[SOLVED] Wont sumbit a query

[blink359]

I have a form that is trying to put account info into a database but when you submit

it thinks this is allways true so that every time you try and make an account it thinks it allready exsists so it doesnt do the query:

		
    $result = mysql_query("SELECT * FROM `accounts` where login='$user'") or die("Error: (" . mysql_errno() . ") " . mysql_error());
	$row = mysql_fetch_array( $result );
	if(isset($row))
	{
            $problemMessage .= "The username already exists <br />"; 
            $success = false; 
	}

here is my PhP code for the form

 

<?php 
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "ascent";
$dbname = "logon";

mysql_connect($dbhost, $dbuser, $dbpass); 
mysql_select_db($dbname); 

$success = true; 
$problemMessage = ""; 
if (isset($_POST['Submit'])) 
{ 
    $user = isset($_POST['user'])?$_POST['user']:false;
    $pass = isset($_POST['pass'])?$_POST['pass']:false;
    $pass2 = isset($_POST['pass2'])?$_POST['pass2']:false;

        if(!$user || !$pass || !$pass2) 
        { 
            $problemMessage .= "Please complete all required data <br />"; 
            $success = false; 
        } 
        if(strlen($user) < 5) 
        { 
            $problemMessage .= "User must be more that 5 characters <br />"; 
            $success = false; 
        } 
        if(strlen($user) > 20) 
        { 
            $problemMessage .= "Your username cannot be longer than 20 characters <br />"; 
            $success = false; 
        } 
        if($pass != $pass2) 
        { 
            $problemMessage .= "Your passwords do not match <br />"; 
            $success = false; 
        } 
        if(strlen($pass2) < 5) 
        { 
            $problemMessage .= "Your password must be more than 5 characters<br />"; 
            $success = false; 
        } 
        if(strlen($pass2) > 20) 
        { 
            $problemMessage .= "Your password cannot be longer than 20 characters <br />"; 
            $success = false; 
        }  

    $result = mysql_query("SELECT * FROM `accounts` where login='$user'") or die("Error: (" . mysql_errno() . ") " . mysql_error());
	$row = mysql_fetch_array( $result );
	if(isset($row))
	{
            $problemMessage .= "The username already exists <br />"; 
            $success = false; 
	}
         
        if ($success) 
        { 
            echo "Valid New Account Entry<br />";
            $result = mysql_query("INSERT INTO `accounts` VALUES ('', '$'user', '$pass2', '', '', '0', '9', '', '', 'flag', 'enUS', '0', '0', null);") or die("Error: (" . mysql_errno() . ") " . mysql_error());

        } 

	$row = mysql_fetch_array( $result );

} 
  
?> 
          <?php 
if ($success == false) { 
    echo $problemMessage; 
} 
?>
          Account Creation Page<br> 
          Username: 
          <input name="user" type="text"/>
          <br>  
          Password: 
          <input name="pass" type="text"/>
          <br>  
          Repeat Password: 
          <input name="pass2" tpye="text"/>
          <br />
          Pre TBC
          <input type="radio" name="flag" value="0">
            <br>
          TBC Enabled
          <input type="radio" name="flag" value="8">
          <br>
          Wotlk Enabled
          <input type="radio" name="flag" value="24">
          <br>
          
          <input name="Submit" type="submit" value="submit" /> 
            </p>
        <imput name="reset" type="reset" value="reset" /> 


</form>

Thanks

Blink359

[wildteen88]

This is incorrect:

$row = mysql_fetch_array( $result );

      if(isset($row))

      {

            $problemMessage .= "The username already exists <br />";

            $success = false;

      }

isset checks for variable existence. You telling PHP to check to see if the variable $row exists. As you have defined it in the line above it'll always return true.

 

What you should of done is use mysl_num_rows to check to see if any results where returned from your query

if(mysql_num_rows( $result ) > 0)
      {
            $problemMessage .= "The username already exists <br />";
            $success = false;
      }

[blink359]

Right thanks for that

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/nathan/wotlk/register.php on line 96

 

                $row = mysql_fetch_array( $result );

 

Hopefully this is last thing i need to ask

Reason im having problems is i wrote it to work on my computer but it doesnt work on others so i got it edited slightly and some of the code i dont know what it does i really need to start reading my "Sam's teach yourself Mysql, PhP and apache" book i got somewhere

 

[gevans]

Show us the lines before, the query and the assigning of $result

[wildteen88]

What is line 96 in register.php? Post lines 90 to 100 here. That error is normally produced when a query fails.

[blink359]

        if ($success)
        {
            echo "Valid New Account Entry<br />";
            $result = mysql_query("INSERT INTO `accounts` VALUES ('', '$user', '$pass2', '', '', '0', '9', '', '', 'flag', 'enUS', '0', null);") or die("Error: (" . mysql_errno() . ") " . mysql_error());

        }

                $row = mysql_fetch_array( $result );

}

?>

[wildteen88]

Your closing your if prematurely. This line

$row = mysql_fetch_array( $result );

 

Should be after:

$result = mysql_query("INSERT INTO `accounts` VALUES ('', '$user', '$pass2', '', '', '0', '9', '', '', 'flag', 'enUS', '0', null);") or die("Error: (" . mysql_errno() . ") " . mysql_error());

 

Like so

if ($success)
        {
            echo "Valid New Account Entry<br />";
            $result = mysql_query("INSERT INTO `accounts` VALUES ('', '$user', '$pass2', '', '', '0', '9', '', '', 'flag', 'enUS', '0', null);") or die("Error: (" . mysql_errno() . ") " . mysql_error());
            $row = mysql_fetch_array( $result );
        }

[blink359]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/nathan/wotlk/register.php on line 93

 

if ($success)
        {
            echo "Valid New Account Entry<br />";
            $result = mysql_query("INSERT INTO `accounts` VALUES ('', '$user', '$pass2', '', '', '0', '9', '', '', 'flag', 'enUS', '0', null);") or die("Error: (" . mysql_errno() . ") " . mysql_error());
            $row = mysql_fetch_array( $result );
        }


} 
  

[gevans]

Look at the query, you're doing an INSERT

[blink359]

Right thanks guys i got that work

now to test it on other peoples computers