problem with if and else ?

gammaman · April 17, 2008

I have the following which is pulled from a form.

If the passwords do not match it displays "passwords do not match"

If the user name and two passwords match and the user does not exist it adds a new user.

and say "New user added"

However if the user already exists and the passwords match instead of saying "user already exists"

It says "New User Added" but does not add it.

I think it is with the if else statments

   <body>
<?php

$Conn=mysql_connect("localhost","fierm","13183");

if(!$Conn){ 
   echo "failed";
  } 
  else{ 
mysql_select_db("fierm");


  $User=$_POST["user"];
  $Pass=$_POST["pass"];
  $Conf=$_POST["confirm"];

  if($Pass==$Conf){ 
  
  $result=mysql_query("select usr_name, FROM user WHERE usr_name='$User'");
  $cou=mysql_num_rows($result);
  
  
  
     if($cou>0) 
     { 
       echo "user already exists";  
  }else{ 
     mysql_query("Insert into user (usr_name,pwd) Values ('$User','$Pass')");
    
     echo "New User Was Added";
  
  }

}   
else{
    echo "passwords do not match";
  } 
}     
?>
</body>

marcus · April 17, 2008

Confirm that you're passing the correct POST variables. Your code looks correct, but the form may have an issue.

gammaman · April 17, 2008

Yes Iam using POST, like I said all the other parts work. It only does not work when I enter an existing user with an existing user name and password.

marcus · April 17, 2008

Ok, well show us the full code, I can tell you're using POST, but are the input names the same as the ones you're passing? Your code is correct. You could also use error checking

mysql_query("whatever") or die(mysql_error());

jonsjava · April 17, 2008

   <body>
<?php

$Conn=mysql_connect("localhost","fierm","13183");

if(!$Conn){ 
   echo "failed";
  } 
  else{ 
mysql_select_db("fierm");


  $User=$_POST["user"];
  $Pass=$_POST["pass"];
  $Conf=$_POST["confirm"];

  	if($Pass == $Conf){ 
  		$result=mysql_query("SELECT * FROM user WHERE usr_name='$User' AND `password`='{$Pass}' LIMIT 1;");
  		$cou=mysql_num_rows($result);
  		if($cou>0) 
    	{ 
    		echo "user already exists";  
  		}
  		else{ 
  			$result = mysql_query("SELECT * FROM user WHERE usr_name='{$User}' LIMIT 1;");
  			if (mysql_num_rows($result) != 1){
     			mysql_query("Insert into user (usr_name,pwd) Values ('$User','$Pass')");
    			echo "New User Was Added";
  			}
  		}
	}   
else{
    	echo "passwords do not match";
  	}	 
}     
?>
</body>

gammaman · April 17, 2008

Still does not work if the user already exists. I tried to echo $cou but nothing was displayed so it is possible that it is automatically falling through to the else

marcus · April 17, 2008

Show us your FULL code, even the form.

jonsjava · April 17, 2008

I agree. I'm betting that your variables are not named "pass" and "confirm"

gammaman · April 17, 2008

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">

<html>
<head>
<title>Change A Department</title>
</head>
<?php

?>
<body>

<form action="checknew.php" method="post">

UserName:<input name="user" type="text"><br />
Password:<input name="pass" type="text"><br />
ReEnter Password:<input name="confirm" type="text"><br />
    
    	<br />
    	 <input name="Submit1" type="submit" value="submit" /><br />

</form>

</body>
</html>

gammaman · April 17, 2008

Oh and by the way it is on two seperate pages

jonsjava · April 17, 2008

try this slightly modified version of my last attempt:

   <body>
<?php

$Conn=mysql_connect("localhost","fierm","13183");

if(!$Conn){ 
   echo "failed";
  } 
  else{ 
mysql_select_db("fierm");


  $User=$_POST["user"];
  $Pass=$_POST["pass"];
  $Conf=$_POST["confirm"];

  	if($Pass == $Conf){ 
  		$result=mysql_query("SELECT * FROM user WHERE usr_name='$User' AND `password`='{$Pass}' LIMIT 1;") or die("It failed on the first query :".mysql_error());
  		$cou=mysql_num_rows($result);
  		if($cou == 1) 
    	{ 
    		echo "user already exists";  
  		}
  		else{ 
  			$result = mysql_query("SELECT * FROM user WHERE usr_name='{$User}' LIMIT 1;") or die("It failed on the second query :".mysql_error());
  			if (mysql_num_rows($result) != 1){
     			mysql_query("Insert into user (usr_name,pwd) Values ('$User','$Pass')");
    			echo "New User Was Added";
  			}
  		}
	}   
else{
    	echo "passwords do not match";
  	}	 
}     
?>
</body>   <body>
<?php

$Conn=mysql_connect("localhost","fierm","13183");

if(!$Conn){ 
   echo "failed";
  } 
  else{ 
mysql_select_db("fierm");


  $User=$_POST["user"];
  $Pass=$_POST["pass"];
  $Conf=$_POST["confirm"];

  	if($Pass == $Conf){ 
  		$result=mysql_query("SELECT * FROM user WHERE usr_name='$User' AND `password`='{$Pass}' LIMIT 1;") or die("It failed on the third query :".mysql_error());
  		$cou=mysql_num_rows($result);
  		if($cou == 1) 
    	{ 
    		echo "user already exists";  
  		}
  		else{ 
  			$result = mysql_query("SELECT * FROM user WHERE usr_name='{$User}' LIMIT 1;");
  			if (mysql_num_rows($result) != 1){
     			mysql_query("Insert into user (usr_name,pwd) Values ('$User','$Pass')");
    			echo "New User Was Added";
  			}
  		}
	}   
else{
    	echo "passwords do not match";
  	}	 
}     
?>
</body>

***EDITED to add error checking

marcus · April 17, 2008

<body>
<?php

$Conn = mysql_connect('localhost', 'fierm', '13183');

if (!$Conn) {
    echo "failed";
} else {
    mysql_select_db('fierm', $Conn);


    $User = $_POST['user'];
    $Pass = $_POST['pass'];
    $Conf = $_POST['confirm'];

    if ($Pass == $Conf) {
        $result = mysql_query("SELECT * FROM `user` WHERE usr_name='" . $User . "'") or
            die(mysql_error());
        $cou = mysql_num_rows($result);
        if ($cou > 0) {
            echo "user already exists";
        } else {
            mysql_query("INSERT INTO `user` (`usr_name`,`pwd`) Values ('".$User."','".$Pass."')");
            echo "New User Was Added";
        }
    } else {
        echo "passwords do not match";
    }
}
?>
</body>

gammaman · April 17, 2008

That does not work either, now it does not display anything when the user already exists. The other parts work fine. Thanks for your help

gammaman · April 17, 2008

It works now, your code was good except that

you put

result=mysql_query("SELECT * FROM user WHERE usr_name='$User' AND `password`='{$Pass}' LIMIT

the field name is pwd not password

Thanks again, that was driving me crazy

marcus · April 17, 2008

You do realize the possibilities of a user creating an account with the same username and password are very slim. You only need to check the username.

gammaman · April 17, 2008

Yes that is true but it does not hurt to check both

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problem with if and else ?

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