Question

我目前正在为我的网站注册页面。我只是需要一些帮助，因为我不确定为什么当我输入注册详细信息（只有2个字段）时，它会转到感谢页面，但是当我检查我的数据库时，USERS下没有新的记录。以下是我的代码：

<?php

$host="*********"; // Host name 
$username="**********"; // Mysql username 
$password="**********"; // Mysql password 
$db_name="arihealthinfo"; // Database name 
$tbl_name="USERS"; // Table name 

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// username and password sent from form
$ParticipantID=$_POST["ParticipantID"];
$password=$_POST["UserPass"];

$sql = "SELECT ID FROM USERS WHERE ID = '$participantID'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

if($count==0)
    {
        $sql = "INSERT INTO USERS (ID, PASSWORD) VALUES ('$participantID', '$password')"; 
        echo "Thank you for registering, you can now login:";
        ?>
        <a href= http://www.arihealth.info/index.php>Login Page.</a>
        <?php
    } else {
        echo "Your ID has already been registered:";
        ?>
        <a href= http://www.arihealth.info/registerphp.php>Register Here.</a>
        <?php
    }
?>

Answer 1

因为您只是编写查询而不执行它。使用mysql_query()

$sql = "INSERT INTO USERS (ID, PASSWORD) VALUES ('$participantID', '$password')"; 
mysql_query($sql);

另外

$participantID!=$ParticipantID

将您的选择查询更改为

$sql = "SELECT ID FROM USERS WHERE ID = '$ParticipantID'";

您的查询已打开以进行sql注入检查How can I prevent SQL injection in PHP?

不要将普通密码存入数据库检查

http://php.net/manual/en/function.password-hash.php

http://php.net/manual/en/faq.passwords.php

注意： - 不推荐使用mysql而是使用mysqli或PDO

php注册页面不会注册用户

1 个答案: