Question

您好，我需要帮助来进行登录检查thourght db，任何人都可以对我说错误在哪里？这是代码

if (!empty($_POST['user']) && !empty($_POST['password']))
{
$user=stripslashes(trim($_POST['user']));
$password=stripslashes(trim($_POST['password']));
mysql_connect("localhost","root","");
mysql_select_db("project");
$check=mysql_query("SELECT * FROM utenti WHERE nome='$user' AND password='$password'");
if(mysql_num_rows($check)!0)
{
$details=mysql_fetch_array($check);
$_SESSION['display_name']=$details[0];
$_SESSION['username']=$details[1];
$_SESSION['password']=$details[2];
print "Login succesful. <p>  Level access: " . $details["type"] ;
}
else
{
print "Error";
}
}
else
{
print "Not all fields are compiled" ;
}

if ($details["type"] == "admin" )
{
    $admn = 1;
}
else
{
    $admn = 0;
}

我不知道为什么不起作用。预先感谢。

Answer 1

更改此：

if(mysql_num_rows($check)!0)

为此：

if(mysql_num_rows($check) != 0)

登录表单数据库检查错误

1 个答案: