3.5、【wnyxy001】作业提交 - 已锁定
3.5、后端开发基础php（下） <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta charset="utf-8"> <title>登录</title> </head> <body> <form method="post" action="login.php"> <div> <div style="height: 30px"> <span>账号</span> <input name="username" type="text" /> </div> <div style="height: 30px"> <span>密码</span> <input name="password" type="password" /> </div> <div style="height: 30px;"> <input type="submit" value="登录" /> </div> </div> </form> </body> </html>
<?php
$dbname="testdb";
$conn=mysqli_connect("localhost","root","root",$dbname);//链接数据库
if(!$conn)
{
echo "数据库连接错误！";
return;
}
$username="";
$password="";
//验证用户名密码
if(empty($_POST['username'])){
echo "用户名不能为空";
return;
} else if(empty($_POST['password'])){
echo "密码不能为空";
return;
}else{
//过滤单引号
$username=str_replace("'","‘",$_POST['username']);
$password=str_replace("'","‘",$_POST['password']);
}
//判断用户是否存在
$strsql="select 1 from admin where username='" . $username . "'" ;
$result=mysqli_query($conn,$strsql);
$rows=mysqli_num_rows($result);
if($rows)
{
//判断用户名密码是否正确
$strsq2="select 1 from admin where username='" . $username . "' and password = '" . $password . "'";
$result2=mysqli_query($conn,$strsq2);
$rows2=mysqli_num_rows($result2);
if($rows2)
{
echo "登录成功！";
}
else
{
echo "密码不正确！";
}
}
else
{
echo "用户名不存在";
}
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发表于 2018-7-12
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