[SOLVED] problems selecting/creating a table...

Derleek · May 7, 2008

i'm completely stumped... granted i'm not all that experienced with php/mySQL but i can't figure out what exactly is wrong with my code here.

i'm sure there is a bunch though.

the object of this program is to take user input/varify it. then if it is valid store it into a database (moto) under the table fans... the table doesn't seem to be created correctly (obviously an issue with my code)

so here it is... sorry if its sloppy...

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en">
<head>
<title>Sign Up!</title>
</head>
<body>
<p>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
<label for="name">Name:  </label>
<input id="name" type="text" name="realName" maxlength="25" /><br />

<label for="name">Screen Name:  </label>
<input id="name" type="text" name="screenName" maxlength="25" /><br />

<label for="name">Password:  </label>
<input id="name" type="password" name="password" maxlength="25" /><br />

<label for="name">Confirm Password:  </label>
<input id="name" type="password" name="cPassword" maxlength="25" /><br />

<input id="submit" type="submit" value="Sign UP!" /><br />
</p>
<?php
function sanityCheck($string, $type, $length){

  // assign the type
  $type = 'is_'.$type;

  if(!$type($string))
    {
    return FALSE;
    }
  // to check if there is anything in the string
  elseif(empty($string))
    {
    return FALSE;
    }
  // to check how long the string is
  elseif(strlen($string) > $length)
    {
    return FALSE;
    }
  else
    {
    // if all is well, we return TRUE
    return TRUE;
    }
}

function checkSet(){
  return isset($_POST['realName'], $_POST['screenName'], $_POST['password'], $_POST['cPassword']);
}
if(checkSet() != FALSE)
{
if(empty($_POST['realName'])==FALSE && sanityCheck($_POST['realName'], 'string', 25)!=FALSE && empty($_POST['screenName'])==FALSE && sanityCheck($_POST['screenName'], 'string', 25)!=FALSE && empty($_POST['password'])==FALSE && sanityCheck($_POST['password'], 'string', 25)!=FALSE)
                {
        //If all is well  the value of the POST is assigned to a variable...
                $realName = $_POST['realName'];
			$screenName = $_POST['screenName'];
			$password = $_POST['password'];
			$cPassword = $_POST['cPassword'];
			if($password !=$cPassword)
				{
				echo 'your passwords don not match';
				exit();
				}
			$link = @mysql_connect('localhost', 'root', 'zOinks12');
			if (!$link) 
			    {
			    die('Not connected : ' . mysql_error());
			    }

			$db_selected = mysql_select_db('moto', $link);
			if (!$db_selected)
			    {
			    die ("Database not selected : " . mysql_error());
			    }

				$input = "CREATE TABLE fans (
				realName VARCHAR( 25 ) NOT NULL ,
				screenName VARCHAR( 100 ) NOT NULL ,
				password VARCHAR( 25 ) NOT NULL ,
				)";
			mysql_query($input,$link);
			// Query
			$uInput = sprintf("INSERT INTO fans (realName,screenName,password)
			    VALUES( '%s', '%s','%s')",
				mysql_real_escape_string($realName),
			    mysql_real_escape_string($screenName),
			    mysql_real_escape_string($password));
			// run the query
			if(!mysql_query($uInput))
			    {
			    echo 'Query failed '.mysql_error();
			     exit();
			    }
			  else
			    {
					echo "welcome to Pro Moto Fan bench racing!!!";
			    }
                }
		else
                {
                echo 'You did not fill out all of the fields!!';
                exit();
                }
}
else
{
	echo "Please fill out all of the above!";
}
?>
</body>

</html>

fenway · May 7, 2008

That's not helpful.

What sql statement doesn't work?

Derleek · May 7, 2008

the error that is spat out is... "Query failed Table 'moto.fans' doesn't exist"

fenway · May 8, 2008

the error that is spat out is... "Query failed Table 'moto.fans' doesn't exist"

That's a good start... which query? And please echo the plaintext sql statement, NOT php code.

Derleek · May 15, 2008

sorry, new to this php/mySQL stuff...

here is the query

$uInput = sprintf("INSERT INTO fans (realName,screenName,password)
			    VALUES( '%s', '%s','%s')",
				mysql_real_escape_string($realName),
			    mysql_real_escape_string($screenName),
			    mysql_real_escape_string($password));
			// run the query
			if(!mysql_query($uInput))
			    {
			    echo 'Query failed '.mysql_error();
			     exit();
			    }
			  else
			    {
					echo "welcome to Pro Moto Fan bench racing!!!";
			    }
                }

i believe this is what you're looking for?

Derleek · May 15, 2008

well, i checked out phpMyAdmin and the it appears that the table isn't even being created... here is that code that is right before the previously posted query code.

$input = "CREATE TABLE fans (
				realName VARCHAR( 25 ) NOT NULL ,
				uID bigint(20) NOT NULL AUTO_INCREMENT,
				screenName VARCHAR( 25 ) NOT NULL ,
				password VARCHAR( 25 ) NOT NULL ,
				PRIMARY KEY (`uID`),
				)";
			mysql_query($input,$link);

this code does not appear to be working...

Derleek · May 15, 2008

solved... syntax error i hate myself... hahah

how come i can't modify my old posts to say this issue was solved?

fenway · May 15, 2008

Just clicked the "TOPIC SOLVED" button on the bottom-right of the button set on the bottom of the page (near the quick reply area).

We specifically don't want old posts changed.

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[SOLVED] problems selecting/creating a table...

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