更新SQL表中的数据

Question

I have created a webpage that gets the user to login, if they havent got an account then they can register and is creates a new user for them.

This is the table users in the database (user_registration)

    user_id    username   password    email                 wage
    1          johnsmith  jsmith99    [e-mail]      0
    2          davidscott dscott95    [e-mail]  0

When a new user registered, the default value for the wage is 0. I want them to be able to edit this through the use of a form - this is the HTML code for the form:-

<form method="post" action="<?php $_PHP_SELF ?>">
  <input name="new-wage" type="text" id="new-wage" class="textbox" placeholder="New Wage">
  <input name="update" type="submit" id="update" value="Update" class="btn">
</form>

PHP Code: (note- the php and the html form are all in the same file(index.php) )

 <?php
 if(isset($_POST['update']))
 {
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
   die('Could not connect: ' . mysql_error());
}

$user_id = $_SESSION['MM_Username'];;
$wage = $_POST['new-wage'];

$query = "UPDATE users ".
       "SET wage = '$wage' ".
       "WHERE user_id = '$user_id'" ;

mysql_select_db('user_registration');
$retval = mysql_query( $query, $conn );
if(! $retval )
{
die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully
";
mysql_close($conn);
}
else
{
?>

When i fill in this form and hit the update button, it reloads the webpage and displays Updated data successfully which is exactly what should happen. BUT - when I refresh the table in PHP My Admin, it keeps the wage as 0, and not what i entered in the form.

Has anyone got any ideas what might be wrong with my code?

Thanks in advance for any answers.

PS.- I know that I have used the functions mysql_* and not mysqli_, simply because i dont know how to convert it, can you also help me with this??

Answer 1

Based on what you said in the comments:

This:

$user_id = $_SESSION['MM_Username'];
$wage = $_POST['new-wage'];

$query = "UPDATE users ".
       "SET wage = '$wage' ".
       "WHERE user_id = '$user_id'" ;

Should be this:

$user = $_SESSION['MM_Username'];
$wage = $_POST['new-wage'];

$query = "UPDATE users ".
       "SET wage = '$wage' ".
       "WHERE username = '$user'";

As you wanted to update the records based on the user id but you passed in the username as the parameter!

However, I'm assuming you want to update by the user_id, as that is unique for each user, and thus a more robust design.

To do that you'd have to add another $_SESSION variable that stores the user id when they log in. Call it $_SESSION['MM_Id'] or whatever. The name is arbitrary.

You also have an unclosed else statement at the end of your code:

else
{
?>

Either close it or remove it.

As Mark said, your code is susceptible to SQL injection attacks, a commonly abused mistake. I'd recommend looking at PDO (my personal favorite). If you're new to object-oriented programming, it may be difficult, but it offers more power, flexibility, and security. Plus, object-oriented programming is an important concept anyway.

To protect your current code from injection, use mysql_real_escape_string()...

$user = mysql_real_escape_string($_SESSION['MM_Username']);
$wage = mysql_real_escape_string($_POST['new-wage']);

$query = "UPDATE users ".
       "SET wage = '$wage' ".
       "WHERE username = '$user'";

This prevents people form putting in special characters (such as quotes) to attack your queries. I gotta run, but if you read up on SQL injection, you'll understand!