Database always receiving same IDs - php

Question

I hope this is not a bad question. I've been trying to understand what I'm doing wrong but I can't.
I'm pretty new to php and mysql so I'm really confused...
I have this database (I will attach a mysql workbench model screenshot)
And I'm trying to insert the sales into sale and print_for_sale tables. The queries seem to be working and the data shows up in phpmyadmin. No error pops up. However the sale user in the sale table is always the same id. Even if I use a different user login. And in the print_for_sale table the fk_sale_id is always the same id. And the price_print_for_sale is always the same as well. That is not supposed to happen. What am I doing wrong?
Can anyone help me?
Thank you!
Here is my php code:
<?php
session_start();
$user_id = $_SESSION['user_id'];
print_r($_POST);
$id_print= $_POST['print_id'];
include('database.php');
$bought_sizes=$_POST['sizes'];
$number_of_bought_sizes= count($bought_sizes);
//header('location:index.php?area=printstore');
$sqlinsertsale = "insert into sale
(fk_sale_user,
fk_payment_id)
values(".$user_id.", 1)";
//payment is not yet defined so I'm using 1 just to try.
mysql_query($sqlinsertsale);
for($i=0; $i< $number_of_bought_sizes; $i++){
$selectmultiple = "select *
from print_has_size
inner join size on fk_size_id = size_id
inner join print on fk_print_id = print_id
where print_id =".$id_print."";
$resultmultiple = mysql_query($selectmultiple);
$linemultiple = mysql_fetch_assoc($resultmultiple);
$size_price = $linemultiple["size_price"];
$selectsale = "select *
from sale";
$resultsale = mysql_query($selectsale);
$linesale = mysql_fetch_assoc($resultsale);
$sale_id = $linesale["sale_id"];
//$sale_id = mysql_insert_id();
/*PARA CADA 1 DOS TAMNHO*/
$sqlinsertprintforsale = "insert into print_for_sale
(fk_sale_id,
price_print_for_sale)
values(".$sale_id.", ".$size_price.")";
mysql_query($sqlinsertprintforsale);
}
?>
I'm also going to attach a screenshot of the selection page so you can see the markup in case it helps.
Edit:
(I'm adding the php code from where I check the user login)
<?php
session_start();
include('database.php');
$user=mysql_real_escape_string($_POST['user_login']);
$pass=mysql_real_escape_string($_POST['user_pass']);
$sql="select user_id, user_name
from user
where
user_login='".$user."'
and user_pass = MD5('".$pass."')";
echo $sql;
$result = mysql_query($sql);
$num_of_regs = mysql_num_rows($result);
echo "<br>".$num_of_regs;
if($num_of_regs!=1) {
header('location:index.php?login=done');
}
else {
$line = mysql_fetch_assoc($result);
$user_name = $line['user_name'];
$user_id = $line['user_id'];
$_SESSION['user_name'] = $user_name;
$_SESSION['user_id'] = $user_id;
header('location:index.php');
}
?>
And I did a log out system too.
<?php
session_start();
session_destroy();
header('location:index.php');
?>
Now I noticed that the sale table is not receiving data..
Only the print_for sale is. Wrong data still. Same IDs... Why? :(
This is the only error message that I get when I used the code
ini_set('display_errors', true); error_reporting(E_ALL);
Deprecated: mysql_connect(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead in /Applications/XAMPP/xamppfiles/htdocs/printstore/database.php on line 7
Added
echo $_SESSION['user_id'];
And this is my output for user 1:
And for user 2:
it seems to be ok, it recognizes user 1 and 2.
Table sale is not being refreshed when I make a "new purchase" and table print_for_sale is refreshed with the same sale id. Always 3 (as shown in the screenshot)
I deleted every row from sale in phpmyadmin and tried again. It's working. Sale table seems to be working fine. The only problem now is the table print_for_sale which even when I use a different user (and it shows up ok in the sale table), it still shows the same sale_id and the same price_print_price which is always 25). And in this case I never selected anything costing 25. -

Answer 1

Give this a shot. It seems that you're getting the same result because you never loop through your results. Also make sure that your field names are exactly the same as your field names in your DB.
$selectsale = "select *
from sale";
$resultsale = mysql_query($selectsale);
while($linesale = mysql_fetch_assoc($resultsale))
{
$sale_id = $linesale["sale_id"];
}

Answer 2

when you logout you should destroy your sessions
I assume you have a logout page like logout.php. In it you should set below lines. If you don't do that session values can't change.
<?php
session_start();
session_destroy();
?>
Also if you store your user id on a session variable, only closing all browser windows than reopen and login will help you.
Extra info for you future code life: WHY PDO http://code.tutsplus.com/tutorials/why-you-should-be-using-phps-pdo-for-database-access--net-12059