklogin.php

<?php
session_start();
require_once("../include/class/Mysql.class.php");
require_once("../include/class/FuncAdmin.class.php");
require_once("../include/class/Users.class.php");

$db = new Mysql();
$db->getConnection(1);
$db->select_db();
$db->query("set names gb2312");

$f = new FuncAdmin();

if(!empty($_POST['Submit'])){
//判断登录
$user_name = $_POST['user_name'];
$user_pwd = $_POST['user_pwd'];
$yzm = $_POST['yzm'];

//加强验证
//此处很重要但没写出呢cxmcook

if(empty($user_name) || empty($user_pwd)){
$f->alert_back("用户名与密码不能为空!");
}

$user = new Users();
$sql = $user->get_select_sql(" where user_name='".$user_name."' and user_pwd='".$user_pwd."' ");
$rs = $db->query($sql);
$row = $db->fetch_array($rs);
//var_dump($row);
//die("<hr>");
$rownum = $db->num_rows($rs);
//die('=========='.$rownum);
if( $rownum >0 ){
//$row = $db->fetch_array($rs);
$_SESSION['admin'] = 'admin';

$_SESSION['user_id'] = $row['user_id'];
$_SESSION['user_name'] = $row['user_name'];
$_SESSION['user_realname'] = $row['user_realname'];
$_SESSION['user_last_logintime'] = $row['user_last_logintime'];
$_SESSION['user_last_loginIp'] = $row['user_last_loginIp'];
$_SESSION['user_loginnum'] = $row['user_loginnum'];

//header("Location:index.php");
echo "<script>
alert('登录成功!".$_SESSION['user_realname']."');
location='index.php';
</script>";

}else{
$f->admin_tip("用户名或密码错误!");
}

}

?>
温馨提示:答案为网友推荐,仅供参考
第1个回答  2019-03-23
login.html
<form
id="form1"
name="form1"
method="post"
action="checklogin.php">
<table
width="400">
<tr>
<th
width="27%"
scope="row">用户名</th>
<td
width="73%"><input
name="user_name"
type="text"
id="user_name"
/></td>
</tr>
<tr>
<th
scope="row">密码</th>
<td><input
name="user_pwd"
type="password"
id="user_pwd"
/></td>
</tr>
<tr>
<th
scope="row">请填写答案
</th>
<td><input
name="yzm"
type="text"
id="yzm"
size="5"
/>
123</td>
</tr>
<tr>
<td
colspan="2"
scope="row"><div
align="center">
<input
type="submit"
name="Submit"
value="
登录
"
/>
</div></td>
</tr>
</table>
</form>
checklogin.php
<?php
session_start();
require_once("../include/class/Mysql.class.php");
require_once("../include/class/FuncAdmin.class.php");
require_once("../include/class/Users.class.php");
$db
=
new
Mysql();
$db->getConnection(1);
$db->select_db();
$db->query("set
names
gb2312");
$f
=
new
FuncAdmin();
if(!empty($_POST['Submit'])){
//判断登录
$user_name
=
$_POST['user_name'];
$user_pwd
=
$_POST['user_pwd'];
$yzm
=
$_POST['yzm'];
//加强验证
//此处很重要但没写出呢cxmcook
if(empty($user_name)
||
empty($user_pwd)){
$f->alert_back("用户名与密码不能为空!");
}
$user
=
new
Users();
$sql
=
$user->get_select_sql("
where
user_name='".$user_name."'
and
user_pwd='".$user_pwd."'
");
$rs
=
$db->query($sql);
$row
=
$db->fetch_array($rs);
//var_dump($row);
//die("<hr>");
$rownum
=
$db->num_rows($rs);
//die('=========='.$rownum);
if(
$rownum
>0
){
//$row
=
$db->fetch_array($rs);
$_SESSION['admin']
=
'admin';
$_SESSION['user_id']
=
$row['user_id'];
$_SESSION['user_name']
=
$row['user_name'];
$_SESSION['user_realname']
=
$row['user_realname'];
$_SESSION['user_last_logintime']
=
$row['user_last_logintime'];
$_SESSION['user_last_loginIp']
=
$row['user_last_loginIp'];
$_SESSION['user_loginnum']
=
$row['user_loginnum'];
//header("Location:index.php");
echo
"<script>
alert('登录成功!".$_SESSION['user_realname']."');
location='index.php';
</script>";
}else{
$f->admin_tip("用户名或密码错误!");
}
}
?>
第2个回答  2013-04-10
BSPHP网络验证系统专业验证系统

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