[mdhall]

I am setting up a form to let people join a site. The script checks for missing form fields, sees if the username and email are already used, and inserts the new members data into a table. The following script works, including the missing form data and the insert, except that it isn't giving the error message when a username or email address is already being used. It simply adds the data into the table. I've seen this type of usage before in other scripts, but apparently I added it here incorrectly. Any suggestions?

Heres the script....

PHP Code:

$ip=$_SERVER['REMOTE_ADDR'];
$first_name=$_POST['first_name'];
$last_name=$_POST['last_name'];
$email=$_POST['email'];
$user=$_POST['user'];
$pass=$_POST['pass'];
                
 if((!$first_name) || (!last_name) || (!$email) || (!$user) || (!$pass)){
 echo 'You did not submit the following required information. Use your browsers "back" button to complete your form. <br />';
 if(!$first_name){
  echo "Your first name is required. <br />";
 }
 if(!$last_name){
 echo "Your last name is required. <br />";
 }
 if(!$email){
 echo "Your email is required. <br />";
 }
 if(!$user){
 echo "A username is required. <br />";
 }
 if(!$pass){
 echo "A password is required. <br />";
 }
 }
 else
 {
              
include("dbinfo.inc.php");
$conn = mysql_connect("$location","$username","$password");
if (!$conn) die ("Could not connect MySQL");
mysql_select_db($database,$conn) or die ("Could not open database");
$user_check = ("SELECT user FROM users");
$user_result=mysql_query($user_check);
              
while($user_taken=mysql_fetch_array($user_result))
{
$username_taken=$user_taken[user];
}
if ($username_taken)
{
echo "That username is already taken. Please select another username.";
}
$email_check = ("SELECT email FROM users");
$email_result=mysql_query($email_check);
while($email_taken=mysql_fetch_array($email_result))
{
$email_used=$email_taken[email];
}
if ($email_used)
{
echo "That email is already taken. Please select another email address.";
}
$query = "INSERT INTO users VALUES('','$first_name','$last_name','$email','$user','$pass','$ip')";
$result=mysql_query($query);
mysql_close();
               
header ("Location: index.php");
   } 

[YourPHPPro]

Originally Posted by mdhall

The following script works, including the missing form data and the insert, except that it isn't giving the error message when a username or email address is already being used.

Instead of getting all users from the database, you could try something like this:

Code:

select count(*) as Count from users where user='$username'

That way you only need to see if the number that is returned is >0 - if so, then you know there is an existing username.

The same thing can be done with the email. It could also be arranged like:

Code:

select count(*) as Count from users where user='$username' OR email='$email'

That would check for either a username conflict or an email conflict. If Count>0 then send visitor to generic error page.

[YourPHPPro]

Here is some code that might help you out:

Code:

mysql_select_db($database,$conn) or die ("Could not open database");
$combined_check = mysql_fetch_assoc(mysql_query("select count(*) as Count from users where user='$user' or email='$email'"));

if ($combined_check["Count"]>0) {
	header ("Location: error_page.php");
	quit;
}

// Results have been checked, now lets insert data into database

[mdhall]

I tried this, and It still does the same thing, i.e., add the info even though its already taken.

$combined_check = mysql_fetch_assoc(mysql_query("select count(*) as Count from users where
user='$user' or email='$email'"));
if ($combined_check["Count"]>0) {
echo "That username or password is in use";
quit;
}


I have had trouble getting sessions to work on members-type pages I set up. Basically, I want a system for viewers to become members of a site, so that only members can post material, however anyone can view what is there already. If you have any suggestions or sample scripts I could try out, I would really appreciate it.

[YourPHPPro]

Can you edit your previous post to reflect that latest code changes that you made ?

[mdhall]

Originally Posted by YourPHPPro

Can you edit your previous post to reflect that latest code changes that you made ?

Heres the full code i tried...

PHP Code:

 
$ip=$_SERVER['REMOTE_ADDR'];
$first_name=$_POST['first_name'];
$last_name=$_POST['last_name'];
$email=$_POST['email'];
$user=$_POST['user'];
$pass=$_POST['pass'];
                
 if((!$first_name) || (!last_name) || (!$email) || (!$user) || (!$pass)){
 echo 'You did not submit the following required information. Use your browsers "back" button to complete your form. <br />';
 if(!$first_name){
  echo "Your first name is required. <br />";
 }
  if(!$last_name){
  echo "Your last name is required. <br />";
 }
 if(!$email){
  echo "Your email is required. <br />";
 }
 if(!$user){
  echo "A username is required. <br />";
 }
if(!$pass){
  echo "A password is required. <br />";
 }
 }
else
{
              
include("dbinfo.inc.php");
$conn = mysql_connect("$location","$username","$password");
mysql_select_db($database,$conn) or die ("Could not open database");
$combined_check = mysql_fetch_assoc(mysql_query("select count(*) as Count from users where user='$user' or email='$email'"));
if ($combined_check["Count"]>0) {
 echo "That username or password is in use";
 quit;
}

$query = "INSERT INTO users VALUES('','$first_name','$last_name','$email','$user','$pass','$ip')";
$result=mysql_query($query);
mysql_close();
               
header ("Location: index.php");
   } 

[NeverMind]

hmm ..
why are you using quit; ? and what is it exactly?
what I know is die; and exit; are the ones that terminate script execution!!

anyway, I suggest that you use the or die() part in your mysql_query to check if the query was excuted or not ..
so something like this might help:

PHP Code:

$check = mysql_query("select count(*) as Count from users where user='$user' or email='$email'")or die(mysql_error());



$combined_check = mysql_fetch_assoc($check); 

[mdhall]

Thnx Nevermind..that did the trick.