Question

我试图使用phpmyadmin数据库和xampp服务器在html中创建一个简单的登录但是这个错误显示* Parse错误：语法错误，意外的'$ password'（T_VARIABLE）*

这是我的代码。

<?php

$username = $_POST['user']
//*here's the Erro// $password = $_POST['pass']

$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);

mysql_connect("localhost", "root", "");
mysql_select_db("Login");

$result = mysql_query("select * from users where username = '$username' and password = '$password'") or die("Failed to query Database".mysql_error());

$row = mysql_fecth_array($result);

if ($row['username'] == $username && $row['password'] == $password ){
    echo "Login Succesful! Welcome" =.$row['username'];
}else {
    echo "Login Failed"
}

?>

Answer 1

你忘记了分号。

$username = $_POST['user'];

你也忘了这里。

echo "Login Failed";

Answer 2

您有多个错误：

<?php

$username = $_POST['user']; //<-------put ';'
//*here's the Erro// $password = $_POST['pass']


$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);



mysql_connect("localhost", "root", "");
mysql_select_db("Login");


$result = mysql_query("select * from users where username = '$username' and password = '$password'") or die("Failed to query Database".mysql_error());

$row = mysql_fecth_array($result);

if ($row['username'] == $username && $row['password'] == $password ){
echo "Login Succesful! Welcome =".$row['username']; //<-------change this line
}else {
echo "Login Failed";//<-------put ';'

}

?>

解析错误：语法错误，意外的'$ password'（T_VARIABLE）第4行

2 个答案: