简体   繁体   中英

php register page won't register user

I am currently working on a register page for my website. I just need some help as i'm unsure why when i enter the registration details (only 2 fields) it goes to the thank you page, but when i check my database there is not new record under USERS. Below is my code:

<?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
    }
?>

Because you just write your query not execute it. Use mysql_query()

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

Also

$participantID!=$ParticipantID

Change your select query to

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

Your query is open for sql injection check How can I prevent SQL injection in PHP?

Don't store plain password in to database check

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

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

Note :- mysql is deprecated instead use mysqli OR PDO

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM