Hello everyone,
I have a problem. I have a test database (used to learn MySQL) with a
table called customers and a field called email. The email field is set
to be unique, so I know that by using: $query = "select email from
customers where email = $email"; I will get one cell with the e-mail
address that matches the email address specified by the variable
$e-mail. Basically, I want to check to see if the e-mail address set in
the variable $email matches the email address in the field named email.
But I get an error at the end.
Here's the source:
Assumptions:
- $email is set from a previous page
- a database connection is open and the database to be used is selected
- session settings will send $error back to the form page if result of
if statement below is true
$query = "select email from customers" // Check for duplicate
entry
. "where email = $email";
$query = stripslashes($query);
$result = mysql_query($query);
$num_results = mysql_num_rows($result); // Get the number of
rows in database (integer)
for ($i = 0; $i < $num_results; $i++) {
$row = mysql_fetch_array($result); // Return
results in an associative array
}
if($row["email"] == $email) { // Check to see
if $email matches in database
$error = "$email has already been registered";
header("Location: signup.php?<?=SID?>");
exit;
}
And this produces the following error:
Warning: Supplied argument is not a valid MySQL result resource in
e:\localhost/book-o-rama/admin/signup_do.php on line 34.
What am I doing wrong? Any help would be appreciated...
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
