Question

我正在尝试为我的网站制作登录系统。

我已经尝试/看了几个视频来解决我的问题，但我仍然不确定我做错了什么。

为了更好地查看我的代码，请查看此链接：http://prntscr.com/2jeye5

解析错误：syntax error, unexpected '$myusername' (T_VARIABLE) in C:\xampp\htdocs\login.php on line 10

第10行以下面代码中的$myusername = $_POST['user'];开头。

<?php
    $dbhandle = mysql_connect($hostname, $password) or die("Could not connect to database");

    $selected = mysql_select_db("login", $dbhandle)

    $myusername = $_POST['user'];
    $mypassword = $_POST['pass'];

    $myusername = stripslashes($myusername);
    $mypassword = stripslashes($mypassword);

    $query = "SELECT * FROM users WHERE Username='$myusername' and Password=$mypassword'";
    $result = mysql_query($query);
    $count = mysql_num_rows($result);

    if($count==1){
        echo 'It worked';
    }
?>

Answer 1

终止第8行，您在$mypassword之前错过了单引号。

Answer 2

缺少半结肠

$selected = mysql_select_db("login", $dbhandle)

应该是

$selected = mysql_select_db("login", $dbhandle);

由于用户名和密码可能是字符串，因此应将其括在引号

中

$query = "SELECT * FROM users WHERE Username='$myusername' and Password=$mypassword'";

应该是

$query = "SELECT * FROM users WHERE Username='$myusername' and Password='$mypassword'";

Answer 3

试试这个：

$dbhandle = mysql_connect($hostname, $password) or die("Could not connect to database");

$selected = mysql_select_db("login", $dbhandle);

if(isset($POST['user']) && isset($POST['pass']))
{
  $myusername = stripslashes($POST['user']);
  $mypassword = stripslashes($POST['pass']);

   $query = "SELECT * FROM users WHERE Username='$myusername' and Password='$mypassword'";
   $result = mysql_query($query);
   $count = mysql_num_rows($result);

   if($count==1){
   echo 'It worked';
}
else
{
    echo 'Not worked';
}

第10行的mySQL解析错误

3 个答案: