Question

我现在重新安装了XAMPP，我摆脱了大多数错误。

但我还有一个错误。这一个：

Warning: mysql_connect(): Access denied for user 'admin'@'localhost' (using password: YES) in D:\xampp\htdocs\login_form\login.php on line 7
Could not connect to database.

SQL DATABASE：

login > users > 
Id = 1
Username = admin
Password = admin

Id = 2
Username = login_system
Password = login_system

index.php：

<!DOCTYPE html>
<html>
  <body>
    <form action="login.php" method="POST">
      <p>Username : </p><input type="text" name="user">
      <p>Password : </p><input type="password" name="pass">
      <br>
      <input type="submit" value="Login">
    </form>
  </body>
</html>

的login.php

<?php
  //$ = variables?
  $username = "admin";
  $password = "admin";
  $hostname ="localhost";

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

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

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

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

  //Checks if the mySQL database matches the user inputs.
  $query = "SELECT * FROM users WHERE Username='".$myusername."' and Password='".$mypassword."'";

  $result = mysql_query($query,$database_handle);
  $count = mysql_num_rows($result);

  if ($count==1){
    echo 'Databse connection and code was successful!';
  }
 ?>

它说我需要添加更多我不能的信息。

这里的随机文字..

Answer 1

请尝试这个并告诉我这是否有效

<?php
  //$ = variables?
  $username = "root";
  $password = "test123";
  $hostname ="localhost:8080";

  $database_handle = mysql_connect($hostname, $username, $password) or die ("Could not connect to databse. Check your internet connection.");

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

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

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

  //Checks if the mySQL database matches the user inputs.
  $query = "SELECT * FROM users WHERE Username='".$myusername."' and Password='".$mypassword."'";

  $result = mysql_query($query,$database_handle);
  $count = mysql_num_rows($result);

  if ($count==1){
    echo 'Databse connection and code was successful!';
  }
 ?>

PHP - mySQL - 表单不会工作

1 个答案: