help

[affordit]

Does anyone know whats wrong with this I am getting a blank page

 

mysql_connect(mysql,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

 

$query = "Select name, email from test Where name = '$name' OR email = '$email'";

$results = mysql_query($query) or die(mysql_error()."<br /><br />".$query);

$num=mysql_numrows($results);

if ($_POST['name'] = ($results['name'])); {

print "Name exists";

 

}elseif{

 

($_POST['email'] = ($results['email'])); {

print "Email exists";

exit();

}

else{

 

mysql_connect(mysql,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

 

$query = "INSERT INTO `test` (`id`, `company`, `name`, `address`, `city`, `state`, `zip`, `phone`, `acct_num`, `email`, `start_date`, `end_date`, `remind_date`, `lost_password`, `lost_password_answer`) VALUES (' ','$company','$name','$address','$city','$state','$zip','$phone','$acct_num','$email','$start_date','$end_date','$remind_date','$lost_password','$lost_password_answer')";

mysql_query($query) or die ("<table align='center' width='300'><td align='center'><tr><b>Email exists</b></td></tr></table>");

 

#  if the user submitted a password

if( isset($psword) )

{

  #  get the user id 

  $id = mysql_insert_id( );

 

  #  and insert the login details

  $query = "INSERT INTO login VALUES ('','$name','$psword','$email','$lost_password_answer','$end_date','$valid')";

  mysql_query($query) or die ("The username you entered already exists, please choose another");

 

$to = "$email";

$subject = "Your Registration";

$message="Thank You for registering with us $name your password is $psword.";

$headers = "From: webmaster@[email protected]";

$sent = mail($to, $subject, $message, $headers) ;

if($sent)

{print "<br><table align='center' width='300'><td align='center'><tr><b>Your information has been saved and we have sent you an mail to $email with your password. To activate your account just login with your username $name and the password included in the Email.</b></td></tr></table>";

print "<table align='center' width='300'><td align='center'><tr><b><font color='#FF6600'>Your registration begins

$start_date2<br>and will expire on $end_date2.<br>The first time you log in<a href ='first_login.php'>Do it here</a></b></td></tr></table>";

}

else

{print "We encountered an error sending your mail"; }

 

}

  }

mysql_free_result( $result );

mysql_close();

?>

 

 

 

 

 

[haku]

You should try indenting your code differently, you would see the problem instantly. It lies here:

 

if ($_POST['name'] = ($results['name'])); {
print "Name exists";

}elseif{

($_POST['email'] = ($results['email'])); {
print "Email exists";
exit();
}
else{

 

Lets indent that code differently:

 

if ($_POST['name'] = ($results['name']));
{
    print "Name exists";
}
elseif
{
    ($_POST['email'] = ($results['email']));
    {
        print "Email exists";
        exit();
    }
    else
    {

 

You didn't give any condition for your 'elseif' statement, which means that it always enters that. Then for the condition you did give, you used ($_POST['email'] = $results['email']). This (attempts) to assign the value of $results['email'] to $_POST['email'] (and fails, as this is not possible) rather than checking for equality. Use two equals signs to check for equality, not one.

[psychowolvesbane]

Are you sure you want the semi colons on the if statements as well???

 

...
if ($_POST['name'] = ($results['name'])); //<--Here
{
    print "Name exists";
}
elseif ($_POST['email'] = ($results['email'])); //<-- and Here
{
     print "Email exists";
     exit();
}
else
{
...

 

Try removing them.

[haku]

heh, good point, I missed that altogether!

 

Thats more likely your problem.

[affordit]

No Still a blank page. I been working at it still can't get it

[revraz]

Do you have error reporting turned on in your php.ini?

[affordit]

I believe it is when I screw other things up I get errors. I'm not sure and I don't know how to check the php is on the yahoo server

[revraz]

Try adding these two lines to the top of your code

 

ini_set ('display_errors', 1);

error_reporting (E_ALL);

 

[affordit]

Still a blank page

[revraz]

Your IF's only have one =

 

if ($_POST['name'] = ($results['name'])); {

 

should be

 

if ($_POST['name'] == ($results['name'])); {

 

Check your other one's as well.

[affordit]

OK added the ++ but still got a blank page so I took out the ; after each IF statement and goty these errors

 

Notice: Use of undefined constant mysql - assumed 'mysql' in /sharon/submit_registration.php on line 72

 

Notice: Undefined variable: acct_num in /sharon/submit_registration.php on line 94

 

Notice: Undefined variable: remind_date in /sharon/submit_registration.php on line 94

 

Email exists

 

[affordit]

OK I meant == not ++ But now I got no errors but when I put in an email and Username that I know both exist in one of the records it goes past the if name part and goes right to Email exists.

