php - Check if email exists in sql database and output thereafter (regarding running if inside if) -

so have form running on page i'm experimenting with. atm have 1 input , email. want register , if person tries add same email again return error msg , otherwise add it.

so code i've been posting with:

<?php     $username="user";     $password="pass";     $database="test";     $email = mysql_real_escape_string( $_post['email'] );      mysql_connect(localhost,$username,$password);     @mysql_select_db($database) or die( "unable select database");      $query = "select * enroll email='$email' ";     $result = mysql_query($query) or die(mysql_error());      if (mysql_num_rows($result) ) {         print 'user in table';     }     else {         $query = "insert enroll values ('','$email')";         $result = mysql_query($query) or die(mysql_error());                print 'user added';     } ?> 

that works fine! if add email exists says "user in table" , if not says "user added".

problem before used run php script validated input , upon submit added db. says "user in table" before i've pressed submit.

i this, when posting doesn't work @ , can't understand what's wrong.

<?php     //if form submitted     if(isset($_post['submit'])) {          // required field names         $required = array('email');          // loop on field names, make sure each 1 exists , not empty         $error = false;         foreach($required $field) {             if (empty($_post[$field])) {             $error = true;             }         }              if ($error) {                 echo 'you forgot fill in email';             } else {                 $username="user";                 $password="pass";                 $database="test";                 $email = mysqli_real_escape_string( $_post['email'] );                  mysqli_connect(localhost,$username,$password);                 mysqli_select_db($database) or die( "unable select database");                  $query = "select * enroll email='$email' ";                 $result = mysqli_query($query) or die(mysqli_error());                  if (mysqli_num_rows($result) ) {                     print 'user in table';                 }                 else {                     $query = "insert enroll values ('','$email')";                     $result = mysqli_query($query) or die(mysqli_error());                     print 'user added';                 }             }     } 

?>

in world, checks if email supplied, checks if exists in db , if not adds it. if exists come error.

what doing wrong?

edit2: @expertsystem code i'm going per recommendation:

<?php          if(isset($_post['submit'])) {              // required field names             $required = array('email');              // loop on field names, make sure each 1 exists , not empty             $error = false;             foreach($required $field) {                 if (empty($_post[$field])) {                 $error = true;                 }             }                  if ($error) {                     echo 'you forgot fill in email';                 } else {                     $username="root";                     $password="root";                     $host="localhost";                     $database="test";                     $email = mysqli_real_escape_string( $_post['email'] );                      $link = mysqli_connect(localhost, $username, $password);                     mysqli_select_db($link, $database) or die("unable select database");                      $query = "select * enroll email='$email'";                     $result = mysqli_query($link, $query) or die(mysqli_error($link));                      if (mysqli_num_rows($result) ) {                         print 'user in table';                     }                     else {                         $query = "insert enroll (email) values ('$email')";                         $result = mysqli_query($link, $query) or die(mysqli_error($link));                          print 'user added';                     }                 }         } ?> 

since using procedural form of mysqli_ functions, need pass mysqli link identifier first argument in mysqli_select_db(), mysqli_real_escape_string(), mysqli_query() , mysqli_error() (see documentation more details). aforementioned mysqli link identifier object returned mysqli_connect().
modified code should this:

... $link = mysqli_connect(localhost, $username, $password); mysqli_select_db($link, $database) or die("unable select database"); $email = mysqli_real_escape_string($link, $_post["email"]); ... $result = mysqli_query($link, $query) or die(mysqli_error($link)); ... if...     ... } else {     ...     $result = mysqli_query($link, $query) or die(mysqli_error($link));     ... ...