忘记密码脚本PHP mysqli数据库

嗨，我试图使忘记密码脚本，并成功地完成，但我得到一个问题。在forget.php中，当用户输入电子邮件时，脚本检查数据库中的电子邮件是否匹配，然后将激活码保存在数据库中，并将激活码发送到他的电子邮件地址

收到电子邮件后，点击链接，他需要在resetpass.php中重置密码表单。首先，它会检查激活代码是否与数据库中的代码匹配，如果匹配，则用户将输入新密码，并重置密码，但问题是密码没有更改，输入他的电子邮件的人会更改密码其他人的密码：D。我不知道这个剧本出了什么问题

忘记。php

resetpass.php

我在resetpass.php中有一个bug

您首先必须使用$\u GET['code']获取激活码并将其存储在

这是经过修改的代码，应该可以使用

<?php

if(isset($_POST['pass'])){
$pass = $_POST['pass'];
$acode=$_POST['code'];

$con=mysqli_connect("xxx","xxx","xxx","xxx");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = mysqli_query($con,"select * from login where activation_code='$acode'")
or die(mysqli_error($con)); 

if (mysqli_num_rows ($query)==1) 
{
$query3 = mysqli_query($con,"update login set Password='$pass' where activation_code='$acode'")
or die(mysqli_error($con)); 

echo 'Password Changed';
}
else
{
echo 'Wrong CODE';
}
}
?>

<form action="resetpass.php" method="POST">
<p>New Password:</p><input type="password" name="pass" />
<input type="submit"  name="submit" value="Signup!" />
<input type="hidden" name="code" value="<?php echo $_GET['code'];?>" />
</form>

---

您需要将操作从resetpass.php更改为resetpass.php？代码=

否则，当您提交表单时，代码将丢失

例如：不是没有错误的

<?php
if(isset($_GET['code'])) $acode = $_GET['code'];
else die("No code!");

$con=mysqli_connect("xxx","xxx","xxx","xxx");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
} else {
    $acode = mysqli_real_escape_string($con, $acode);
    $query = mysqli_query($con,"select * from login where activation_code='$acode'")
    or die(mysqli_error($con)); 
    if(mysqli_num_rows($query) == 0) {
        echo "Wrong code";
        die();
    } elseif (mysqli_num_rows ($query)==1 && isset($_POST['pass'])) {
        $pass = mysqli_real_escape_string($con, $_POST['pass']);
        $query3 = mysqli_query($con,"update login set Password='$pass' where activation_code='$acode'")
        or die(mysqli_error($con)); 

        echo 'Password Changed';
    }
}

?>

    <form action="resetpass.php?code=<?php echo $_GET['code'];?>" method="POST">
    <p>New Password:</p><input type="password" name="pass" />
    <input type="submit"  name="submit" value="Signup!" />
    </form>

但想想看：

您的代码非常不安全，最好尝试uniqidrand； 使用此代码，两个条目可能获得相同的代码 有人可以尝试所有的代码缺陷

---

在forget.php中，$code=$\u GET['activation\u code']；是什么？？我猜您正在生成随机代码：$code=rand100999$代码=100999兰特；那么，有什么可以阻止任何人通过脚本运行所有的可能性来清除/更改您的所有密码呢？此外，您还提供了登录详细信息mysqli_connectmysql.3gwebhosters.com、u777946695_root、melody、123、u777946695_用户；现在到internet，可能想更改它们。您可以使用mysql\u real\u escape\u string来防止SQL注入攻击：@Ravi mysql\u real\u escape\u string与SQL注入无关attack@Ravi不管它在这里是否相关，您可能希望使用mysqli\u real\u escape\u字符串，因为mysql*命令已被弃用。一些额外的解释可能会有所帮助。

<?php

if(isset($_POST['pass'])){
$pass = $_POST['pass'];
$acode=$_POST['code'];

$con=mysqli_connect("xxx","xxx","xxx","xxx");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = mysqli_query($con,"select * from login where activation_code='$acode'")
or die(mysqli_error($con)); 

if (mysqli_num_rows ($query)==1) 
{
$query3 = mysqli_query($con,"update login set Password='$pass' where activation_code='$acode'")
or die(mysqli_error($con)); 

echo 'Password Changed';
}
else
{
echo 'Wrong CODE';
}
}
?>

<form action="resetpass.php" method="POST">
<p>New Password:</p><input type="password" name="pass" />
<input type="submit"  name="submit" value="Signup!" />
<input type="hidden" name="code" value="<?php echo $_GET['code'];?>" />
</form>

<?php
if(isset($_GET['code'])) $acode = $_GET['code'];
else die("No code!");

$con=mysqli_connect("xxx","xxx","xxx","xxx");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
} else {
    $acode = mysqli_real_escape_string($con, $acode);
    $query = mysqli_query($con,"select * from login where activation_code='$acode'")
    or die(mysqli_error($con)); 
    if(mysqli_num_rows($query) == 0) {
        echo "Wrong code";
        die();
    } elseif (mysqli_num_rows ($query)==1 && isset($_POST['pass'])) {
        $pass = mysqli_real_escape_string($con, $_POST['pass']);
        $query3 = mysqli_query($con,"update login set Password='$pass' where activation_code='$acode'")
        or die(mysqli_error($con)); 

        echo 'Password Changed';
    }
}

?>

    <form action="resetpass.php?code=<?php echo $_GET['code'];?>" method="POST">
    <p>New Password:</p><input type="password" name="pass" />
    <input type="submit"  name="submit" value="Signup!" />
    </form>

<?php
if(isset($_GET['code'])) $acode = $_GET['code'];
else die("No code!");

$con=mysqli_connect("xxx","xxx","xxx","xxx");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
} else {
    $acode = mysqli_real_escape_string($con, $acode);
    $query = mysqli_query($con,"select * from login where activation_code='$acode'")
    or die(mysqli_error($con)); 
    if(mysqli_num_rows($query) == 0) {
        echo "Wrong code";
        die();
    } elseif (mysqli_num_rows ($query)==1 && isset($_POST['pass'])) {
        $pass = mysqli_real_escape_string($con, $_POST['pass']);
        $query3 = mysqli_query($con,"update login set Password='$pass' where activation_code='$acode'")
        or die(mysqli_error($con)); 

        echo 'Password Changed';
    }
}

?>

    enter code here
    <form action="resetpass.php?code=<?php echo $_GET['code'];?>" method="POST">
    <p>New Password:</p><input type="password" name="pass" />
    <input type="submit"  name="submit" value="Signup!" />
    </form>