Question

这是用于将用户登录到我正在构建的网站的PHP代码，但每当我运行此代码时，我得到以下输出。请帮帮我。

我正在使用WAMP服务器，启用了短打开标记。

输出

密码错误＆＃39 ;; }} else {echo＆＃39;无效的用户名＆＃39 ;; }＆gt; `

这是PHP代码

<?php
if(isset($_POST['name'])){

//connecting to data base 

$con=mysqli_connect("localhost:3306","root","","recommender");

// Check connection
if (mysqli_connect_errno())
{

echo "Failed to connect to MySQL: " . mysqli_connect_error();

}


  $q = "select * from user where name ='".mysql_real_escape_string($_POST['name'])."'";
  $res = mysql_query($q) or die("wrong query");
    $row = mysql_fetch_assoc($res);
  if(!empty($row))
{
    if($_POST['password']==$row['password'])
    {
        $_SESSION['name']=$_POST['name'];
        header('Localhost/home.html');
    }
    else
    {
        echo '<centre><font color="red">wrong password</font></centre>';

    }
}
else
{
    echo '<centre><font color="red">invalid username</font></centre>';

}

}
?>

Answer 1

上有语法错误

$q = "select * from user where name ='.mysql_real_escape_string($_POST['name'])."'";<br>
                                     ^^ lost "

用

解决这个问题

$q = "select * from user where name = '". mysql_real_escape_string($_POST['name']) . "'";

Answer 2

更改此行：

$q = "select * from user where name ='.mysql_real_escape_string($_POST['name'])."'";<br>

到

$q = "select * from user where name = `".mysql_real_escape_string($_POST['name'])."` LIMIT 1";

php代码以html格式打印

2 个答案: