About to give up - unexpected ELSE error?

[tracdoor]

Resolved 🙂

[NogDog]

You cannot have consecutive else's within the same if block. I believe this is what you want, but give it a look-over to make sure it looks like the logic you want:

<?php

include ("database.php");


if(isset($_POST['recemails']))
{

   $recemails = '1';

}
else
{

   $recemails = '0';

}

if($_GET['action'] == 'register')
{

   if(isset($_POST['accepttos']))
   {

  if(isset($_POST['password']))
  {

     $result = mysql_query('SELECT * WHERE username = $_POST[username]');

     if(!$result)
     {

        echo "This username is already in use, please use another -<a href=http://sam.exofire.net/register.php>try again</a>.";

     }
     else
     {

        mysql_query("INSERT INTO users
    (username, password, country, recemails) VALUES ('$_POST[username]', '$_POST[password]', '$_POST[country]', '$recemails' ) ") or
           die(mysql_error());

     }
  }
  else
  {

     echo ("Please choose a password. <a href=http://sam.exofire.net/register.php>back</a>");

  }
   }
   else
   {

  echo ("You Did Not Accept The Terms And Conditions. <a href=http://ihome.sam.exofire.net/login/index.php?action=register>Try Again</a>.");

   }
}
else
{

   echo ("Please Declare A Valid Action.");

}

mysql_close($con);

?> 

[tracdoor]

That got rid of the error message - I worked out what I had done wrong in the morning, but now whenever I submit I get the "This username is already in use, please use another - try again." echoed, anyone help me?

[leatherback]

Tracdoor,

You just need to look through the logic of things. Sit down with a piece of paper, and work out what should happen in which case. In this case, the error is easily spotted, if you just take 2 minutes to look at the logic of it.

Hint: It is in this statement:

if(!$result)
         { 

PS: Don't feel that I am being harsh. I am trying to teach you that you can help yourself. 😃

[tracdoor]

I've found it! it shouldn't be if its returned false (using exclamation mark) it should be if its returned true, trouble is, i'm new to this and I don't know how to do it, i've just tried modifying the mysql_query bit with this code:

$result = mysql_query("SELECT * WHERE username = '%s'");
			mysql_real_escape_string($_POST['username']);

but it still returns with the username is already taken, even though its not in the database 🙁

Can anyone tell me how it should be (only executes if the if statement is true? I think thats what i've got wrong.

[laserlight]

If the SQL statement succeeds, mysql_query() returns true. But the SQL statement succeeds if it can be executed, regardless of the number of rows returned in the result set. So what you should do is:

$result = mysql_query("SELECT COUNT(*) WHERE username = '%s'",
                      mysql_real_escape_string($_POST['username']));
if (mysql_result($result, 0) == 0) {
    // Username is not taken.
}

Actually, even better would be to declare the username column UNIQUE, then attempt to insert, and catch a UNIQUE constraint violation.

[tracdoor]

Ok, I replaced that code, now another error has come up:

Parse error: syntax error, unexpected T_ELSE in /home/tracdoor/public_html/login/index.php on line 35

here are those few lines:

if(isset($_POST['password'])) 
      { 

     $result = mysql_query("SELECT COUNT(*) WHERE username = '%s'");
                  mysql_real_escape_string($_POST['username']) or die(mysql_error());

				if (mysql_result($result, 0) == 0);  {

        echo ("This username is already in use, please use another - <a href=http://sam.exofire.net/register.php>try again</a>.");

		} else { 

        mysql_query("INSERT INTO users 
    (username, password, country, recemails) VALUES ('$_POST[username]', '$_POST[password]', '$_POST[country]', '$recemails' ) ") or 
           die(mysql_error()); 

     } 
  }

Anything to get rid of this? so many errors on just a simple registration form!

[laserlight]

That's because you have typo errors, and you did not read our code examples carefully. A possible fix:

if(isset($_POST['password'])) {
    $query = sprintf("SELECT COUNT(*) WHERE username = '%s'",
                     mysql_real_escape_string($_POST['username']));
    $result = mysql_query($query) or die(mysql_error());

if (mysql_result($result, 0) == 0) {
    echo 'This username is already in use, please use another - <a href="http://sam.exofire.net/register.php">try again</a>.';
} else {
    $query = sprintf("INSERT INTO users (username, password, country, recemails)
                      VALUES ('%s', '%s', '%s', '%s')",
                     mysql_real_escape_string($_POST['username']),
                     mysql_real_escape_string($_POST['password']),
                     mysql_real_escape_string($_POST['country']),
                     mysql_real_escape_string($recemails));
    mysql_query($query) or die(mysql_error());
}
}

Notice that my code example is well indented. I use sprintf() with mysql_real_escape_string() to sanitise database input.

[tracdoor]

Thanks for the help, really appreciate your help 🙂

BUT, now I get an error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE username = 'truytiyuiuyi'' at line 1

I suspect this could be a problem with my host? I know very little about PHP and I think i'll go read a decent tutorial after i've got this working...

[HalfaBee]

It is hard to see what is wrong with the query without seeing the whole thing.

I normally do die( mysql_error()."<br>$query" );

So you get the error message and the whole query.

[leatherback]

you seem to keep forgetting to put in the table you are selecting from. Is this the case here too?

[tracdoor]

Thanks - just one thing wrong now, the code that checks whether a password has been specified doesn't seem to work, any idea's? (code in first post).

[laserlight]

How does it not work?

[tracdoor]

Don't worry, I fugured it out, when the form was sent the $_POST['password'] was set as nothing, but i fixed it with this code:

if ($_POST['password'] && (strlen($_POST['password']) > 0))

Now to work on my login script! 🙂

I somehow have a feeling i'll be back.

[laserlight]

Oh. Instead of:

if ($_POST['password'] && (strlen($_POST['password']) > 0))

use:

if (isset($_POST['password']) && strlen($_POST['password']) > 0)

[Weedpacket]

tracdoor wrote:

(code in first post).

Not any more 🙁. Why decapitate a perfectly good thread?