Php 解析mysql\u real\u escape\u string（）的最佳方法：无需关闭警告通知
为什么在我尝试登录我的页面时会出现这些警告 ？？我尝试浏览，大多数论坛都建议关闭通知，但我不认为这是一种好的编程方式。我只想了解如何修复此问题或警告，下面我将介绍这些错误Php 解析mysql\u real\u escape\u string（）的最佳方法：无需关闭警告通知,php,Php,为什么在我尝试登录我的页面时会出现这些警告 ？？我尝试浏览，大多数论坛都建议关闭通知，但我不认为这是一种好的编程方式。我只想了解如何修复此问题或警告，下面我将介绍这些错误 Warning: mysql_real_escape_string(): Access denied for user ''@'localhost' (using password: NO) in C:\xampp\htdocs\sot\sot.php on line 12 Warning: mysql_real_escap
Warning: mysql_real_escape_string(): Access denied for user ''@'localhost' (using password: NO) in C:\xampp\htdocs\sot\sot.php on line 12
Warning: mysql_real_escape_string(): A link to the server could not be established in C:\xampp\htdocs\sot\sot.php on line 12
Warning: mysql_real_escape_string(): Access denied for user ''@'localhost' (using password: NO) in C:\xampp\htdocs\sot\soft.php on line 13
Warning: mysql_real_escape_string(): A link to the server could not be established in C:\xampp\htdocs\sot\sot.php on line 13
Warning: mysql_query(): Access denied for user ''@'localhost' (using password: NO) in C:\xampp\htdocs\soft\sot.php on line 14
Warning: mysql_query(): A link to the server could not be established in C:\xampp\htdocs\sot\sot.php on line 14
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\sot\sot.php on line 15
<?php
session_start();
include_once 'dbconfig.php';
if(isset($_SESSION['user'])!="")
{
header("Location: admin.php");
}
if(isset($_POST['btn-login']))
{
$email = mysql_real_escape_string($_POST['email']);
$upass = mysql_real_escape_string($_POST['pass']);
$res=mysql_query("SELECT * FROM users WHERE email='$email'");
$row=mysql_fetch_array($res);
if($row['password']==md5($upass))
{
$_SESSION['user'] = $row['user_id'];
header("Location: admin.php");
}
else
{
?>
<script>alert('wrong details');</script>
<?php
}
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>ALEXORG</title>
<link rel="stylesheet" href="css/bootstrap.css" type="text/css" />
<link rel="stylesheet" href="style.css" type="text/css" />
</head>
<body id="login">
<div class="container">
<form class="form-signin" method="post">
<h2 class="form-signin-heading text-center">Muligence 1.0</h2>
<label for="inputEmail" class="sr-only">Email address</label>
<input type="text" name="email" id="inputEmail" class="form-control" placeholder="Enter Your email" required autofocus>
<label for="inputPassword" class="sr-only">Password</label>
<input type="password" name="pass" id="inputPassword" class="form-control" placeholder="Password" required>
<div class="checkbox">
<label>
<input type="checkbox" value="remember-me"> Remember me
</label>
</div>
<button class="btn btn-lg btn-primary btn-block" type="submit" name="btn-login">Sign in</button>
<a href="register.php">Sign Up Here</a>
</form>
</div> <!-- /container -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script src="js/bootstrap.js"></script>
</body>
</html>
警报（“错误的详细信息”）；
阿列克索格
多源1.0
电子邮件地址
暗语
记得我吗
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<?php
$DB_host = "localhost";
$DB_user = "root";
$DB_pass = "ca@19";
$DB_name = "zack";
try
{
$DB_con = new PDO("mysql:host={$DB_host};dbname={$DB_name}",$DB_user,$DB_pass);
$DB_con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(PDOException $e)
{
echo $e->getMessage();
}
include_once 'clo.crud.php';
$crud = new crud($DB_con);
?>
您正在使用PDO连接到数据库，但随后尝试使用ext/mysql（既不兼容又存在）来转义数据 您需要选择一个数据库API（PDO是一个不错的选择）并坚持使用它 演示如何使用PDO转义数据
是的，在尝试使用
mysql\u real\u escape\u字符串之前先连接到数据库。该问题是因为无法建立数据库连接。Oh和mysql_*
函数在PHP5.5中不推荐使用，您应该使用mysqli_*
替代方法。您可能正在连接mysqli_*将永远无法工作。无效语法。请向我们显示您的dbconfig.php
代码
$stmt = $pdo->prepare('SELECT * FROM employees WHERE name = :name');
$stmt->execute(array('name' => $name));