 

Here is the updated code

 

mysql_connect(mysql,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

 

$query = "Select * from test Where name = '$name' OR email = '$email'";

$results = mysql_query($query) or die(mysql_error()."<br /><br />".$query);

$num=mysql_numrows($results);

if ($_POST['name'] == ($results['name'])) //<--Here

{

    print "Name exists";

exit();

}

elseif ($_POST['email'] == ($results['email'])) //<-- and Here

{

    print "Email exists";

exit();

}

else

{

[revraz]

You need to use mysql_fetch_assoc in order to get the info from the DB.

 

Also, this line is wrong

 

$num=mysql_numrows($results);

 

should be

 

$num=mysql_num_rows($results);

 

But you are not even checking $num so it's not doing anything.

 

[affordit]

I am not familiar with mysql_fetch_assoc Iguess I need to go and take a look at some documentation

[trq]

Also, you use two variables $name and $email in your query. Where are these defined?

[affordit]

They are set on a form. What it is is a registration form but when a new person registers I want to insure that the username  and email are not duplicated in the login or clients tables.

 

What happens is I insert into the clients table all the information and then insert into the login table the login info. The username and email are set to unique in the tables.

[revraz]

If the form is not on the same page, then they are not being passed so they won't get inserted into the DB.

[trq]

If they are sent via a form using the POST method then the variables will be within the $_POSTarray, they don't just magically appear.

 

What it is is a registration form but when a new person registers I want to insure that the username  and email are not duplicated in the login or clients tables.

 

Then your code ought to be....

 

<?php

 // connect to database

 if (isset($_POST['submit'])) {
   
   $name = mysql_real_escape_string($_POST['name']);
   $email = mysql_real_escape_string($_POST['email']);
   
   $sql = "SELECT name FROM test WHERE name = '$name' || email = '$email'";
   if ($result = mysql_query($sql)) {
     if (mysql_num_rows($result)) {
       echo "username or email already in use";
     } else {
       // do registration
     }
   }
 }

?>

[affordit]

I have no prob geting them to insert with this part of the code:

 

$query = "INSERT INTO `test` (`id`, `company`, `name`, `address`, `city`, `state`, `zip`, `phone`, `acct_num`, `email`, `start_date`, `end_date`, `remind_date`, `lost_password`, `lost_password_answer`) VALUES (' ','$company','$name','$address','$city','$state','$zip','$phone','$acct_num','$email','$start_date','$end_date','$remind_date','$lost_password','$lost_password_answer')";

mysql_query($query) or die ("<table align='center' width='300'><td align='center'><tr><b>Email exists</b></td></tr></table>");

 

#  if the user submitted a password

if( isset($psword) )

{

 #  get the user id  

 $id = mysql_insert_id( );

 

 #  and insert the login details

 $query = "INSERT INTO login VALUES ('','$name','$psword','$email','$lost_password_answer','$end_date','$valid')";

 mysql_query($query) or die ("The username you entered already exists, please choose another");

 

$to = "$email";

$subject = "Your Registration";

$message="Thank You for registering with us $name your password is $psword. You must go to affordit.us/sharon/first_login.php to log in so that your Email will be verified. Then you will be able to login on the normal login page. Thank you again and if you have any questions or concerns please feel free to contact us through our contact page. Patrick St. Charles Affordable Everything";

$headers = "From: [email protected]";

$sent = mail($to, $subject, $message, $headers) ;

if($sent)

{print "<br><table align='center' width='300'><td align='center'><tr><b>Your information has been saved and we have sent you an mail to $email with your password. To activate your account just login with your username $name and the password included in the Email.</b></td></tr></table>";

print "<table align='center' width='300'><td align='center'><tr><b><font color='#FF6600'>Your registration begins

$start_date2<br>and will expire on $end_date2.<br>The first time you log in<a href ='first_login.php'>Do it here</a></b></td></tr></table>";

}

else

{print "We encountered an error sending your mail"; }

 

}

 }

mysql_free_result( $result );

mysql_close();

?>

 

That part works fine but when I try to add this to the top of the page:

 

include("sharons_dbinfo.inc.php");

mysql_connect(mysql,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

 

$query = "Select * from test Where name = '$name' OR email = '$email'";

$results = mysql_query($query) or die(mysql_error()."<br /><br />".$query);

$num=mysql_numrows($results);

if ($_POST['name'] == ($results['name'])) //<--Here

{

   print "Name exists";

exit();

}

elseif ($_POST['email'] == ($results['email'])) //<-- and Here

{

    print "Email exists";

exit();

}

else

{

 

and enter values I know exist into the form it skips over the part where it checks the name and displays the resuut from the email check "EMAIL EXISTS" I don't get it.

 

 

[trq]

As oyu yourself said, you best have a look at how mysql_fetch_assoc() works. mysql_query does not return a record.

 

And please, can you use

tags when posting code in the future?

[affordit]

Yes I will thanks