Question

Okay What im looking to do return all the data a user has already submitted on one page. So basicly they are already loged in so I should know there email address and I just want to pull all the mysql data and display all the data under that email address

So basicly mysql has email as first colum and then and bunch of junk and I want to display all the junk if the email colume equals the same email that they are loged in as

here is my code im on the right track but not quite there

 <?php
    session_start();
    include 'config.php';
    $_SESSION['Email']=$Email;

    mysql_select_db("DATABASEJUNK") or die ("Couldn't find database.");

    $query = mysql_query("SELECT * FROM DATABASELIST WHERE Email='$Email'");

    $numrows = mysql_num_rows($query);

    while ($row = mysql_fetch_assoc($query))
      echo $row['Url'];

 ?>

Answer 1

i think what you need is something like that

while ($row = mysql_fetch_assoc($query)){
  foreach($row as $value){
     echo $value;
  }
}

if course you need to put this at the correct part of your script. or save the output of $value to an array

$allUsers = new Array();
while ($row = mysql_fetch_assoc($query)){
  $user = new Array();
  foreach($row as $key => $value){
     $user[$key] = $value;
  }
  $allUsers[] = $user;
}

this gives you an array with all users … and each user has its own data as key! if you want just 1 user forget about the $allUser arrays and keep just the rest (if mysql_num_rows() == 1)