Question

MySQL不会从表中获取数据;为什么我的代码没有工作？

我试图检查密码是否正确但是由于某种原因它不会得到密码。我还应该注意，这段代码适用于Windows VPS，但现在我使用的是Ubuntu 14，它无法正常工作。我尝试过chmod 777，但这也没有帮助。关于我能做什么的任何想法？

if(isset($_COOKIE['user'])){
    $username =  mysql_real_escape_string($_COOKIE['user']);
    mysql_connect($dbhost, $dbuser, $dbpass)or error("Could not connect: ".mysql_error());
    mysql_select_db($dbname) or error(mysql_error());
    $result = mysql_query("SELECT `id`, `username`, `password`, `ranks`, `active` FROM $table WHERE username='$username'") or die (mysql_error()); 
    list($id,$username,$password,$rank,$active) = mysql_fetch_row($result);
    if($_COOKIE['pass'] == $password) {
        $display = "Hello, $username!";
    } else {
        $display = "Invalid Password, attempted post data fail?";
    }
} else {
    $display = "You are not logged in!";
}

Answer 1

尝试：

$result = mysql_query("SELECT `id`, `username`, `password`, `ranks`, `active` FROM $table WHERE username='{$username}'") or die (mysql_error());

如果它不起作用，请告诉我如何在用户表中插入内容。您没有在PASSWORD()上使用password功能吗？

如果你这样做，试试这个：

$pass = $_COOKIE['pass'];
$result = mysql_query("SELECT `id`, `username`, `password`, `ranks`, `active` FROM $table WHERE username='{$username}' and password=PASSWORD('{$pass}')") or die (mysql_error()); 
if($result) {
    list($id,$username,$password,$rank,$active) = mysql_fetch_row($result);
    $display = "Hello, $username!";
} else {
    $display = "You are not logged in!";
}

MySQL不会从表中获取数据

1 个答案: