Question

由于某种原因，我的代码更新了玩家密码，甚至没有看到是否有旧密码是正确的我先在这里检查＆＃34;否则if（$ pass！= mysql_result（$ result，0））＆＃34;但它没有用？

PHP

 <?php
            session_start();
        if(!isset($_SESSION["sess_user"])){
            header("location: index");
        } else {
            $my_player = $_SESSION['sess_user'];
        }
        if(isset($_POST["sumbit"])){

            $link= mysql_connect ("localhost:8889","root","root")or die("Could not connect: ".mysql_error());
            mysql_select_db("register") or die(mysql_error());
            $pass = $_POST['pass'];
            $newpass = $_POST['newpass'];
            $confirmnewpass = $_POST['confirmnewpass'];
            $result = mysql_query("SELECT password FROM login WHERE username='$my_player'");
            if(!$result) {
            $alert = "Failure ";
            }
            else if($pass!= mysql_result($result, 0)) {
            $alert = "incorect password";
            }
            if($newpass==$confirmnewpass)
            $sql=mysql_query("UPDATE login SET password='$newpass' where username='$my_player'");
            if($sql) {
            $alert = "You just changed your password to $newpass";
            }
            else {
            $alert = "newpassword field not entered";
            }
        }
?>

HTML

<div class="container">
<form role="form" class="form-signin" action="" method="POST">
    <h1 class="text-left">Change Password</h1><p>for <?=$my_player;?></p>
    <input type="password" name="pass" placeholder="Enter your password" class="form-control" autofocus required><br/>
    <input type="password" name="newpass" placeholder="Enter new password" class="form-control" required><br/>
    <input type="password" name="confirmnewpass" placeholder="Re-Enter new password" class="form-control" required><br/>
    <input type="submit" name="sumbit" value="Update Password" class="btn btn-lg btn-primary btn-block"><br/>
    <?=$alert;?>
</form>
</div>

Answer 1

试试这个：

添加：

$opass = mysql_fetch_array($result);

之后：

$result = mysql_query("SELECT password FROM login WHERE username='$my_player'");

并替换：

else if($pass!= mysql_result($result, 0)) {

人：

else if($pass!= $opass['password']) {

此外，定义警报时，您的代码不会停止。您需要替换：

 $alert = "Alert";

人：

die("Alert");

如果不是，即使发生错误，您的代码也会继续。

Answer 2

抛出错误后代码仍然继续。如果你把它放到功能中，并且在每个错误消息之后放置return: false;它将起作用

php代码不检查密码是否正确

2 个答案: