[SOLVED] Value in database is 1 brings error, value 2 brings authorization

[Anxious]

<?php 	  
      /* checks if user is activated or not */
  $q = "SELECT activated FROM ".TBL_USERS." WHERE username = '$username'";
  if($activated == "1")
      {
         return 3;
      }  ?>

 

Is this right to be seeing if the value in database is "1" that it brings up the error field (result 3). If its not 1, and its 2 instead, it doesn't bring up error field.

 

I've tried everything I can find of. Just not sure

Can you help?

[jOE :D]

You need to actually submit the query and get the results. You're not submitting the query and you're never assigning $activated to anything, so that if statement will never be true. You'd need something more like this:

 

<?php
$results = mysql_query("SELECT activated FROM ".TBL_USERS." WHERE username = '$username' LIMIT 1");
$row = mysql_fetch_array($results);

if($row['activated'] == 1) {
     return 3;
}
?>

[Anxious]

It still doesn't work, it still logs me in, when in the database under column "activated" says 1.

[jOE :D]

You could always try something like:

 

<?php
$results = mysql_query("SELECT COUNT(*) as count FROM ".TBL_USERS." WHERE `username`='$username' AND `activated`='1' LIMIT 1");
$row = mysql_fetch_array($results);

if($row['count'] > 0) {
     return 3;
}
?>

 

If this function is made for checking if the user is activated or not, you may want to try returning a boolean true/false instead of an integer.

[Anxious]

No change, still the same.

[jOE :D]

Could there be something wrong somewhere else? This part is pretty straight forward...

[Anxious]

The whole 'login' code, the that has the error fields on, nothing to do with database.

 

<?php
   function login($subuser1, $subpass1, $subremember){
      global $database, $form;  //The database and form object

      /* Username error checking */
      $field = "user1";  //Use field name for username
      if(!$subuser1 || strlen($subuser1 = trim($subuser1)) == 0){
         $form->setError($field, "* Username not entered");
      }
      else{
         /* Check if username is not alphanumeric */
         if(!eregi("^([0-9a-z])*$", $subuser1)){
            $form->setError($field, "* Username not alphanumeric");
         }
      }

      /* Password error checking */
      $field = "pass1";  //Use field name for password
      if(!$subpass1){
         $form->setError($field, "* Password not entered");
      }
      
      /* Return if form errors exist */
      if($form->num_errors > 0){
         return false;
      }

      /* Checks that username is in database and password is correct */
      $subuser1 = stripslashes($subuser1);
      $result = $database->confirmUserPass($subuser1, md5($subpass1));

      /* Check error codes */
      if($result == 1){
         $field = "user1";
         $form->setError($field, "* Username not found");
      }
      else if($result == 2){
         $field = "pass1";
         $form->setError($field, "* Invalid password");
      }
  else if($result == 3){
     $field = "activate";
	 $form->setError($field, "* Account not activated");
  }
      
      /* Return if form errors exist */
      if($form->num_errors > 0){
         return false;
      }

      /* Username and password correct, register session variables */
      $this->userinfo  = $database->getUserInfo($subuser1);
      $this->username  = $_SESSION['username'] = $this->userinfo['username'];
      $this->userid    = $_SESSION['userid']   = $this->userinfo['userid'];
      $this->userlevel = $this->userinfo['userlevel'];
      
      /* Insert userid into database and update active users table */
      $database->addActiveUser($this->username, $this->time);
      $database->removeActiveGuest($_SERVER['REMOTE_ADDR']);

      if($subremember){
         setcookie("cookname", $this->username, time()+COOKIE_EXPIRE, COOKIE_PATH);
	 setcookie("cookpass", $this->password, time()+COOKIE_EXPIRE, COOKIE_PATH);
	 setcookie("cookact",  $this->activated,   time()+COOKIE_EXPIRE, COOKIE_PATH); 
         setcookie("cookid",   $this->userid,   time()+COOKIE_EXPIRE, COOKIE_PATH);
      }

      /* Login completed successfully */
      return true;
   }
?> 

 

Ths following little snippet is what is involved in the code above, which is the error field.  

<?php
else if($result == 3){
     $field = "activate";
	 $form->setError($field, "* Account not activated");
  }
?>

 

[Anxious]

If I did...

<?php
      $results = "SELECT activated FROM ".TBL_USERS." WHERE username = '$username' LIMIT 1";
      $row = mysql_fetch_array($results, $this->connection);
      if($row['activated'] == 2) {
          return 0; /* user logged in successfully */
  }
          else
  {
	 return 3;  /* indicates user not activated */
      } ?>

 

It always says my accounts not activated, even when it is.

 

If I swapped it round to this..

<?php
      $results = "SELECT activated FROM ".TBL_USERS." WHERE username = '$username' LIMIT 1";
      $row = mysql_fetch_array($results, $this->connection);
      if($row['activated'] == 1) {
          return 3; /* indicates user not activated */
  }
          else
  {
	 return 0;  /* user logged in successfully */
      } ?>

 

It then always logs me in even if its not activated.

 

I'm confused as to why it does this.

[Anxious]

anybody with any suggestions ?

[Yesideez]

Try dumping some stuff to the browser so you can see what's being used...

 

<?php
      $results = "SELECT activated FROM ".TBL_USERS." WHERE username = '$username' LIMIT 1";
echo 'QUERY:'.$results.'<br>';
      $row = mysql_fetch_array($results, $this->connection);
vardump($row);
      if($row['activated'] == 1) {
          return 3; /* indicates user not activated */
     }
          else
     {
       return 0;  /* user logged in successfully */
      } ?>

 

Try that - you might have to right-click and view source to read the var_dump() output clearly.

[Anxious]

Done, I now have these errors.

 

Warning: Missing argument 4 for Session::login(), called in /home/myveeco/public_html/process.php on line 50 and defined in /home/myveeco/public_html/include/session.php on line 93

QUERY:SELECT activated FROM users WHERE username = 'JeanieTallis' LIMIT 1

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/myveeco/public_html/include/database.php on line 61

 

Fatal error: Call to undefined function vardump() in /home/myveeco/public_html/include/database.php on line 62

 

session.php

line 93 is

 <?php
   function login($subuser1, $subpass1, $subremember, $activated){
?>

 

Database.php

Line 61 and 62 is

<?php
      $row = mysql_fetch_array($results, $this->connection);
vardump($row);
?>

 

Thank you so much for trying to help.

[Yesideez]

oops my bad - typo - meant to type var_dump($row);

 

Is the query as it should be? Is there data in the database with "JeanieTallis" as the username?

 

The invalid resource is saying your $this->connection isn't a valid connection (mysql_connect() didn't work or mysql_select_db() didn't work)

[Anxious]

it's a valid connection, as its used else where. It's defined by the database connections

[Yesideez]

$results = mysql_query("SELECT activated FROM ".TBL_USERS." WHERE username = '$username' LIMIT 1");

 

Now just remove that echo I added and it should work.

[Anxious]

Nice. It works, thanks.

 

Can you tell me what I had to actually do?

 

      $q = "SELECT password FROM ".TBL_USERS." WHERE username = '$username'";
      $result = mysql_query($q, $this->connection);
      if(!$result || (mysql_numrows($result) < 1)){
         return 1; //Indicates username failure
      }

 

Thats to check if the user entered in the textbox is in the database, I followed the same format for both the two top lines, and then did..

if($row['activated'] == 1) {
          return 3; /* indicates user not activated */
     }

So it should of worked like it should do.

 

So what was it that I actually needed..

Just to help me understand.

[Yesideez]

mysql_fetch_array(), mysql_fetch_object(), mysql_fetch_assoc() and the like all require the RESULT from a mysql_query() call to work.

 

You were skipping the mysql_query() call and passing it a query in text form which it doesn't understand.